WBBSE Solutions For Class 9 Physical Science Chapter 6 Equivalence Of Work And Heat Latent Heat

Chapter 6 Equivalence Of Work And Heat Latent Heat Synopsis

  1. If work can be fully converted into heat, then work done and heat emitted are directly proportional to each other. This is Joule’s law.
  2. If H amount of heat is generated due to performance of work W, then W ∝ H or, W = JH; J being the mechanical equivalent of heat.
  3. Mechanical equivalent of heat is that amount of work, which is to be performed to generate one unit of heat. Mechanical equivalent of work, J = 4.2 x 107 erg/cal = 4.2 J/cal, Value of mechanical equivalent of heat in SI is 1.
  4. A material may exist in different forms or physical states. These forms are physically distinct from each other but have the same chemical composition. By application of heat or extraction of heat or by any other mechanical method, forms of a material can be changed. Each form of the material is known as a distinct state or phase.

On the basis of the absorption or release of heat, change of state is of two types:

  1. Higher change of state and
  2. Lower change of state.

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A change of state of a material due to absorption of heat is called higher change of state.

Examples: Melting or fusion, vaporisation.

On the other hand, a change of state of material due to release of heat is called lower change of state.

Example: solidification, condensation.

The quantity of heat released or absorbed per unit mass of a substance during its change from one state to another by keeping the temperature constant is called the latent heat of the substance.

Units of latent heat in CGS system and SI are cal/g and J/kg, respectively.

Here, 1 cal/g = 4200 J/kg.

Dimensional formula of latent heat is L2T-2.

Chapter 6 Equivalence Of Work And Heat Latent Heat Short And Long Answer Type Questions

Question 1. Write and explain Joule’s law regarding equivalence of work and heat.

Answer:

Joule’s Law Regarding Equivalence Of Work And Heat:-

If work can be fully converted to heat, then work done and heat produced are directly proportional to each other. This is Joule’s law. Suppose, W amount of work is done to produce H amount of heat. According to Joule’s law,

W ∝ H or, W = JH

where J is a constant and is called mechanical equivalent of heat.

Question 2. Define mechanical equivalent of heat. What is its value of erg/cal?

Answer:

Mechanical Equivalent Of Heat And Value Of erg/cal.

  1. Mechanical equivalent of heat is defined as the amount of work that has to be done to produce one unit of heat.
  2. Value of J in CGS system is 4.2 x 107 erg/cal, this may be expressed as 4.2 J/cal.

Question 3. Heat is produced when work is done give an example.

Answer:

Example For Heat Is Produced When Work Is Done:-

When a wooden door or any material is polished with a piece of cloth by applying some pressure, we find that both our hand and the cloth warm up considerably. Polishing means reducing the roughness of the surface to make it more smooth. Work done against friction during polishing gets converted into heat.

Question 4. Water in a waterfall falls from a height to the ground. Why is the temperature of water at the ground level slightly more than that at the top?

Answer:

When water from the top of the waterfall falls towards the ground, its potential energy decreases. This decrease of potential energy is converted into kinetic energy. After water strikes the ground, a portion of this kinetic energy is converted to heat energy.

Now, a portion of this generated heat remains confined to water to increase its temperature. For this reason, temperature at the ground level is slightly higher than that at the top.

Question 5. If a chunk of ice is dropped from a distant height, then why does a portion of it melt after striking the ground?

Answer:

At normal temperature, if a chunk of ice at 0°C is dropped from a distant height to the ground, its potential energy decreases. This decrease of potential energy is converted into kinetic energy. After it strikes the ground, a portion of this kinetic energy is converted to heat energy.

Apart from this, heat is also generated due to its friction with air. Now a part of this produced heat remains confined to ice and melts a portion of the ice.

Question 6. What do you mean by the state or phase of a material?

Answer:

State Or Phase Of A Material Means

A material may exist in different forms. These forms are physically distinct from each other but have the same chemical composition. By absorption or release of heat or by any other mechanical means, the forms of a material can be changed. Each form of the material is known as a distinct state or phase.

Question 7. What is the change of state of a material? How many types of change are envisaged on the basis of absorption or release of heat? What are those changes?

Answer:

Change of state of a material is defined as the phenomenon of transformation of a material from one state to another due to either absorption or release of definite amount of heat to or from a material.

Two types of change are envisaged. The changes are:

  1. Higher change of state and
  2. Lower change of state.

Question 8. What do you mean by higher change of state?

Answer:

Higher Change Of State Means:-

A change of state of a material due to application of heat is called higher change of state.

A solid material can be converted to a liquid material and a liquid material can be converted to a gaseous material by absorption of heat. Transformation from solid to liquid state is called melting and transformation from liquid to gaseous state is called vaporisation. Thus, melting and vaporisation are higher change of state.

Question 9. What do you mean by lower change of state?

Answer:

Lower Change Of State Means:-

A change of state of a material due to release of heat is called lower change of state.

A liquid material can be converted to a solid material and a gaseous material can be converted to a liquid material by release of heat. Transformation from liquid to solid state is called solidification and transformation from gaseous to liquid state is called condensation. Thus, solidification and condensation are lower change of state.

Question 10. With the help of an experiment, demonstrate that temperature remains constant during the time of change of state.

Answer:

Some amount of ice is taken in a vessel. Now a thermometer is clamped by its side and its bulb is immersed in ice. Reading of the thermometer is now 0°C. Now the vessel is heated slowly. Ice starts melting due to absorption of heat.

It is observed that till the total amount of ice melts away, the reading of the thermometer remains constant at 0°C. After the total amount of ice melts in the water, it is observed that the reading of the thermometer starts increasing. It is inferred from this experiment that the temperature remains constant during the time of change of state.

WBBSE Solutions For Class 9 Physical Science Chapter 6 Equivalence Of Work And Heat Latent Heat Experiment Demonstrate That Temperature Remains Constant

Question 11. What do you mean by latent heat?

Answer:

Latent Heat:-

During the period of change of state of any material, whether there is any absorption or release of heat, there is not any change of temperature. The amount of heat released or absorbed per unit mass of a substance during the change from one state to another by keeping the pressure constant is called the latent heat of the substance for the constant change of state.

Question 12. What are the units of latent heat in the CGS system and SI? Establish a relationship between these two units.

Answer:

  1. Units of latent heat in the CGS system and SI are cal/g and J/kg, respectively.
  2. 1 cal/g = \(\frac{4.2 \mathrm{~J}}{10^{-3} \mathrm{~kg}}\) = 4200 J/kg

Question 13. Calculate the dimensional formula of latent heat.

Answer:

Dimensional Formula Of Latent Heat:-

= \(\frac{\text { dimenșional formula of heat }}{\text { dimensional formula of mass }}\)

= \(\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{M}}=\mathrm{L}^2 \mathrm{~T}^{-2}\)

Question 14. What do you understand by the statement: Latent heat of melting of ice is 80 cal/g

Answer:

The latent heat of melting of ice is 80 cal/g means that under normal pressure, 80 cal of heat is required to convert 1 g of ice at 0°C to 1 g of water at 0°C.

Question 15. What do you understand by the statement: The latent heat of vaporization of water is 537 cal/g.

Answer:

The latent heat of vaporization of water is 537 cal/g means that under normal pressure, 537 cal of heat is required to convert lg of water at 100°C to 1 g of vapor at 100°C.

Question 16. Give a graphical explanation of latent heat.

Answer:

Graphical Explanation Of Latent Heat:-

Solid material is taken. Suppose the initial temperature of the material is t1. Point P indicates the initial temperature of the material. If heat is supplied to the material at a definite rate, its temperature increases.

This is represented by the line PQ in the graph. Then the material starts melting. Point Q indicates the beginning of melting. The temperature of point Q is the melting temperature tm. Now, if heat is given continuously, the material melts and till the melting is complete, temperature remains constant.

Line QR in the graph represents the melting of the material. Melting is complete at point R where the entire solid material has been converted to liquid. Now if further heat is given, the temperature of the liquid increases. Line RS in the graph represents this increase.

In the same way, we may take a liquid material with an initial temperature of t2 and start releasing heat at a definite rate. Change of temperature with time is depicted. where point T denotes the initial temperature of the liquid.

The reduction of the temperature of the liquid is expressed by line TU. At point U, the solidification of liquid starts and it ends at point V.

WBBSE Solutions For Class 9 Physical Science Chapter 6 Equivalence Of Work And Heat Latent Heat Graphical Explanation Of Latent Heat

The temperature remains unchanged from U to V. At point V, the entire liquid material has been converted to solid material. After this, if heat release goes on, the temperature of the solid is decreased which is represented by line VW.

QR portion and UV portion represent the latent heat of absorption and latent heat of release, respectively. These portions of the lines are straight lines parallel to the time axis because in these two cases, the temperature remains unchanged

Question 17. Why does ice at 0°C feel colder than water at 0°C?

Answer:

Ice At 0°C Feel Colder Than Water At 0°C:-

Under normal pressure; 1 g of ice at 0°C is obtained after the release of 80 cal of heat from 1 g of water at 0°C. This means that 1 g of ice at 0°C contains 80 cal less heat than 1 g of water at 0°C. Therefore, ice at 0°C feels colder.

Question 18. Why does steam at 100°C feel warmer than water at 100°C?

Answer:

Steam At 100°C Feel Warmer Than Water At 100°C:-

Under normal pressure, 1 g of steam at 100°C is obtained after applying 537 cal of heat to lg of water at 100°C. This means that lg of Steam at 100°C contains 537 cal more heat than lg of water at 100°C. For this reason, steam at 100°C feels warmer than water at 100°C.

Question 19. Why is the latent heat of vaporization higher than the latent heat of melting?

Answer:

Yes, Latent Heat Of Vaporisation is Higher Than the Latent Heat Of Melting:-

There are strong intermolecular forces working amongst the molecules in a solid matter. Compared to these, intermolecular forces in liquid are less, and in gases, these forces are significant. When converted from liquid to gaseous state, intermolecular distances between the molecules increase more than when it is converted from solid to liquid state.

As a result, during the transformation from liquid to gaseous state, more work is to be done against these intermolecular attractive forces, requiring more heat. For this reason, the latent heat of vaporization is higher than the latent heat of melting.

Question 20. Why is water kept in an ice tray at 0°C not converted into ice?

Answer:

Water is kept in an ice tray. If the temperature of the water is more than 0°C, then that water releases heat to become water at 0°C and some amount of ice melts in the process. At this stage, as the temperature of ice and water are at 0°C, there is no exchange of heat. So, when water is kept in an ice tray, it does not get converted to ice.

Chapter 6 Equivalence Of Work And Heat Latent Heat Very Short Answer Type Questions Choose The Correct Answer

Question 1. The amount of heat required to convert 100 g of water at a temperature of 100°C to 100 g of steam at a temperature of 100°C is (the latent heat of vaporization = 540 cal/g)

  1. 5400 cal
  2. 540000 cal
  3. 27000 cal
  4. 54000 cal

Answer: 4. 54000 cal

Question 2. If the mechanical equivalent of heat, J = 4.2 J/cal, how much work has to be performed to produce 5 cal of heat?

  1. 20J
  2. 42J
  3. 21J
  4. 8.4J

Answer: 3. 21J

Question 3. During the time of higher change of state

  1. Heat is absorbed
  2. Heat is released
  3. Heat is neither applied nor released
  4. Temperature increases

Answer: 1. Heat is absorbed

Question 4. During the time of lower change of state

  1. Heat is absorbed
  2. Heat is released
  3. Heat is neither absorbed nor released
  4. Temperature decreases

Answer: 2. Heat is released

Question 5. The value of the mechanical equivalent of heat in the CGS system is

  1. 4.2 J/cal
  2. 4.2 x 107 erg/cal
  3. 4.2 x 106 erg/cal
  4. 4.2 x 105 erg/cal

Answer: 2. 4.2 x 107 erg/cal

Question 6. If the latent heat of vaporization of water is 540 cal/g, then its value in SI is

  1. 22.68 x 105 J/kg
  2. 22.68 x 106 J/kg
  3. 5.4 x 106 J/kg
  4. 5.4 x 105 J/kg

Answer: 1. 22.68 x 105 J/kg

Question 7. The correct relationship between work (W) and heat (H) is

  1. W = \(\frac{H}{J}\)
  2. \(\frac{W}{H}=\frac{1}{J^2}\)
  3. H = \(\frac{W}{J}\)
  4. W = \(\frac{J}{H}\)

Answer: 3. H = \(\frac{W}{J}\)

Question 8. At any point in time during the absorption of heat

  1. Only change of state takes place
  2. Only a change of temperature takes place
  3. Change of state or temperature takes place
  4. Both change of state and temperature take place

Answer: 3. Change of state or temperature takes place

Question 9. Due to the release of heat, the electrical conductivity of a body

  1. Decreases
  2. Increases
  3. Remains constant
  4. Sometimes increases, sometimes decreases

Answer: 3. Remains constant

Question 10. According to scientists Rumford and Davy, which energy is converted to heat energy?

  1. Chemical energy
  2. Electrical energy
  3. Mechanical energy
  4. Potential energy

Answer: 2. Electrical energy

Question 11. When mechanical energy is converted to heat energy, then

  1. the kinetic energy of the body increases
  2. The kinetic energy of atoms and molecules of the body increases
  3. The water equivalent of the body increases
  4. Specific heat of the body increases

Answer: 2. Kinetic energy of atoms and molecules of the body increases

Question 12. The amount of heat required to melt the ice of 1 g into the water at 0°C is

  1. 100 cal
  2. 500 cal
  3. 4 cal
  4. 80 cal

Answer: 4. 80 cal

Question 13. The amount of heat required to convert 1 g of water into vapour at 100°C is

  1. 537 cal
  2. 100 cal
  3. 80 cal
  4. 4 cal

Answer: 1. 537 cal

Question 14. On the upper surface of an uncovered liquid

  1. Condensation = vaporisation
  2. Condensation > vaporisation
  3. Condensation < vaporisation
  4. Data incomplete

Answer: 3. Condensation < vaporisation

Question 15. If there is no change of state, then which of the following quantities is unnecessary for the calculation of absorption or release of heat?

  1. Mass
  2. Latent heat
  3. Specific heat
  4. Change of temperature

Answer: 2. Latent heat

Question 16. If heat is applied to dry ice (solid carbon dioxide), then it is directly converted to gaseous carbon dioxide. This phenomenon is called

  1. Vaporization
  2. Boiling
  3. Sublimation
  4. Melting

Answer: 3. Sublimation

Question 17. If heat is applied to water at 0°C, then

  1. Density increases at first and then decreases
  2. Volume increases at first and then decreases
  3. Volume increases continuously
  4. Mass increases at first and then decreases

Answer: 1. Density increases at first and then decreases

Question 18. A piece of ice is floating in a glass of water at 3°C. If this ice melts completely, the level of water in the glass

  1. Decreases
  2. Increases
  3. Remains same
  4. Is unpredictable

Answer: 2. Increases

Question 19. A piece of ice is floating in a glass of water at 5°C. Melting of the piece of ice decreases the temperature of water to 4°C. At this condition, the level of water

  1. Decreases
  2. Increases
  3. Remains same
  4. Data incomplete

Answer: 1. Decreases

Question 20. On the upper surface of a liquid kept in a closed vessel

  1. Rate of condensation > rate of vaporization
  2. Rate of condensation < rate of vaporization
  3. Rate of condensation = rate of vaporization
  4. Rate of condensation ≥ rate of vaporization

Answer: 3. Rate of condensation = rate of vaporization

Question 21. Vaporization of water takes place

  1. At 0°C
  2. At 15°C
  3. Only at 100°C
  4. At any temperature from 0°C to 100°C

Answer: 4. At any temperature from 0°C to 100°C

Question 22. At what temperature latent heat is required for the conversion of water to vapor?

  1. At 0°C
  2. At 14.5°C
  3. Only at 100°C
  4. At a temperature between 0°C and 100°C

Answer: 4. At a temperature between 0°C and 100°C

Question 23. A phenomenon or state that does not depend on the amount of water vapour present in air is

  1. Falling of dew
  2. Formation of fog
  3. Freezing of water to form ice
  4. Relative humidity

Answer: 3. Freezing of water to form ice

Question 24. Latent heat of vaporization of water

  1. Depends on the amount of heat absorbed
  2. Depends on the mass of water
  3. Depends on the volume of water
  4. Does not depend on anything

Answer: 4. Does not depend on anything

Question 25. It takes 10 minutes to raise the temperature of some amount of water from 0°C to 100°C and to vaporize the same amount, it takes another 55 minutes with the help of a heater. What is the latent heat of vaporization in this case?

  1. 530 cal/g
  2. 540 cal/g
  3. 550 cal/g
  4. 560 cal/g

Answer: 3. 550 cal/g

Question 26. 30 g of steam at temperature 100°C is supplied to 100 g of water at 20°C. How much gram of steam is condensed due to this process? [Latent heat of condensation of steam =540 cal/g]

  1. 17.8 g
  2. 16.8 g
  3. 15.8 g
  4. 14.8 g

Answer: 4. 14.8 g

Question 27. Which of the following is a higher change of state?

  1. Solidification
  2. Vaporization
  3. Condensation
  4. None of these

Answer: 2. Vaporisation

Question 28. Which of the following is a lower change of state?

  1. Melting
  2. Vaporization
  3. Condensation
  4. None of these

Answer: 3. Condensation

Question 29. The dimensional formula of latent heat is

  1. LT-2
  2. MT-2
  3. L2T-2
  4. ML2T-2

Answer: 3. L2T“2

Question 30. A lead bullet speeding with a velocity of 200 m/s hits a wall. If 25% kinetic energy of the bullet is converted into heat, what is the increase in temperature of lead? (specific heat of lead = 0.03 cal • g-1 • °C-1)

  1. 29.67°C
  2. 39.67°C
  3. 49.67°C
  4. 59.67°C

Answer: 2. 39.67°C

Chapter 6 Equivalence Of Work And Heat Latent Heat Answer In Brief

Question 1. Some amount of heat is required for a change of state of a material. What is that heat called?

Answer: The heat that is required for the change of state of a material is called latent heat.

Question 2. What is the unit of latent heat in the CGS system?

Answer: The unit of latent heat in the CGS system is cal/g.

Question 3. What is the unit of latent heat in SI?

Answer: The unit of latent heat in SI is J/kg.

Question 4. What type of change of state is represented by melting and vaporization?

Answer: Melting and vaporization represent a higher change of state.

Question 5. What type of change of state is represented by solidification and condensation?

Answer: Solidification and condensation represent a lower change of state.

Question 6. What is the value of the mechanical equivalent of heat in a CGS system?

Answer: The value of the mechanical equivalent of heat in the CGS system is 4.2 x 107 erg/cal.

Question 7. What is the value of the mechanical equivalent of heat in SI?

Answer: The mechanical equivalent of heat in SI is 1.

Question 8. If there is no change of temperature of a body in spite of heating, can there be a change of state of the body?

Answer: Yes, there may not be a change in the temperature of a body but due to the application of heat, there may be a change in the state of the body.

Question 9. What is sublimation?

Answer: Sublimation is the process by which a solid material is converted directly to its gaseous state due to the absorption of heat.

Question 10. What is the water equivalent of a body if its heat capacity is 50 cal/°C?

Answer: The water equivalent of the body is 50g.

Question 11. If work done is fully converted to heat, what is the relationship between work done and heat produced?

Answer: Work done and heat produced are directly proportional to each other.

Question 12. Name four fundamental states of matter.

Answer: The four fundamental states of matter are solid, liquid, gaseous, and plasma state.

Question 13. Write the unit of J in SI.

Answer: J has no unit in SI.

Question 14. What amount of heat will produce if 1J of work is fully converted into heat?

Answer: The amount of heat produced = 1/4.2 = 0.238 cal.

Question 15. To produce 1 kcal heat how much work is needed?

Answer: The amount of work needed to produce 1 kcal of heat is = 1000 x 4.2 J = 4200 J.

Question 16. What is the value of latent heat of fusion of ice in the CGS system?

Answer: The value of latent heat of fusion of ice in the CGS system is 80 cal/g.

Question 17. What is the latent heat of vaporization of water in cal • g-1?

Answer: The value of latent heat of vaporization of water in cal • g-1 unit is 540.

Question 18. What is the specific heat of water during boiling at 100°C?

Answer: During boiling at 100°C specific heat of water is infinite.

Question 19. What amount of heat is needed to convert 100 g of water at 100°C to steam at 100°C?

Answer: The latent heat of vaporization of water is 537 cal/g.

∴ The amount of heat required = 537 x 100 = 53700 cal.

Question 20. What is the latent heat of condensation?

Answer: The amount of heat extracted from the unit mass of a vapor to change it into its liquid state at a constant temperature, is the latent heat of condensation or the vapor.

Question 21. Why steam at 100°C is a better warming agent than water at 100°C?

Answer: 1 g of steam at 100°C can transfer 537 cal more heat than 1 g of water at 100°C. That is why steam at 100°C is a better warming agent than water at 100°C

Question 22. Under what conditions can heat be supplied to a body without causing any change in temperature?

Answer: During the change from one state to another the substance absorbs heat without any change. in the temperature.

Chapter 6 Equivalence Of Work And Heat Latent Heat Fill in The Blanks

Question 1. _______ is the amount of work that needs to be done to produce one unit of heat.

Answer: Mechanical equivalent of heat

Question 2. Dimension of L in the dimensional formula of latent heat is ________

Answer: 2

Question 3. On the basis of absorption or release of heat, change of state of any material is of _______ types.

Answer: Two

Question 4. Water falls from top to bottom in a waterfall. Compared to the water at the top, the temperature is slightly __________  at the bottom.

Answer: Higher

Question 5. Water kept in the ice tray at 0°C temperature _________ convert into ice.

Answer: Does not

Question 6. Matter requires latent heat for any change Of ________

Answer: State

Question 7. Latent heat of melting of ice is _________

Answer: 80 cal/g

Question 8. An amount of 8.4 J work has to be done to produce ________ of heat.

Answer: 2 cal

Question 9. _______ of liquid takes place at any temperature.

Answer: Vaporisation

Question 10. Rate of vaporization _________ with increase of temperature of liquid.

Answer: Increases

Question 11. Unit of water equivalent in SI is __________

Answer: kg

Chapter 6 Equivalence Of Work And Heat Latent Heat State Whether True Or False

Question 1. The latent heat of melting is greater than the latent heat of vaporization.

Answer: True

Question 2. Heat is the transformed state of the kinetic energy of a particle.

Answer: True

Question 3. In the steam engine heat is converted into mechanical work.

Answer: True

Question 4. Steam at 100°C causes more severe burns than water at 100 °C.

Answer: True

Question 5. The gradient of temperature vs time graph of a substance has – ve value when heat is supplied to it.

Answer: False

Question 6. The gradient of temperature vs time graph of a substance has – ve value when heat is extracted from it.

Answer: True

Question 7. Sublimation is the direct conversion from solid to gaseous state.

Answer: True

Chapter 6 Equivalence Of Work And Heat Latent Heat Numerical Examples

Useful information

If W amount of work is fully converted into heat and H is the amount of heat produced, then according to Joule’s law.

W = JH, J is a constant, called the mechanical equivalent of heat.

In CGS J = 4.2 x 107 erg/cal and in SI J = 1 and J = 4.2 J/cal

If Q amount of heat to be supplied or given out to change the state of mass m of a substance, then, Q = mL or, L = \(\frac{Q}{m}\)

L = Latent heat of the substance

If a piece of ice of mass m hits the earth’s surface from a height h, if total kinetic energy is converted into heat the amount of energy released H = \(\frac{mgh}{J}\)

g = acceleration due to gravity.

Question 1. what is the amount of work done to produce heat of 50 cal?

Answer:

Heat produced, H = 50 cal

As the mechanical equivalent of heat, J = 4.2 J/cal, the amount of work done is given by W = JH = 4.2 x 50 = 210 J

Question 2. A block is dragged 4 m on a rough floor. If the frictional force between the block and the floor is 40 N and the work done is fully transformed into heat, then how much heat is produced? (Mechanical equivalent of heat, J = 4.2 J/cal)

Answer:

The frictional force between the block and the floor, F = 40 N

Displacement of the block, s = 4 m

Work done against friction,

W = Fs = 40 N x 4 m = 160 J

Now if the work done against friction is fully converted into heat, then the heat produced is given by

H = \(\frac{W}{J}=\frac{160}{4.2}\) = 38.1 cal

Question 3. The temperature of a piece of lead is 27°C. What is the minimum velocity with which it should strike a wall so that the heat point of lead is 327°C, the specific heat of lead is 0.03 cal • g-1 • °C-1, latent heat of melting is 5 cal/g and the mechanical equivalent of heat, J = 4.2 J/cal.)

Answer:

Suppose, the mass of the piece of lead = m g and its minimum velocity = v m/s

So the kinetic energy of the piece of lead,

E = 1/2 mv2 erg

Mechanical equivalent of heat,

J = 4.2 J/cal = 4.2 x 107 erg/cal

Now if the kinetic energy of lead is fully transformed into heat after it strikes a wall, then the heat produced is given by

H = \(H=\frac{E}{J}=\frac{m v^2}{2 \times 4.2 \times 10^7} \mathrm{cal}\)

To melt the piece of lead, its temperature is to be raised from 27°C to 327°C.

So the heat required to increase the temperature is given by H1 = m x 0.03 x(327 – 27) = 9m cal

and the required latent heat for melting the piece of lead, H2 = m x 5 = 5m cal

According to the condition, H =H1 + H2

or, \(\frac{m v^2}{2 \times 4.2 \times 10^7}=9 m+5 m=14 m\)

or, \(v^2=14 \times 2 \times 4.2 \times 10^7=11.76 \times 1-10^8\)

v = \(\sqrt{11.76 \times 10^3}=3.43 \times 10^4 \mathrm{~cm} / \mathrm{s}\)

Question 4. Calculate the amount of work to be done to convert 100 g of ice at 0°C to water of temperature 100°C. (Latent heat of melting of ice =80 cal/g, the mechanical equivalent of heat, J = 4.2 J/cal.)

Answer:

Latent heat required for melting of ice,  H1 = 100 x 80 = 8000 cal

Necessary heat is required for raising 100 g of water at 0°C to a temperature of 100°C,

H2 = 100 x 1 x(100 – 0) = 10000 cal

So the heat required to convert 100 g of ice at 0°C to water of temperature 100°C is given by H =H1 + H2

= 8000 + 10000 = 18000 cal

Hence the required work to be done,

W = JH = 4.2 X 18000 = 75600 J

Question 5. What is the height from which a chunk of ice at 0°C has to be dropped so that it melts completely due to its impact on the ground?

Answer:

Suppose the mass of the chunk of ice = m g and h is the height from where it has been dropped.

Now, just before the moment it strikes the ground, its kinetic energy is equal to the loss of potential energy = mg.

This kinetic energy is transformed into heat energy due to its impact on the ground.

Now if J is the mechanical equivalent of heat, then heat produced, H = \(\frac{mgh}{J}\).

Latent heat of melting of ice = 80 cal/g, i.e., the heat required to melt mg of ice = 80m cal.

so, 80m = \(\frac{mgh}{J}\)

or h = \(\frac{80 \times J}{9}=\frac{80 \times 4.2 \times 10^7}{980}\)

∴ h = 34.286 x 105 cm = 34.286 cm

Question 6. What is the temperature difference between the top and the bottom of a water fall of 100 m height if 50% of heat produced remains confined in water?

Answer:

Height of the waterfall, h = 100 m = 100 x 100 cm = 1 x 104 cm

Suppose, water is falling from the top to the bottom at the rate of m g/s.

Now, just before the moment it strikes the ground, its kinetic energy is equal to the loss of potential, energy = mgh.

This kinetic energy is converted fully into heat energy after striking the ground. So, heat produced = \(\frac{mgh}{J}\), where mechanical equivalent of heat, J = 4.2 x 107 erg/cal.

Now, heat confined in water = \(\frac{50}{100} \times \frac{m g h}{J}=\frac{m g h}{2 J}\)

and specific heat of water, s = 1 cal • g-1 • °C-1

If the temperature of water increases by t°C due to its fall from the top of the waterfall, then

mst = \(\frac{m g h}{2 J} \text { or, } t=\frac{g h}{2 \mathrm{sJ}}\)

∴ t = \(\frac{980 \times 1 \times 10^4}{2 \times 1 \times 4.2 \times 10^7}=0.117^{\circ} \mathrm{C}\)

Question 7. While passing through the earth’s atmosphere, the velocity of a meteorite of mass 42 g is reduced from 10 km/s to 5 km/s. If the heat of 3.75 x 108 cal is produced due to this change of kinetic energy, calculate the mechanical equivalent of heat.

Answer:

According to the question, mass of the meteorite, m = 42kg; initial velocity, u = 10 km/s = 10 x 103 m/s and final velocity, v = 5 km/s = 5 x 103 m/s.

∴ loss of kinetic energy by the meteorite

= \(\frac{1}{2} m u^2-\frac{1}{2} m v^2=\frac{1}{2} m\left(u^2-v^2\right)\)

= \(\frac{1}{2} m(u+v)(u-v)\)

= \(\frac{1}{2} \times 42 \times\left(10 \times 10^3+5 \times 10^3\right) \times\left(10 \times 10^3-5 \times 10^3\right)\)

= 21 x 15 x 103 x 5 x 103

= 21 X 75 x 106 J

Now if J is the mechanical equivalent of heat, then the heat produced is given by

H = \(\frac{\text { loss of kinetic energy }}{J}\)

or, \(3.75 \times 10^8=\frac{21 \times 75 \times 10^6}{J}\)

∴ J = \(\frac{21 \times 75 \times 10^6}{3.75 \times 10^8}=4.2 \mathrm{~J} / \mathrm{cal}\)

Question 8. What happens if 4g of steam at a temperature of 100 °C is passed through 100 g of water at 20°C? (Latent heat of condensation of steam = 540 cal/g.)

Answer:

If 4 g of steam is transformed into water of temperature 100°C, then heat released, Q: = 540 X 4 = 2160 cal.

If temperature of 100 g water at 20°C is to be raised to 100°C, then required heat, Q2 = 100 x 1 x (100 – 20) = 8000 cal

As Q1 < Q2, the entire amount of steam is transformed into water.

Suppose, final temperature = t°C.

According to basic principle of calorimetry, heat absorbed by water = latent heat released by steam + heat released by water at 100°C

or, 100 x 1 x (t- 20) = 540 x 4 + 4 x 1 x (100 – t)

or, 100t – 2000 = 21600 + 400 – 4t or, 100t + 4t = 2160 + 400 + 2000 or, 104t = 4560

∴ t = \(\frac{4560}{104}\) = 43.85°C

Question 9. What amount of steam in gram is converted to water if 40 g of steam at 100°C passes through 100 g of water at 20°C? (Latent heat of condensation of steam =540 cal/g)

Answer:

If 40 g of steam is transformed into water of temperature 100°C, then heat released, Q1 = 540 X 40 = 21600 cal

If temperature of 100 g of water is to be raised from 20°C to 100°C,then the required heat, Q2 = 100 x 1 x (100 – 20) = 8000 cal

As Q2 < Q1, entire amount of steam is not transformed into water.

Suppose, m g steam is transformed to water. From the basic principle of calorimetry, 540m = Q2 or, 540m = 8000

∴ m = \(\frac{8000}{540}\) = 14.81 g

Question 10. Calculate the amount of heat required to convert 10 g of ice at -5°C into steam. (Latent heat of melting of ice = 80 cal/g, latent heat of valorisation of water = 537 cal/g and specific heat of ice = 0.5 cal • g-1 • °C-1)

Answer:

Mass of ice = 10 g.

Heat required to increase the temperature of ice from -5°C to 0°C, H1 = 10 x 0.5 {0-(-5)} =25 cal

Necessary heat for melting of ice, H2 = 10 x 80 = 800 cal

Required heat to increase the temperature of 10 g of water from 0°C to 100°C, H3 = 10 x 1 x (100 – 0) = 1000 cal

Heat required for vaporisation of water, H4 = 537 x 10 cal = 5370 cal

total heat required is given by H = H1 + H2 + H3 + H4

= (25 + 800 + 1000 + 5370) cal = 7195 cal

Question 11. Temperature of 200 g of water has to be raised from 24°C to 90°C by passing steam through it. Calculate the mass of steam required for it. (Latent heat of condensation of steam = 540 cal/g)

Answer:

Suppose, mass of steam = m g

When m g of steam at 100°C condenses and transforms to water at 100°C, then the heat released.

H1 = m x 540 cal = 540m cal

Again, when temperature of m g of water at 100°C is reduced to 90°C, then heat released is given by H2 = m x 1 x (100 – 90) cal = 10m cal

∴ total heat released = H1 + H2 = 540m + 10m = 550m cal

Heat required to increase the temperature of 200 g of water from 24°C to 90°C, H3 = 200 x 1 x (90 – 40) cal = 13200 cal

According to the condition, H1 + H2 = H3 or, 550m = 13200

∴ m = \(\frac{13200}{550}\) = 24 g

Question 12. Equal amounts of water and ice at 0°C temperature are mixed together, if the entire quantity of ice melts to form water at 0°C, then what is the initial temperature of water? (Latent heat of melting of ice = 80 cal/g)

Answer:

Let the mass of water and ice be m g and initial temperature of water =t°C.

Latent heat required for melting of entire ice = 80m cal.

Heat released when water temperature comes down form t°C to 0°C = m x 1 x (t – 0) = m cal.

According to the basic principle of calorimetry, mt = 80m

∴ t = 80°C

Question 13. What is the result if 2 x 104 cal of heat is extracted for 0.4 kg of water at 30°C temperature? What is the final temperature?

Answer:

Mass of water, m = 0.4 kg = 40 kg = 400 g

Specific heat of water, s = 1 cal • g-1 • °C-1

Total heat released, Q = 2 x 104 cal = 20000 cal

Now, heat released when total water at 30°C is converted to water at 0°C, Q1 = 400 x 1 x (30 – 0) = 12000 cal

Amount of absorbed heat, Q2 = Q- Q1 = 20000 – 12000 = 8000 cal

Now, the amount of released heat when entire amount of water of 0°C temperature is transformed to ice of 0°C is given by Q3 = 400 x 80 = 32000 cal

As Q2 < Q3, total water is not transformed to ice.

Let m1 g water is transformed to ice.

So, m1 = Q2 or, 80m1 = 8000

∴ m1 = 100 g

Hence, if 2 x 104 cal of heat is extracted from 0.4 kg of water at 30°C temperature, final temperature is 0°C, in which 100 g of ice and (400 – 100) g = 300 g of water is available.

Question 14. How do you divide 1 kg of water at 5°C temperature in two parts, such that if one part is transformed into ice, then the released heat is sufficient to transform the other part into vapour? (Latent heat of melting of ice = 80 cal/g and latent heat of vaporisation of water = 540 cal/g .)

Answer:

Mass of water = 1 kg = 1000 g

Specific heat of water = 1 cal • g-1 • °C-1

Suppose, Mg of water becomes ice and (1000 -M) g of water is vaporised.

Now when M g of water at 5°C is transformed to ice at 0°C, then the heat released is given by

Q1 = M x 1 x (5 – 0) + 80M = 5M + 80M = 85Mcal

Again, heat required to transform (1000 – M) g of water at 5°C to steam of 100°C,

Q1 = (100 – M) x 1 x (100 – 5) + 540 x (1000 – M) = 635 x (1000 -M) cal

According to the condition, Q1 = Q2

or, 85 M = 635 X (1000 – M)

or, 85M = (635 x 1000) – 635M

or, SSM + 635M= 635 X 1000

or, 720M = 635 x 1000

∴ M = 881.94 g

and 1000 – M = 1000 – 881.94 = 118.06 g

Hence, the amount of heat that is released when 881.94 g of water is transformed into ice is sufficient to transform 118.06 g of water into vapor.

Question 15. 168.75g of water is produced when steam at 100°C is passed through a chunk of ice at 0°C. The mass of ice before the passing of steam is 450 g and after the passing of steam the mass becomes 300 g. What is the latent heat of the vaporization of steam? [Latent heat of melting of ice = 80 cal/g]

Answer:

The mass of ice before the passing of steam is 450 g and after the passing of steam the mass becomes 300 g.

Therefore, mass of melted ice = (450 – 300) = 150 g

So, heat required for melting of 150 g of ice, Q1 = 150 x 80 = 12000 cal

Mass of condensed steam = (168.75- 150) g = 18.75 g

Let the latent heat of vaporization of steam = L cal/g.

Now when 18.75 g of steam at 100°C is converted to water at 0°C, then the heat released is given by

Q2 = 18.75 L + 18.75 x 1 x (100 – 0)

= 18.75 L+ 1875 cal

According to the question,

Q1 = Q2 or, 12000 = 18.75L + 1875

or, 18.75L  = 12000 – 1875

or, 18.75L = 10125

∴ L = \(\frac{10125}{18.75}\) = 540 cal/g

Question 16. What amount of heat is required to convert 10 g of ice at 0°C to 10 g of water at 10°C?

Answer:

The total amount of heat is required = heat is required for melting of ice + heat is required to increase temperature from 0°C to 10°C

= m • L + m • s • (t2 – t1)

= 10 x 80 + 10 x 1 x (10 – 0)

= 800 + 100 = 900 cal

Question 17. What amount of heat is required to convert 10 g of water at 0°C to water vapor at 100°C.

Answer:

The total amount of heat is required = heat is required to increase the temperature from 0°C to 100°C + heat is required for vaporization

= m • s • (t2 – t1) + m • L

= 10 • 1 • (100 – 0) + 10 X 537 = 1000 + 5370 = 6370 cal

WBBSE Solutions For Class 9 Physical Science Chapter 7 Some Properties Of Sound And Characteristics Of Sound

Chapter 7 Some Properties Of Sound And Characteristics Of Sound Synopsis

Laws Of Reflection Of Sound:

  1. The incident sound wave, the reflected sound wave, and the normal to the reflector at the point of incidence, all lie on the same plane.
  2. The angle of incidence is equal to the angle of reflection.
  3. Regular reflection of light waves occurs from a small-sized smooth reflector, whereas sound waves are not reflected by it. Again, regular reflection of sound waves occurs from a large-sized rough reflector.
  4. If the frequency of sound is within the range of 20 Hz to 20 kHz, it produces a sensation of hearing in the human ear and is known as audible sound.
  5. Sound having a frequency lower than 20 Hz is known as infrasonic sound.
  6. Sound having frequency higher than 20 kHz is known as ultrasonic sound.
  7. The wavelength of audible sound approximately ranges from 1.5 cm to 16 cm, hence a large reflector is required for the reflection of such sound waves.
  8. If a sound produced from a source is reflected by a reflector and reaches the listener in addition to the original sound, then the reflected sound is called an echo of the original sound.
  9. If we hear any sound, its trace remains in the brain for approximately 0.1s. During this time duration of 0.1s, any other sound entering the ears cannot be distinguished. This time duration is called persistence of hearing.
  10. If the velocity of sound in air at 27°C is 348 m/s, the minimum distance between the source and the reflector for hearing an echo of an inarticulate sound is 17.4 m. The minimum distance between the source and the reflector for hearing the echoes of monosyllabic, bisyllabic, and trisyllabic sounds are 34.8m, 69.6m, and 104.4m respectively.
  11. The persistence of sound within a closed space after its source ceases to produce sound, caused by multiple reflections of the sound from walls, roof, etc. is known as reverberation.
  12. SONAR is an acronym for SOund Navigation And Ranging. With the help of ultrasonic sound, SONAR is used for the detection of ruins of a sunken ship or existence of a mountain deep inside an ocean.
  13. When sound of a single frequency is emitted from a tuning fork or different musical instruments, then the emitted sound is known as tone. Sound due to a mixture of more than one frequency is called a note.
  14. The sound which is produced by regular and periodic vibrations of the source and appears euphonious to our ears is known as musical sound. Sound of guitar, sitar, flute, tuning fork etc. are examples of musical sound.
  15. The sound which is produced by irregular and non-periodic vibrations of the source and appears cacophonous to our ears is known as noise. Sound of crackers, automobiles, uproar of crowd etc. are examples of noise.

There Are Three Characteristics Of A Musical Sound. These Are:

  1. Loudness,
  2. Pitch and
  3. Quality or timbre.

Read and Learn More WBBSE Solutions for Class 9 Physical Science and Environment

  1. Amount of sound energy passing through an unit area around a point in one second, perpendicular to the direction of sound is called intensity of the sound at that point.
  2. The dimensional, formula of intensity of sound is MT-3 and its unit in SI is W/m2.
  3. The characteristic property of a musical sound due to which a listener can differentiate between a high-pitched and a low-pitched sound is known as pitch of the sound.
  4. Frequency is the cause, pitch is its effect. Frequency is a measurable physical quantity. But pitch is a sensation or feeling which is not measurable.
  5. The characteristic property by which sound of same intensity and pitch emitted by different sources can be distinguished in known as the quality or timbre of the musical sound.

Chapter 7 Some Properties Of Sound And Characteristics Of Sound Short And Long Answer Type Questions

Question 1. Write down the laws of reflection of sound.

Answer:

The Laws Of Reflection Are:

  1. The incident sound wave, the reflected wave and the normal to the reflector at the point of incidence, all lie on the same plane.
  2. The angle of incidence is equal to the angle of reflection.

Question 2. Demonstrate with an experiment that during reflection of sound, angle of incidence = angle of reflection. Or, Demonstrate by an experiment that sound wave is reflected similar to a light wave.

Answer:

The Angle Of Incidence = Angle Of Reflection:-

Two hollow long pipes, open at both ends, are taken. Also, one table clock and one raised partition are taken. One long table is kept touching the wall and at a little distance from the wall, the partition is kept in a perpendicular way.

Now, the pipes are kept on the two sides of the partition in such a way that the axes of both the pipes intersect at the point O of the wall.

Now, a table clock is kept at the end A of the left pipe. The right pipe is now rotated slowly with respect to the point O so that a position is obtained, where tick-tock sound of the clock is heard loudly. Suppose, sound is heard loudly at the point D.

This is the reflected sound. The tick-tock sound of the clock passes through pipe AB, reflected at point O, and comes to the ear through the pipe CD and is heard. Now pencil marks are made at the points O, A, D, and E.

Next, pipe and partition are removed, and A,0; D,0; E,0 are connected. If ∠AOE and ∠DOE are measured, it is found that ∠AOE = ∠DOE, i.e., angle of incidence = angle of reflection.

So, it can be concluded that sound wave is reflected similar to a light wave.

WBBSE Solutions For Class 9 Physical Science Chapter 7 Some Properties Of Sound And Characteristics Of Sound Experiment During Reflection Of Sound

Question 3. How is the property of reflection of sound utilised in a stethoscope?

Answer:

Property Of Reflection Of Sound Utilised In A Stethoscope:-

  1. The stethoscope is a medical instrument using which doctors examine the heart or lungs of a patient. There are two rubber pipes in this instrument which are attached to a round metallic disc covered with a thin diaphragm.
  2. If the two open ends of the rubber pipes are fitted in the ears and the disc is placed on the chest of the patient, sound of the heart of the patient is reflected repeatedly through the rubbers tubes and reaches the ears of the doctor.

WBBSE Solutions For Class 9 Physical Science Chapter 7 Some Properties Of Sound And Characteristics Of Sound Reflection Of Sound utilised In Stethoscope

Question 4. Why do we curve our palms by the side of our ear to clearly hear a faint sound from a distant place?

Answer:

We Curve Our Palms By The Side Of Our Ear To Clearly Hear A Faint Sound From A Distant Place:-

To clearly hear a faint sound from a distant place, we curve our palm by the side of our ear. In addition to the direct faint sound, sound reflected from the palm also enters the ear. As a result, loudness of the sound increases to some extent. Here, palm of the hand works like a concave reflector.

WBBSE Solutions For Class 9 Physical Science Chapter 7 Some Properties Of Sound And Characteristics Of Sound By the Side of Our Ear Sound from A Distance Place

Question 5. Why is the roof of a modern auditorium curved like an arch?

Answer:

The Roof Of A Modern Auditorium Curved Like An Arch Because:-

The roof of a modern auditorium is curved like an arch so that it works as a concave reflector and if the source of sound is kept at its focus, sound spreads evenly throughout the auditorium. As a result, the loudness of the sound increases.

Question 6. How can one talk in a speaking tube?

Answer:

One Talk In A Speaking Tube As Follows:-

If we talk at one end of a long metallic tube, the produced sound advances through the tube without scattering and reaches the other end after repeated reflections. The listener on the other side hears this sound very clearly. This arrangement is called a speaking tube. A funnel is attached to the front side for the speaker.

Question 7. What do you mean by audible sound?

Answer:

Audible Sound Means:-

One can hear sound only when the number of vibrations per second, i.e., frequency of a vibrating body lies within the range of 20 Hz to 20000 Hz. This sound is called audible sound.

Question 8. What do you mean by infrasonic sound and ultrasonic sound? give examples.

Answer:

Infrasonic Sound And Ultrasonic Sound:-

  1. Infrasonic sound is the sound whose frequency is less than 20 Hz.
  2. Ultrasonic sound is the sound whose frequency is more than 20 kHz.

When a pendulum oscillates, its frequency remains less than 20 Hz. Sound produced in this case is infrasonic sound, hence inaudible. Again, when a bat flies, it produces a type of sound from its mouth whose frequency is more than 20kHz, that is, ultrasonic sound.

Question 9. A flying bat makes a sound from its mouth. What type of sound wave does it produce?

Answer:

The sound wave emitted by a flying bat from its mouth has a frequency more than 20 kHz. Hence, it is an ultrasonic sound. If there is any obstacle in front of the sound, sound waves get reflected. By hearing that reflected sound, the bat can locate the position of any obstacle. We can not hear this ultrasonic sound emitted from the mouth of a bat.

Question 10. Give some practical applications of ultrasonic sound.

Answer:

These Are Some Of The Practical Applications Of Ultrasonic Sound:

  1. In the process of ultrasonography, ultrasound is used to detect the location of a tumour in a human body.
  2. Ultrasonic sound is used to clean dirty clothes.
  3. Ultrasonic sound is also used to measure depth of the sea, to locate submarine or a shoal of fish, or to understand the presence of aquatic animals inside sea.

Question 11. Mention two applications of ultrasonic sound in medical science.

Answer:

Applications Of Ultrasonic Sound In Medical Science:-

It has been confirmed by experiments that there are some bacteria which can be destroyed by the application of ultrasonic sound. Also in the process of ultrasonography or USG, ultrasonic sound is used to know more about the internal muscles, different joints, arteries, etc. of the body.

Question 12. Write about a few animals in nature who normally make use of ultrasonic sound in their manner of living.

Answer:

Animals In Nature Who Normally Make Use Of Ultrasonic Sound In Their Manner Of Living Are Given Below:-

It has been observed that animals like bats, dolphins, whales, etc. utilise ultrasonic sound. Bats emit ultrasonic sound of very high frequency (100 kHz – 200 kHz) from their mouth at the time of flying.

This sound is reflected back from some obstacle in front of it and this helps the bat in recognising the obstacle. It thus flies freely without any impediment. Whales, dolphins, etc. living deep inside the sea communicate with each other through the use of ultrasound.

Question 13. Light wave is reflected regularly from a small and smooth reflector whereas sound wave is not reflected. Why?

Answer:

LightWave Is Reflected Regularly From A Small And Smooth Reflector Whereas Sound Wave Is Not Reflected

Reflection of a wave is possible only when the size of the reflector is larger than its wavelength. Again, whether reflection of a wave is regular or scattered, depends on the roughness of the surface of the reflector. If the wavelength of the wave is greater than the roughness of the surface of the reflector, there is regular reflection, otherwise the reflection is scattered.

The wavelength of sound in the audible range is from 1.5 cm to 16 m and wavelength of light in the visible range is from 1000 A (or 4 x 10-5 cm) to 8000 A (or 8 x 10-5 cm ). The size of a small, smooth reflector is comparatively larger than the wavelength of visible light.

So, in this case, light is reflected and the value of roughness being less than the wavelength of light, the reflection is regular. But as the size of a reflector is smaller than the wavelength of a sound wave, there is no reflection of sound wave.

Question 14. What are the differences between reflection of sound and reflection of light?

Answer:

The Differences Between The Reflection Of Sound And Reflection Of Light Are:

WBBSE Solutions For Class 9 Physical Science Chapter 7 Some Properties Of Sound And Characteristics Of Sound Difference Between Reflection Of Sound And Light

Question 15. What are the differences between sound wave and light wave?

Answer:

The Differences Between Sound Wave And Light Waves Are:

WBBSE Solutions For Class 9 Physical Science Chapter 7 Some Properties Of Sound And Characteristics Of Sound Difference Between Sound Wave And Light Wave

Question 16. What is the persistence of hearing?

Answer:

Persistence Of Hearing:-

If we hear any sound, its trace remains in the brain for approximately 0.1s. During this time duration of 0.1 s, more than one sound entering the ears cannot be distinguished. This time duration is called persistence of hearing.

Question 17. Calculate the minimum distance between the source and the reflector for hearing an echo of a short-lived sound; taking time of persistence of hearing as 0.1 second. [Given, velocity of sound in air = 348 m/s.]

Answer:

  1. As the velocity of sound in air is 348 m/s, it traverses a distance of 348 x 0.1 m = 34.8 m in 0.1s.
  2. So, the minimum distance between the source and the reflector for hearing an echo of a short-lived sound = \(\frac{34.8}{2}\) = 17.4 m.

Question 18. Calculate the minimum distance between the source and the reflector to hear an echo for a syllabic and intelligible sound, taking time of persistence of sound as 0.1 s. [Given that velocity of sound in air is 348 m/s and a human being cannot pronounce more than five syllables in one second.]

Answer:

Intelligible sound consists of one or more than one syllable. A human being cannot pronounce more than five syllables in one second. So, sound has to reach the ear after 1/5 s = 0.20 s, in order to hear an echo of a monosyllabic word. As the velocity of sound in air is 348 m/s, sound traverses a distance of 348 x 0.2 m * 69.6 m during this time.

In this case, minimum distance between the source and the reflector = \(\frac{69.6}{2}\) m = 34.8 m

Similarly, minimum distances between the source and the reflector, to hear an echo for a bisyllabic and trisyllabic sound are 34.8 x 2 = 69.6 m and 34.8 x 3 = 104.4 m respectively.

Question 19. What do you mean by an echo?

Answer:

Echo:-

If a sound produced from a source is reflected by a reflector situated at some distance and reaches the listener in addition to the original sound, then the reflected sound is called an echo of the original sound.

Question 20. State two conditions for the echo to be heard.

Answer:

Echo Can Be Heard If:-

  1. The minimum distance between the source of sound and the obstacle will be 17 m if velocity of sound in air be 340 m/s.
  2. The obstacle will be rigid, hard, and big in size (compare to wavelength of sound wave).

Question 21. Echo can’t be heard in a small room. Explain.

Answer:

Echo Can’t Be Heard In A Small Room:-

The minimum distance between source of sound and the obstacle to hear an echo should be 17 m if the velocity of sound in air be 340 m/s. Therefore echo can’t be heard in a small room.

Question 22. How is the depth of a sea measured with the help of echo?

Answer:

Depth Of A Sea Measured With The Help Of Echo As Follows:-

The depth of a sea can be measured with the help of echo. From the bottom of the ship, a source of sound (S) and a hydrophone (H) (a hydrophone is a microphone designed to be used underwater for recording or listening to underwater sound) are hung from the two sides at the same depth no.

If an intense and short-lived sound is produced from the source, one sound traverses along the straight path SH in time t1 to reach the hydrophone and another sound reaches the hydrophone in time t2, after getting reflected from the bottom at P. The first sound is the original sound and the second one is the reflected sound. A perpendicular PQ is drawn on SH from P.

WBBSE Solutions For Class 9 Physical Science Chapter 7 Some Properties Of Sound And Characteristics Of Sound Deapth Of A Sea Measured With The Help Of Echo

Now if V is the velocity of sound in water, then SH = Vt1

or, SQ = \(=\frac{1}{2} S H=\frac{V t_1}{2}\)

Again, SP + PH = Vt2

or, SP = \(\frac{V t_2}{2}\) [SP = PH]

∴ PQ = \(\sqrt{S P^2-S Q^2}\)

= \(\sqrt{\left(\frac{V t_2}{2}\right)^2-\left(\frac{V t_1}{2}\right)^2}=\frac{V}{2} \sqrt{t_2^2-t_1^2}\)

As h0 is the depth of the source from the water surface, so depth of sea at that place from the surface of sea = \(h_0+P Q=h_0+\frac{v}{2} \sqrt{t_2^2-t_1^2}\)

Question 23. How is the height of an airplane measured with the help of echo?

Answer:

Height Of An Airplane Measured With The Help Of Echo:-

The height of an airplane can be measured from the surface of the earth with the help of echo. Suppose, an airplane is flying parallel to the earth with a velocity u.

When it comes to position A, an intense sound of very short duration is made and when the airplane reaches the position C, then an echo is heard. If the sound is reflected at point 6, a perpendicular BD is drawn from B to AC.

If f is the difference of time between creating of sound and hearing of echo, then AC = ut.

∴ AD = \(\frac{AC}{2}{/latex] = [latex]\frac{ut}{2}{/latex]

WBBSE Solutions For Class 9 Physical Science Chapter 7 Some Properties Of Sound And Characteristics Of Sound Height Of An Airplane Measured With Help Of Echo

Again, time taken for sound to reach B from A = [latex]\frac{t}{2}{/latex]

∴ if V is the velocity of sound in air, AB = [latex]\frac{Vt}{2}{/latex]

height of the airplane from the surface of the earth,

h = BD = [latex]\sqrt{A B^2-A D^2}\)

= \(\sqrt{\left(\frac{V t}{2}\right)^2-\left(\frac{u t}{2}\right)^2}=\frac{t}{2} \sqrt{V^2-u^2}\)

Question 24. What is reverberation?

Answer:

Reverberation

The persistence of sound within a closed space after its source ceases to produce sound, caused by multiple reflections of the sound from walls, roof, etc. is known as reverberation.

Question 25. What measures are taken to prevent reverberation inside an auditorium?

Answer:

Measures To Take For The Prevention of Reverberation Inside An Auditorium:-

Reverberation takes place if a lot of reflectors are present within a small distance. For this reason, sound persists for a considerable amount of time inside an empty hall even after the original sound stops. This creates great difficulty for the audience as they cannot hear the original sound.

To remove this difficulty, walls of the auditorium are covered with soft pads or rubber-type materials. These work as sound absorbers. As a result, the problem of reverberation does not arise. Furniture and material of the seats are also made of materials which can absorb sound.

Question 26. What is SONAR?

Answer:

SONAR:-

SONAR is an acronym for Sound Navigation And Ranging. With the help of ultrasonic sound, SONAR is used for detection of ruins of a sunken ship or existence of a mountain deep inside an ocean.

Question 27. How can we measure the depth of a sea with the help of SONAR?

Answer:

We Can Measure The Depth Of A Sea With The Help Of SONAR As Follows:-

One sound emitting machine and a detector of SONAR are present in a ship. A very powerful ultrasonic sound wave is sent inside the sea with – the help of the sound emitter. This wave is reflected by the sea bed and comes back to the detector instrument.

Suppose, t is the time difference between the time of sending the wave and time of receiving the echo of the wave. If the velocity of the ultrasonic sound is V, then the depth of sea at that place is \(\frac{Vt}{2}\).

Question 28. What are the differences between a tone and a note?

Answer:

Differences Between A Tone And A Note:-

Tone is the sound of a single frequency, emitted from different musical instruments or tuning fork. In other words, it may be stated that the sound emitted from any musical instrument is a mixture of sounds of different frequencies. Each one of these frequencies is a tone. In the language of music, a summation of several tones is a note. Therefore, the note is a mixture of several frequencies.

Question 29. What do you mean by fundamental tone, overtone, and harmonic?

Answer:

Fundamental Tone, Overtone, And Harmonic:-

Fundamental tone: Fundamental tone is defined as the tone of minimum frequency present among the tones of a note.

Overtone: Overtone is defined as any tone present in a note other than the fundamental tone.

Harmonic: Harmonic is defined as any tone whose frequency is an integral multiple of the fundamental tone.

Question 30. The following tones with given frequencies are mixed in a sound emitted from a musical instrument: 256Hz, 512Hz, 1020Hz, and 1280Hz. Identify them as:

  1. Fundamental tone
  2. First harmonic
  3. First overtone
  4. Second overtone
  5. Second harmonic.

Answer:

  1. Fundamental tone: 256 Hz
  2. First harmonic: 256 Hz
  3. First overtone: 512 Hz
  4. Second overtone: 1020 Hz
  5. Second harmonic: 512 Hz

Question 31. What do you mean by musical sound and noise?

Answer:

Musical Sound And Noise:-

Musical sound: Musical sound is defined as that sound which is produced by regular and periodic vibrations of the source and appears euphonious to our ears. Sound of guitar, sitar and flute, tuning fork etc. are examples of musical sound.

Noise: Noise is defined as that sound which is produced by irregular and non-periodic vibration of the source and appears cacophonous to our ears. Sound of crackers, automobiles, uproar of crowd etc. are examples of noise.

Question 32. What are the characteristics of a musical sound? What is intensity of sound?

Answer:

Characteristics Of A Musical Sound:-

There are three characteristics of a musical sound. These are:

  1. Loudness,
  2. Pitch and
  3. Quality or timbre.

Question 33. Amount of sound energy passing through an unit area around a point in one second, perpendicular to the direction of sound is called intensity of the sound at that point. what is the unit of intensity in SI? Calculate the dimensional formula of intensity.

Answer:

Unity of intensity in SI is W • m-2.

Dimensional formula of intensity = \(\frac{\text { dimensional formula of energy }}{\text { dimensional formula of area}}\) x dimensional formula of time

= \(\frac{M L^2 T^{-2}}{L^2 \times T}=M T^{-3}\)

Question 34. What do you understand by loudness of sound?

Answer:

Loudness Of Sound:-

Among the sounds we hear everyday, some are high-pitched and some are low-pitched. Loudness of sound is such a characteristic which defines the strength with which it reaches the ears of the listener. Loudness is a feeling only and is not measurable. Loudness of sound is measured by its intensity.

Question 35. Intensity is the cause, loudness is its effect—explain.

Answer:

Intensity Is The Cause, Loudness Is Its Effect

When intensity of sound is more, more amount of sound energy reaches our ears, and hence, greater loudness is felt by us. Again, when intensity is less, less amount of sound energy enters our ears, so loudness of sound is less. This is why it may be concluded that if intensity is the cause, then loudness is its effect.

Question 36. Intensity of sound depends on which factors?

Answer:

Intensity of sound depends on several factors

Amplitude of vibration of the source of sound:

Intensity of sound is directly proportional to the square of the amplitude of vibration of the source of sound.

Size of the source:

Larger the size of the source, more is the intensity of the sound.

Distance of the listener from the source:

Intensity of sound is inversely proportional to the square of the distance of the listener from the source.

Density of the medium:

Intensity is more if the density of the medium is more.

Flow of air: Intensity of sound increases if flow of air is in the direction of motion of the sound. But if flow of air is in the opposite direction of motion of sound, intensity gets reduced.

Question 37. On a particular day, wind is blowing from east to west. A song is played through a microphone. A man is standing slightly away from the microphone in the east. Now, this man crosses the microphone and moves to the west side of the microphone at an equal distance. He hears more intense sound in which case? Explain with reasons.

Answer:

Sound spreads in every direction with equal loudness. But if wind blows along the direction of propagation of sound, loudness increases in that direction. According to this question, wind blows from east to west. As a result, the man hears sound with greater intensity while standing in the west.

Question 38. What do you mean by pitch of a sound?

Answer:

Pitch Of A Sound:-

If a male, a female and a child pronounce a sound with the same intensity, this same sound is known to create three different feelings in our ears. The voice of a female appears to be shriller than that of a male, again the voice of a child appears to be shriller than that of a female. This characteristic of sound is called pitch.

Pitch is defined as that characteristic of a musical sound due to which a listener can differentiate between a high-pitched and a low-pitched sound.

Question 39. Explain the statement — frequency is the cause of a sound and pitch is its effect.

Answer:

Frequency Is The Cause Of A Sound And Pitch Is Its Effect

Pitch depends mainly on frequency. If the frequency of a tone is high, its pitch is also high. On the other hand, if frequency is low, pitch is also low. So, it is said that frequency is the cause and pitch is the effect.

But frequency is a physical quantity and is measurable whereas pitch is a feeling which is not measurable.

Question 40. Loudness and pitch of sound change mainly due to which physical quantities?

Answer:

Due to change of intensity of a musical sound, loudness of sound perceived by a person changes. It has been found by experiment that this sensation increases as intensity of sound increases.

On the other hand, due to change in frequency of a musical sound, the pitch of a sound perceived by ear changes. Generally, as the frequency of the source of sound increases, the sound appears to be more high-pitched in the ears of an individual.

Question 41. What do you mean by quality or timbre of a sound?

Answer:

Timbre Of A Sound:-

The characteristic property by which sound of same intensity and pitch emitted by different sources can be distinguished is called the quality or timbre of a musical sound. Due to this characteristic of sound, tones of the same intensity and pitch emitted from harmonium, violin and table, etc. can be easily differentiated by us.

Question 42. The quality or timbre of a note depends on which factors?

Answer:

The quality of a note depends on three factors. These are:

  1. The number of overtones mixed with the fundamental tone;
  2. Frequencies of the overtones and
  3. Relative intensity of the overtones.

Chapter 7 Some Properties Of Sound And Characteristics Of Sound Very Short Answer Type Questions Choos The Correct Answer

Question 1. Due to which characteristic of sound, one can distinguish between the low-pitched and high-pitched sound of same intensity?

  1. Loudness
  2. Amplitude
  3. Frequency
  4. Pitch

Answer: 4. Pitch

Question 2. Which one is the first harmonic among the following frequencies: 500 Hz, 900 Hz, 550 Hz, 750 Hz, 300 Hz, 600 Hz?

  1. 300 Hz
  2. 500 Hz
  3. 600 Hz
  4. 900 Hz

Answer: 1. 300 Hz

Question 3. Out of 256 Hz, 512 Hz, 1020 Hz, and 1280 Hz, the frequency 1280 Hz is the

  1. Last tone
  2. Fourth overtone
  3. Third harmonic
  4. Second harmonic

Answer: 2. Fourth overtone

Question 4. Intensity of sound does not depend on

  1. Shape of the source of sound
  2. Amplitude of vibration of the source of sound
  3. Density of the medium
  4. Number of overtones in note

Answer: 4. Number of overtones in note

Question 5. The sound which contains more number of is more melodious.

  1. Harmonics
  2. Overtones
  3. Fundamental tones
  4. Beats

Answer: 1. Harmonics

Question 6. Characteristic of musical sound is

  1. Loudness
  2. Pitch
  3. Quality
  4. All of these

Answer: 4. All of these

Question 7. The quantity which does not produce any perception is

  1. Loudness
  2. Intensity
  3. Pitch
  4. Quality

Answer: 2. Intensity

Question 8. Among the following pairs, which one represents reciprocal quantities?

  1. Frequency and wavelength
  2. Frequency and time period
  3. Velocity of wave and wavelength
  4. Wavelength and loudness

Answer: 2. Frequency and time period

Question 9. Which of the following cannot reflect sound waves?

  1. Range of mountains
  2. High embankment of a river
  3. A small mirror
  4. Wall of an empty hall

Answer: 3. A small mirror

Question 10. Reverberation of sound may be heard

  1. In a stethoscope
  2. Inside an empty hall
  3. In a distant wall
  4. In vacuum

Answer: 2. Inside an empty hall

Question 11. Reverberation of sound is created

  1. In one reflection
  2. In two reflections
  3. In three reflections
  4. In multiple reflections

Answer: 4. In multiple reflections

Question 12. Rubber pads are fixed on the walls of a cinema hall in order to

  1. Increase reflection of sound
  2. Decrease reverberation of sound
  3. Increase velocity of sound
  4. Increase loudness of sound

Answer: 2. Decrease reverberation of sound

Question 13. If a tuning fork is struck, sound emitted from it has a frequency of 500 Hz. What is the frequency of the emitted sound when the tuning fork is struck with twice the previous force?

  1. 1000 Hz
  2. 500 Hz
  3. 250 Hz
  4. Insufficient data

Answer: 2. 500 Hz

Question 14. If the frequencies of fundamental tone, overtone, and harmonic are denoted by n1, n2, and n3 respectively, then

  1. n1> n2> n3
  2. n1 > n2 but n2 > n3
  3. n1< n2> n3
  4. n1< n2 and n1 ≤ n3

Answer: 4. n1< n2 and n1 ≤ n3

Question 15. Which of the following is essential for a melodious sound?

  1. The presence of more harmonics
  2. The presence of more overtones
  3. Presence of different frequencies
  4. Presence of different musical instruments

Answer: 1. Presence of more harmonics

Question 16. Which of the following statements is correct?

  1. Tone is a sound of single frequency
  2. Note is a sound of single frequency
  3. Tone is a sound composed of different frequencies
  4. Tone is composed of several notes

Answer: 1. Tone is a sound of single frequency

Question 17. If the frequency of the source of sound increases, the sound heard by a person appears to be more

  1. Loud
  2. Discordant
  3. Melodious
  4. Shrill

Answer: 4. Shrill

Question 18. If the velocity of sound in air at 27°C is 348 m/s, then what is the minimum distance between the source and the reflector of sound to hear an echo of transient sound?

  1. 17.4 m
  2. 34.8 m
  3. 69.6 m
  4. 104.4 m

Answer: 1. 17.4 m

Question 19. If the velocity of sound in air at 27°C is 348 m/s, what is the minimum distance between the source and the reflector of sound to hear an echo of a monosyllabic sound?

  1. 17.4 m
  2. 34.8 m
  3. 69.6 m
  4. 104.4 m

Answer: 2. 34.8 m

Question 20. Bats detect the obstacles in their path by receiving the reflected

  1. Electromagnetic waves
  2. Radio waves
  3. Infrasonic wave
  4. Ultrasonic waves

Answer: 4. Ultrasonic waves

Question 21. Infrasonic sound can be heard by

  1. Human beings
  2. Dolphins
  3. Rhinoceros
  4. Monkey

Answer: 3. Rhinoceros

Question 22. Which of the following is used in ultrasonography?

  1. Infrasonic sound
  2. Ultrasonic sound
  3. Audible sound
  4. Radio waves

Answer: 2. Ultrasonic sound

Question 23. Which of the following is used in SONAR?

  1. Infrasonic sound
  2. Micro waves
  3. Ultrasonic sound
  4. Audible sound

Answer: 3. Ultrasonic sound

Question 24. Repeated reflection of sound produces

  1. Reverberation
  2. Stationary wave
  3. Echo
  4. Refraction

Answer: 1. Reverberation

Question 25. A mirror is a good reflector of light wave but poor reflector of sound wave because

  1. Wavelength of sound is bigger than the mirror
  2. Wavelength of light is bigger than the mirror
  3. Velocity of sound is smaller than that of light
  4. Sound can not travel in vacuum

Answer: 1. Wavelength of sound is bigger than the mirror

Question 26. Sound wave transfers a physical quantity ‘X’ from one place to another, where ‘X’ is

  1. Mass
  2. Velocity
  3. Energy
  4. Density

Answer: 3. Energy

Question 27. A sound wave strikes a wall perpendicularly. What is the angle of reflection of the sound wave?

  1. 90°
  2. 45°
  3. 30°

Answer: 2. 0°

Question 28. A supersonic jet plane

  1. Can fly faster than sound
  2. Can fly at the highest velocity of sound
  3. Cannot fly with the velocity of sound
  4. Can fly with the velocity of light

Answer: 1. Can fly faster than sound

Question 29. A jet plane is flying with supersonic speed. Its Mach number is

  1. Equal to one
  2. More than one
  3. Less than one
  4. Equal to one hundred

Answer: 2. More than one

Chapter 7 Some Properties Of Sound And Characteristics Of Sound Answer In Brief

Question 1. What is the range of frequency of audible sound?

Answer: Range of frequency of audible sound is 20 Hz to 20000 Hz .

Question 2. What is the frequency of infrasonic sound?

Answer: Frequency of infrasonic sound is less than 20 Hz.

Question 3. What is the frequency of ultrasonic sound?

Answer: Frequency of ultrasonic sound is more than 20000 Hz.

Question 4. What is a tone?

Answer: Sound of a single frequency is known as a tone.

Question 5. What is a note?

Answer: Note is a sound containing more than one frequency.

Question 6. Write the SI unit of intensity of sound.

Answer: J • s-1 • m-2 or, W • m-2.

Question 7. Give one application where principle of echo is utilised.

Answer: Hearing aid.

Question 8. 500 Hz is written on the body of tuning fork. Is the sound produced from this fork a tone or a note?

Answer: If the tuning fork is vibrated, the sound produced from it is a tone since sound of only 500 Hz frequency is produced.

Question 9. Which property of sound is applied in a doctor’s stethoscope?

Answer: Property of reflection of sound is applied in a doctor’s stethoscope.

Question 10. What type of sound is emitted when a tuning fork is struck?

Answer: Audible sound is emitted when a tuning fork is struck.

Question 11. What type of sound is used in the in a radar?

Answer: Ultrasound is used in a radar.

Question 12. Mention the name of an animal which makes use of ultrasound.

Answer: A bat uses ultrasound while flying.

Question 13. Which characteristic of a musical sound depends on its frequency?

Answer: Pitch of a musical sound depends on its frequency.

Question 14. Which characteristic of a musical sound depends on its intensity?

Answer: Loudness of a musical sound depends on its intensity.

Question 15. Which characteristic of a musical sound changes when the distance between the source of sound and the listener changes?

Answer: In this case, loudness of the musical sound changes.

Question 16. Sound of the same frequency and with the same intensity are emitted from different string instruments. Which characteristic of sound helps us in distinguishing the sound of one musical instrument from the other?

Answer: Quality or timbre of the musical sound helps us in distinguishing the sound of different musical instruments.

Question 17. Which is the first harmonic in a sound consisting of more than one frequency?

Answer: The sound with the minimum frequency is the first harmonic.

Question 18. What do you mean by the intensity of sound?

Answer: Sound energy per unit time per unit area is known as the intensity of sound.

Question 19. What does SONAR stand for?

Answer: SONAR stands for SOund Navigation And Ranging.

Question 20. What is the basic principle with which SONAR works?

Answer: SONAR works on the principle of reflection of waves.

Question 21. State one important use of ultrasound in industries.

Answer: Ultrasounds are used to find faults and cracks in metal.

Question 22. State one important use of ultrasound for medical purpose.

Answer: Diagnosing the diseases in human body.

Question 23. What do you understand by the reverberation?

Answer: The phenomenon of prolongation of original sound due to the multiple reflection of sound wave even after the source stops producing sound is called reverberation.

Question 24. Why is the speed of sound grater in solid than in gas?

Answer: This is because particle of solids are closer than the particles of gases.

Question 25. Which wave property determines loudness?

Answer: Amplitude of the wave determines loudness.

Question 26. Does sound follow the same lows of reflection as light does?

Answer: Yes- Sound wave are reflected just like light wave

Question 27. Why is sound not heard when a bird flies?

Answer: When a bird flies, the frequency of its wings is less than 20 Hz which is in the infrasonic range and thus the sound cannot be heard.

Question 28. How does the velocity of sound in air change when temperature is increased?

Answer: Velocity of sound in air increases when temperature is increased.

Question 29. How does the velocity of sound in air change when humidity of air increases?

Answer: Velocity of sound in air increases when the humidity of air increases.

Question 30. When a pendulum oscillates, its frequency remains less than 20 Hz. What type of sound is produced in this case?

Answer: Infrasonic sound is produced in this case.

Question 31. Which type of sound waves are used to clean dirty clothes?

Answer: Ultrasonic sound waves are used to clean dirty clothes.

Question 32. Which type of sound waves are used to detect a shoal of fish deep inside sea water?

Answer: Ultrasonic sound waves are used to detect a shoal offish deep inside sea water.

Chapter 7 Some Properties Of Sound And Characteristics Of Sound Fill In The Blanks

Question 1. The perception of difference of pitch sound arises due to change of ______ of sound.

Answer: Frequency

Question 2. In the case of  ________ repeated reflection of sound occurs.

Answer: Reverberation

Question 3. Sound having frequency more than the frequency of the audible range is known as _______ sound.

Answer: Ultrasonic

Question 4. Sound having frequency less than the frequency of the audible range is known as ______ sound.

Answer: Infrasonic

Question 5. Stethoscope is the application of the property of ______ of sound.

Answer: Reflection

Question 6. Soft pads fixed to the walls of different auditoriums serve as __________

Answer: Sound absorbers

Question 7. SONAR uses ________ sound to determine the depth of an ocean.

Answer: Ultrasonic

Question 8. _______ of sound increases with the increase of size of the source of sound.

Answer: Intensity

Question 9. ________ of the sound increases with increase of frequency of the source of sound.

Answer: Pitch

Question 10. The intensity of sound __________ if air flows along the direction of motion of the sound.

Answer: Increases

Question 11. The difference in the voices of males and females is due to the ________ of sound.

Answer: Pitch

Question 12. Reflection of a wave is possible only when the size of the reflector is ________ compared to the wavelength of the wave.

Answer: Greater

Question 13. The trace of a sound remains in our brain for almost ________

Answer: 0.1 s

Question 14. The ratio of the velocities of a body and sound in a medium is called __________

Answer: Match number

Question 15. The velocity of a moving body is called ________ if the value of its Mach number is more than 1.

Answer: Super sonic

Chapter 7 Some Properties Of Sound And Characteristics Of Sound State Whether True Or False

Question 1. Echo and reverberation of sound refer to a single phenomenon.

Answer: False

Question 2. SONAR is an acronym for Sound Navigation and Ranging.

Answer: True

Question 3. The tone having the lowest frequency in a note is known as fundamental tone.

Answer: True

Question 4. Ultrasonic sound is used for the purpose of sterilizing of milk.

Answer: True

Question 5. Loudness of sound varies inversely to the square of the amplitude.

Answer: True

Question 6. Pitch of a note depends on its amplitude.

Answer: False

Question 7. A sound which is music to some one may be a noise to others.

Answer: False

Question 8. Sound of one frequency is tone.

Answer: True

Question 9. In a note the sound of least frequency is called the fundamental tone.

Answer: True

Question 10. A continuous noise of high level may cause migraine.

Answer: True

Chapter 7 Some Properties Of Sound And Characteristics Of Sound Numerical Examples

Useful Information

The impression of a sound persist in our brain for about \(\frac{1}{10}\) s.

If velocity of sound in air at a particular temperature = V then,

  1. The minimum distance between listener and the reflector for hearing distinctly the echo of a short sound = \(\frac{1}{2}\) x V x \(\frac{1}{5}\) = \(\frac{V}{20}\)
  2. Our ears can not recognise separately more than 5 syllables in one second, for monosyllabic sound the minimum distance between the listener and the reflector = \(\frac{1}{2}\) x V \(\frac{1}{5}\) = \(\frac{V}{10}\).

For disyllabic word the distance should be = \(\frac{1}{2}\) x V x \(\frac{2}{5}\) = 2 x \(\frac{V}{10}\) and so on.

Measurement of depth of a sea:

If, velocity of sound in water be V, the depth of the source and hydrophone from the surface of the sea be hQ, the depth of the sea with respect to the source be h, time taken by the sound to reach the hydrophone directly be t1, and the time taken by the echo be t2, then depth of the sea from its surface,

h = \(h_0+\frac{V}{2} \sqrt{t_2^2-t_1^2}\)

Measurement of the height of an airplane:

If, an airplane flying horizontally at a height h with a velocity u, produces a loud short sound and heard the echo after time t and velocity of sound in air be V then,

h = \(\frac{t}{2} \sqrt{V^2-u^2}\)

Question 1. A man fires a shot in front of a hill and hears the echo after 3 s. If the velocity of sound in air is 332 m/s, what is the distance between this man and the hill?

Answer:

Let distance of the man from the hill is L m .

2L m is traversed in 3 s.

So, \(\frac{2L}{3}\) = 332 m 3

or, L = \(\frac{332 \times 3}{2}\) = 498 m

Question 2. A pilot fires a shot when the airplane is flying parallel to the earth’s surface at a speed of 72 km/h . If the echo is heard after 4 $ and the velocity of sound in air is 332 m/s, calculate the height of the airplane from the earth’s surface.

Answer:

Velocity of the airplane, u = 72 km/h = 72 x \(\frac{5}{18}\) = 20 m/s

Velocity of sound, V = 332 m/s; time, t = 4 s

Suppose, height of the airplane from the surface of the earth =h

∴ h = \(\frac{t}{2} \sqrt{V^2-u^2}=\frac{4}{2} \sqrt{(332)^2-(20)^2}\) = 662.8

Question 3. A man standing between two parallel mountains hears the first echo after 2 s and the second echo after 5 s of firing a bullet from a gun. If the velocity of sound in air is 332 m/s, what is the distance between the two mountains?

Answer:

Suppose the distance of the man from the nearer mountain is a m and its distance from the distant one is b m.

The distance between the two mountains = (o + b) m .

According to the question, sound traverses a distance 2a in 2 seconds and a distance 2b in 5 seconds.

∴ 2a = 2 x 332 or, a = 332 and 2b = 5 x 332 or, b = 830

∴ distance between the two mountains = 830 + 332 = 1162 m .

Question 4. An airplane is flying at a height of 1200 m with a velocity Of 125 m/s. A high- pitched sound of very short duration is emitted from the plane. How much time does the echo of that sound take to reach the pilot after being reflected from the earth’s surface? Velocity of sound in air = 325 m/s.

Answer:

Suppose, the time gap between emission of sound and its audibility as echo = t.

According to the formula, h = \(=\frac{t}{2} \sqrt{v^2-u^2}\)

t = \(\frac{2 h}{\sqrt{V^2-u^2}}=\frac{2 \times 1200}{\sqrt{325^2-125^2}}\)

= \(\frac{2 \times 1200}{\sqrt{450 \times 200}}=\frac{2 \times 1200}{300}=8 \mathrm{~s}\)

So, pilot hears the echo after 8 s of its emission.

Question 5. An echo repeated four syllables. If the velocity of sound be 335 m/s , find the distance of the reflector.

Answer:

To hear echo for a word with four syllable, the minimum time required = 4 x 1/5 = 4/5 s

∴ The distance of the reflector = \(\frac{\text { distance travels }}{2}=\frac{4 \times 335}{5 \times 2}\) = 134 m

Question 6. Two children are at opposite ends of an aluminium rod. One strikes at one end of the rod with a hammer. Find the ratio of time taken by the sound waves in air and in the aluminium to reach the second child. (Speed of sound in aluminium is 6420 m/s and in air is 338 m/s.

Answer:

Let, l = length of the rod.

Time taken by sound to travel distance l in aluminium rod

⇒ \(t_1=\frac{\text { distance }}{\text { speed of sound in alumimium }}\)

= \(\frac{l}{64210} \mathrm{~s}\)

and time taken by sound to travel distance l in air

t2 = \(\frac{l}{338}\) s

∴ The ratio of time taken by the sound wave in air and in the aluminium

= \(t_2: t_1=\frac{1}{338}: \frac{1}{6420}=3210: 169\)

Question 7. A man standing 12 m away from a high wall makes a short sound. If velocity of the sound in air is 330 m/s. Check whether he hear the echo or not.

Answer:

The persistence of hearing = 0.1 s

∴ The minimum distance for hearing echo for short sound = \(\frac{0.1 \times 330}{2}\) = 16.5 m

According to the problem as the man standing at a distance 12 m from the wall, he can not hear the echo.

Question 8. Sound beyond the human audio range is produced at a place deep inside the sea. After 5 s, echo is detected by a SONAR at the same place. Calculate the depth of the sea from the place of production of sound. Velocity of sound in sea water is 1500 m/s.

Answer:

Suppose, depth of sea = h m

∴ sound travels 2h m in 5 seconds.

So, sound travels \(\frac{2h}{5}\) m in one second.

∴ \(\frac{2h}{5}\) = 1500 or, h = 1500 x \(\frac{5}{2}\) = 3750 m

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Concept Of Mole

Chapter 4 Matter Concept Of Mole Topic A Concept Of Mole And Importance Of Avogadro’s Number Synopsis

  1. The literal meaning of the word ‘mole’ is heap.
  2. One mole of a substance is the amount that contains 6.022 x 1023 constituent particles (atoms, molecules, or ions) of the substance.
  3. Chemical calculations can be done in a simpler and more convenient method by using the concept of mole.
  4. The number of molecules present in one gram-mole of a substance, which may be either an element or a compound (solid, liquid, or gas) is known as Avogadro’s number (NA).
  5. The value of Avogadro’s number (NA) is 6.22x 1023. It is independent of both temperature and pressure.
  6. It is very difficult to grasp the enormousness of Avogadro’s number. NA = 602213670000000000000000.
  7. The value of Avogadro’s constant is 6.022 x 1023 mol-23.
  8. Avogadro’s number correlates with the microscopic and macroscopic world.
  9. Avogadro’s number finds useful applications in calculations related to physics, chemistry, and biological science.

Read and Learn More WBBSE Solutions for Class 9 Physical Science and Environment

Chapter 4 Matter Concept Of Mole Topic A Concept Of Mole And Importance Of Avogadro’s Number Short And Long Answer Type Questions

Question 1. Define mole.

Answer:

Mole:

One mole of a substance is the amount of the substance (element or compound) which contains Avogadro’s number (6.022 x 1023) of fundamental particles (atoms, molecules, or ions).

Question 2. What is Avogadro’s number?

Answer:

Avogadro’s Number:

The number of molecules present in one gram mole of a substance which may be either an element or a compound (solid, liquid, or gas) is known as Avogadro’s number.

Question 3. What is Avogadro’s constant? How is it different from Avogadro’s number?

Answer:

Avogadro’s Constant:

  1. The number of particles present per mole of a substance is called Avogadro’s constant. Thus, Avogadro’s constant is Avogadro’s number/mole. Its value is 6.022 x 1023 mol-1. It is a universal constant.
  2. Avogadro’s number is a pure number. It has no unit but the unit of Avogadro’s constant is mol-1.

Question 4. Why is it necessary to mention the corresponding fundamental particle while using the term ‘mole’?

Answer:

It/is necessary to mention the corresponding fundamental particle while using the term ‘mole’. This is because the amount of a substance entirely depends on the nature of particles present in it. For example, the fundamental particle of oxygen can be atom as well as molecule.

Thus, the term 1 mol oxygen1 does not clearly indicate the amount of oxygen present in that quantity, because 1 mol oxygen represents both 1 mol oxygen molecule and 1 mol oxygen atom.

Now, a 1 mol oxygen molecule contains twice the number of oxygen atoms present in a 1 mol oxygen atom, although the number of particles in both quantities are same.

Question 5. Avogadro’s number creates a correlation between the macroscopic and microscopic world— explain.

Answer:

Molecules or atoms are too small to be seen. So, it is extremely difficult to count the number of atoms or molecules in a given mass of substance. However, by using the mole concept, scientists have been able to successfully calculate the number of atoms, molecules, or ions in a given mass of substance.

According to the definition of mole, 1 mol of a substance contains 6.022 x 1023 number of fundamental particles. This number (6.022 x 1023) is known as Avogadro’s number.

We cannot see 1 molecule of water but, 1 mol of water (which is equal to 18 g water) is visible to us. Now, 1 mol of water (i.e., 18 g water) contains 6.22x 1023 molecules of water.

Thus it can be concluded that, Avogadro’s number creates a correlation between the macroscopic and microscopic world.

Question 6. Discuss the significance of mole concept or Avogadro’s number.

Answer:

Significance Of Mole Concept:

The concept of mole is used in different fields of science. 1 mol of a substance contains Avogadro’s number (6.022 x 1023) of fundamental particles (atoms, molecules or ions) this concept has simplified the chemical calculations to a large extent.

It has also become easier to correlate the different physical quantities with the help of Avogadro’s number and mole concept. Some important applications of Avogadro’s number are illustrated below

1. The number of atoms, molecules or ions present in a given mass of a substance is calculated by using mole concept. For example, 88g CO2 = 88g/44g = 2 mol (as a gram-molecular mass of CO2 is 44 g).

Hence, number of molecules in 88 g CO2 = 2 x 6.022 x 1023 = 12.044 x 1023

2. The actual mass of an atom or a molecule is expressed in atomic mass unit (u).

Now, u = \(\frac{1}{\text { Avogradro’s number }}\) g

Thus, mass of one N2 molecule = 28 u

= 28 x \(\frac{1}{6.022 \times 10^{23}}\) = 28 x 1.6605 x 10-24

3. Using Avogadro’s number, it is possible to calculate the number of molecules present in a definite volume of any gas at STP.

4. The atomic radii of solid metals can be calculated using Avogadro’s number.

5. The value of Boltzmann constant can be calculated using Avogadro’s number.

Question 7. Do 1 mol O2 and 1 mol O represent the same quantity of oxygen?

Answer:

The molecular formula of oxygen is O2 while O represents an atom of oxygen. Hence, 1 mol O2 represents Avogadro’s number of oxygen molecules. On the other hand, 1 mol O2 indicates Avogadro’s number of O-atoms.

The mass of Avogadro’s number of O2 molecules is 32 g and that of Avogadro’s number of O-atoms is 16 g. Thus, 1 mol O2 and 1 mol O do not represent the same quantity of oxygen.

Question 8. Can Avogadro’s constant be considered as universal constant?

Answer:

Yes, Avogadro’s constant may be considered as universal constant.

Reasons:

  1. 22.4 L of any gas at STP contains Avogadro’s number of molecules.
  2. Number of molecules in 1 g-mole or number of atoms in 1 g-atom of any element or compound are equal to Avogadro’s number.
  3. Number of ions present in 1 g-ion of any sample is equal to Avogadro’s number.

Question 9. Write down the importance of Avogadro’s number in biology. Or, Mention an use of Avogadro’s number in biology.

Answer:

In biological science, Avogadro’s number can be used for quantitative calculations of solids, liquids, and gases.

Example:

Chlorophyll contains 2.68% Mg. Using the concept of Avogadro’s number one can find the number of Mg atoms present in 1 g chlorophyll as 6.69 x 1020.

Chapter 4 Matter Concept Of Mole Topic A Concept Of Mole And Importance Of Avogadro’s Number Very Short Answer Type Questions Choose The Correct Answer

Question 1. Which of the following correlates between the microscopic and macroscopic world?

  1. Dalton’s number
  2. Berzelius’ number
  3. Avogadro’s number
  4. Faraday’s number

Answer: 3. Avogadro’s number

Question 2. The meaning of the Latin word ‘moles’ is

  1. Many
  2. Having large volume
  3. Having large mass
  4. Heap

Answer: 4. Heap

Question 3. The word ‘mole’ was first used by

  1. Dalton
  2. Ostwald
  3. Avogadro
  4. Millikan

Answer: 2. Ostwald

Question 4. The value of Avogadro’s number is

  1. 6.024 X 1020
  2. 0.6023 X 1022
  3. 0.6022 X 1024
  4. 0.623 x 1023

Answer: 3. 0.6022 X 1024

Question 5. The amount of Avogadro’s number of fundamental particles (example: electron, proton, atom, molecule, ion) is known as

  1. Mole
  2. Gram-mole
  3. Gram-atom
  4. Gram-ion

Answer: 1. Mole

Question 6. The element for which the number of gram-moles and number of gram-atoms will be equal for any quantity is

  1. Oxygen
  2. Helium
  3. Hydrogen
  4. Chlorine

Answer: 2. Helium

Question 7. Number of atoms in 0.1 mol of a triatomic gas is

  1. 6.022 x1022
  2. 1.806 X 1023
  3. 3.600 x1023
  4. 1.800 x 1022

Answer: 2. 1.806 X 10223

Question 8. If 10-5 mol electron flows per second through a wire, the number of electron flows per second through it is

  1. 6.022 X1028
  2. 6.022 x 1023
  3. 6.022 X10-23
  4. 6.022 x 1018

Answer: 4. 6.022 x 1018

Question 9. 1 mol of oxygen atom means

  1. 6.022 x 1023 number of oxygen atoms
  2. 2 x 6.022 x 1023 number of oxygen atoms
  3. 1/2 x 6.022 x 1023 number of oxygen atoms
  4. 6.022 x 1023 number of oxygen molecules

Answer: 1. 6.022 x 1023 number of oxygen atoms

Question 10. Number of Cl ions obtained from 1 mol of CaCI2 is

  1. 2
  2. 6.022 X 1023
  3. 12.046 x 1023
  4. 3 x 6.022 x 1023

Answer: 3. 12.046 x 1023

Question 11. Unit of Avogadro’s constant is

  1. Mole
  2. Per mole
  3. (mole)2
  4. Dobson

Answer: 2. Per mole

Question 12. Number of molecules in 1 millimol ammonia is

  1. 6.022 x 1026
  2. 6.022 x 1023
  3. 6.022 X 1020
  4. 6.0222 x 10-3

Answer: 3. 6.022 X 1020

Question 13. Which of the following cannot be calculated using Avogadro’s number?

  1. Boltzman constant
  2. Atomic radius of solid metals
  3. Velocity of light in vacuum
  4. Number of molecules in a definite volume

Answer: 3. Velocity of light in vacuum

Chapter 4 Matter Concept Of Mole Topic A Concept Of Mole And Importance Of Avogadro’s Number Answer In Brief

Question 1. Who determined the value of Avogadro’s number?

Answer: Robert Millikan .

Question 2. What is the unit of molar mass?

Answer: g • mol-1 or kg • mol-1.

Question 3. What is the effect of temperature and pressure on Avogadro’s number?

Answer: Mass and number of molecules do not depend on either temperature or pressure. Hence, temperature or pressure has no effect on Avogadro’s number.

Question 4. What is the SI unit of quantity of matter?

Answer: The SI unit of quantity of matter is mole.

Question 5. How many H-atoms are present in 1 mol hydrogen gas?

Answer: 1 mol hydrogen gas contains 6.022 x 1023 molecules.

Hence, number of H-atoms in 1 mol of hydrogen gas = 2 x 6.022 x 1023 = 12.044 x 1023.

Question 6. How many Fe-atoms does 1 gram-atom of iron indicate?

Answer: 6.022 x 1023 atoms of iron.

Question 7. How many moles of oxygen atoms do 6.22x 1024 number of oxygen atoms indicate?

Answer: 6.02 2 x 1024 number of oxygen atoms = 10 x 6.022 x 1024 oxygen atoms = 10 mol of oxygen atoms.

Question 8. How many grams of CO2 does 1 gram-mole of CO2 indicate?

Answer: 1 gram-mole of CO2 indicates 44 g of CO2

Question 9. How many molecules are present in 1 millimole CO2?

Answer: 6.022 x 1020 number of molecules are present in 1 millimol CO2.

Question 10. 2 balloons contain 1.8066 x 1023 number of hydrogen molecules and 2 mol hydrogen gas respectively. Which balloon contains higher number of molecules?

Answer: The balloon having 2 mol of hydrogen gas.

Question 11. What is the relation between the number of constituent particles of a substance, Avogadro’s number, and number of moles?

Answer: Number of moles

= \(\frac{\text { Number of constituent particles of a substances }}{\text { Avogadro’s number }}\)

Question 12. What is the total charge of 1 mol electron?

Answer: The charge of 1 mol electron is 1 Faraday.

Chapter 4 Matter Concept Of Mole Topic A Concept Of Mole And Importance Of Avogadro’s Number Fill In The Blanks

Question 1. In physical science, the 6.022 x 1023 is known as _______ number.

Answer: Avogradro’s

Question 2. In l mol oxygen gas, the number of oxygen atoms is _______

Answer: Fundamental

Question 3. 1 mol represents Avogadro’s number of _______ particles.

Answer: 2 x 6.022 x 1023

Question 4. The unit of Avogadro’s constant is _______

Answer: mol-1

Question 5. 6.22x 1023 number of H+ ions will be produced from __________ gram-mol of H2SO4.

Answer: 0.5

Question 6. When the term ‘mole’ is used, it is necessary to mention the corresponding _______ particle.

Answer: Fundamental

Question 7. 1 millimol = _______ mol.

Answer: 10

Question 8. 12.044 x 1023 number of protons = ________ mol protons.

Answer: 2

Question 9. 1 mol sugar = ________ g sugar.

Answer: 342

Question 10. 1 mol nitrogen molecule = ________ g nitrogen.

Answer: 28

Question 11. 1 millimol oxygen molecule indicates __________ number of oxygen molecules.

Answer: 6.022 x 1020

Question 12. Number of moles = mass of the substance ÷ mass of _________ of the substance.

Answer: 1 mol

Chapter 4 Matter Concept Of Mole Topic A Concept Of Mole And Importance Of Avogadro’s Number State Whether True Or False

Question 1. The value of Avogadro’s constant changes with the change in both temperature and pressure.

Answer: False

Question 2. The given mass of any substance when divided by the number of moles gives the corresponding value of molar mass.

Answer: True

Question 3. The value of Avogadro’s number was determined by Millikan.

Answer: True

Question 4. The unit of Avogadro’s constant is mol-1.

Answer: True

Question 5. The number of atoms in 1 gram-atom of oxygen is 6.022 x 1023

Answer: True

Question 6. 1 mol N2 and 1 mol N signify the same amount.

Answer: False

Question 7. 1 mol of CO2 contains 1 mol oxygen atom.

Answer: False

Question 8. Avogadro’s number is not applicable for microscopic substances.

Answer: False

Question 9. Avogadro’s number is independent of the properties of matter, it only depends upon the volume, pressure, and temperature.

Answer: False

Chapter 4 Matter Concept Of Mole Topic A Concept Of Mole And Importance Of Avogadro’s Number Numerical Examples

Question 1. P and Q flasks contain 0.5 mol oxygen and 0.4 mol ozone gas respectively. In which flask number of oxygen atoms is higher?

Answer:

Number of oxygen atoms in P flask = 2 x 0.5 x 6.022 x 1023 = 6.022 x 1023

Number of oxygen atoms in Q flask = 3 X 0.4 X 6.022 X 1023 = 1.2 x 6.022 x 1023

∴ Q flask contains higher number of oxygen atoms than that of P flask.

Question 2. How many moles of C-atom and how many moles of O-atom are present in 44 gCO2?

Answer:

Molecular mass of carbon dioxide = 12 + 2 x 16 =44

∴ 44 g CO2 = 1 gram-mole of CO2

Now, 1 molecule of CO2 contains 1 C-atom and 2 O-atoms.

∴ 1 gram-mole or 44 g of C02 contains 1 mol of C-atom and 2 mol of O atoms.

Chapter 4 Matter Concept Of Mole Topic B Gram Atom Atomic Mass Unit Gram Mole Molar Volume Of Gases And Formula Mass Synopsis

In hydrogen (1H) scale, the relative atomic mass of an element

= \(\frac{\text { mass of one atom of the element }}{\text { mass of one hydrogen atom }}\)

In 12C scale, the relative atomic mass of an element

= \(\frac{\text { mass of one atom of the element }}{\text { mass of one }{ }^{12} \mathrm{C} \text {-atom } \times \frac{1}{12}}\)

  1. The relative atomic mass of an element is expressed as a ratio of the masses of two atoms. Hence, it has no unit.
  2. The atomic mass of an element expressed in gram is known as the gram-atomic mass of the element.
  3. The actual mass of an atom is expressed in atomic mass unit.
  4. Atomic mass unit (1u) = 1/12 x actual mass of one C atom = 1.6605 x 10-24 g
  5. One gram atom of an element contains Avogadro’s number of atom.
  6. Atomic mass is expressed as a ratio of two masses. So it does not have any unit.
  7. The actual mass of an atom is very small. So, instead of actual mass, relative atomic mass of the element is used during chemical calculations.
  8. The atomic mass of an element is the average of atomic masses of all the naturally occurring isotopes of that element. Consequently, the atomic mass of most of the elements is fractional.
  9. The molecular mass of a substance expressed in gram is known as its gram-molecular mass.
  10. One gram-mole of a substance (element or compound) contains Avogadro’s number of molecules.
  11. Number of moles of a matter = \(\frac{\text { mass of the matter }}{\text { its molar mass }}=\frac{W}{M}\)
  12. At a given temperature and pressure, the volume occupied by 1 mol of a substance (element or compound) is known as its molar volume.
  13. At standard temperature and pressure, the molar volume of all the gases is 22.4 L.
  14. Formula mass of a compound is the sum of the atomic masses of all the atoms present in a formula unit of the compound. In case of ionic compounds, the formula of the compound does not represent the molecule of the compound, so the use of formula mass is more appropriate.

Chapter 4 Matter Concept Of Mole Topic B Gram Atom Atomic Mass Unit Gram Mole Molar Volume Of Gases And Formula Mass Short Ans Long Answer Type Questions

Question 1. Define relative atomic mass with respect to the hydrogen scale.

Answer:

Considering the mass of one hydrogen atom as unity (1), the number of times an atom of an element is heavier than a hydrogen atom indicates the relative atomic mass of the given element.

Thus, the relative atomic mass of an element is defined as the ratio of the mass of one atom of the element to the mass of one atom of hydrogen.

Relative atomic mass of an element = \(\frac{\text { mass of one atom of the element }}{\text { mass of one hydrogen atom }}\)

Question 2. Define relative atomic mass with respect to the carbon (12C) scale.

Answer:

The number of times an atom of an element is heavier than 1/12 th part of the actual mass of an atom of 12C isotope indicates the relative atomic mass of the given element.

Thus, the relative atomic mass of an element is defined as the ratio of mass of one atom of the element to 1/12th part of the mass of an atom of 12C isotope.

Relative atomic mass of an element = \(\frac{\text { mass of one atom of the element }}{\frac{1}{12} \times \text { mass of one }{ }^{12} \mathrm{C} \text {-atom }}\)

= \(\frac{\text { mass of one atom of the element }}{\text { mass of one }{ }^{12} \mathrm{C} \text {-atom }} \times 12\)

Question 3. Why is 12C and not hydrogen considered As the standard element to determine as the standard element to determine the relative atomic mass?

Answer:

  1. Hydrogen is the lightest element. So, while calculating atomic mass with respect to hydrogen, a very small error in measurement causes a large deviation in the actual result. This does not happen when 12C is taken as the standard element.
  2. The atomic masses calculated with respect to hydrogen scale are found to be fractional for most of the elements. However, atomic masses of most of the elements are integers with respect to the 12C scale.

Due to these advantages, nowadays 12C is considered as the standard element in the determination of relative atomic mass.

Question 4. What is gram-atomic mass?

Answer:

Gram-Atomic Mass :-

The gram-atomic mass of an element is defined as the atomic mass of the element expressed in gram. For example, atomic mass of oxygen is 16. Therefore, the gram-atomic mass of oxygen is 16 g.

Question 5. What is gram-atom?

Answer:

Gram-Atom:-

1 gram-atom of an element is defined as the amount of the element expressed in gram, which contains 6.022 x 1023 atoms of the element. For example, 1 gram-atom of nitrogen =14 g nitrogen, because 14g nitrogen contains 6.22x 1023 number of atoms.

Question 6. Relative atomic mass of an element has no unit. Explain

Answer:

The Relative Atomic Mass Of An Element:-

= \(\frac{\text { mass of one atom of the element }}{\frac{1}{12} \times \text { mass of one }{ }^{12} \mathrm{C} \text {-atom }}\)

As relative atomic mass of an element is a ratio of the masses of two atoms, it has no unit.

Question 7. What is atomic mass unit?

Answer:

Atomic Mass Unit:-

Atomic mass unit may be defined as the unit in which the actual mass of an atom is expressed and which is equal to 1/12 th of the actual mass of an atom of 12C isotope.

1 atomic mass unit or lu = 1.6605 x 10-24 g

Question 8. Why is the atomic mass of most elements fractional?

Answer:

The Atomic Mass Of Most Elements Fractional:-

Almost all the naturally occurring elements exist as a mixture of two or more isotopes. The isotopes have the same atomic number but different mass numbers. The relative atomic mass is calculated by taking the average of the mass numbers of different isotopes of that element.

Though the mass number is a whole number, yet the average is taken on the basis of percentage abundance of the isotopes which more or less have a fixed proportion in nature. This is why the atomic masses of most of the elements are found to be fractional.

For example, chlorine has two naturally occurring isotopes, \({ }_{17}^{35} \mathrm{Cl} \text { and }{ }_{17}^{37} \mathrm{Cl}\). The percentage abundance of these two isotopes are 75% and 25% respectively.

Hence, the average atomic mass of chlorine = \(\frac{35 \times 75+37 \times 25}{100}\) = 35.5 (fractional).

Question 9. Write the differences between atomic mass and mass of an atom of an element.

Answer:

Differences Between Atomic Mass And Mass Of An Atom Of An Element:-

  1. The atomic mass of an element is the ratio of mass of one atom of the element to 1/12th part of the mass of an atom of 12C isotope. On the other hand, the mass of an atom of an element means the actual mass of the atom.
  2. Atomic mass is a ratio. So, it has no unit. However, mass of an atom represents a definite mass. Hence, it has a unit of mass.
  3. Mass of an atom = atomic mass of the element x atomic mass unit (u). For example, atomic mass of oxygen is 16 but, mass of 1 oxygen atom = 16 x 1.6605 x 10-24 g = 26.656 x 10-24 g

Question 10. What is gram-molecule or gram-mole?

Answer:

Gram-Molecule:-

1 gram-molecule or gram-mole of a substance (element or compound) is defined as the amount of the substance expressed in gram, which contains 6.022 x 1023 molecules of the substance. For example, 1 gram-mole of oxygen =32 g oxygen, because, 32 g oxygen contains 6.22x 1023 number of molecules.

Question 11. What is meant by molar mass?

Answer:

Molar Mass:-

Molar mass is the mass of 1 mol of a substance. Alternatively, it is defined as the mass of Avogadro’s number of constituent particles (atoms, molecules, or ions) of the substance. For example, the gram-atomic mass of helium is 4g.

Hence, molar mass of helium = 4g • mol-1. Again, the gram-molecular mass of water is 18 g. Hence, molar mass of H2O = 18 g • mol-1.

Question 12. What is gram-molar mass?

Answer:

Gram-Molar Mass:-

The gram-molar mass of a substance (element or compound) is defined as the molar mass of that substance expressed in gram. For example, molar mass of carbon dioxide is 44 g • mol-1. Therefore, gram-molar mass of CO2 is 44 g.

Question 13. What is molar volume?

Answer:

Molar Volume:-

At a given temperature and pressure, the volume occupied by 1 mol of any substance (element or compound) is known as its molar volume at that temperature and pressure. The molar volume of any gas at standard temperature and pressure (STP) is 22.4 L.

Question 4. What can be known from ‘1 mol CO2‘?

Answer:

From ‘1 mol CO2‘ it can be known that

  1. Mass of the CO2 sample is 44 g.
  2. Number of CO2 molecules in the sample is 6.22 x 1023.
  3. Volume of the CO2 gas at STP is 22.4 L.

Question 15. What is formula mass?

Answer:

Formula Mass:-

Formula mass of an ionic compound is the sum of the atomic masses of all the atoms present in a formula unit of the compound. For example, formula mass of NaCI = 23+ 35.5 = 58.5.

Question 16. What is gram-formula mass?

Answer:

Gram-Formula Mass:-

The gram-formula mass of an ionic compound is defined as that amount of the compound expressed in gram numerically equal to its formula mass. For example, formula mass of NaCI is 58.5. Therefore, the gram-formula mass of NaCI is 58.5g.

Question 17. What is meant by gram-formula unit?

Answer:

Gram-Formula Unit:-

The amount of a substance in gram calculated from its formula unit is called 1 gram-formula unit of the substance.

Thus, 1 gram-formula unit = \(\frac{\text { given mass of the substance in gram }}{\text { gram-formula mass of the substance }}\)

For example, 58.5 g of NaCI = \(\frac{58.5g}{58.5g}\)= 1 gram-formula unit of sodium chloride.

Question 18. Find the formula mass of CaCI2.

Answer:

Formula Mass Of CaCI

Atomic mass of calcium = 40; atomic mass of chlorine = 35.5.

Hence, formula mass of CaCI2 = 40 + 35.5 x 2 = 111

Question 19. Use of formula mass instead of molecular mass is more appropriate in case of ionic compounds. Explain.

Answer:

Use Of Formula Mass Instead Of Molecular Mass Is More Appropriate In Case Of Ionic Compounds

For an ionic compound, there is no existence of any discrete molecule. The formula of the compound represents the ratio of different ions in the compound.

For example, although sodium chloride is expressed by the formula NaCI, yet there is no existence of individual sodium chloride molecule. Actually, the compound contains Na+ ions and Cl ions in the ratio of 1:1 which is represented by the formula.

In the crystalline form of sodium chloride, each Na+ ion is surrounded by 6 Cl ions and each Cl ion is further surrounded by 6 Na+ ions forming a network structure called crystal lattice.

Thus, in the crystal lattice of sodium chloride, there is no presence of molecules. Hence, it is not appropriate to use the term molecular mass in case of ionic compounds rather, the term formula mass should be used.

Question 20. Molecular mass and formula mass are not always the same. Explain with examples.

Answer:

Molecular Mass And Formula Mass Are Not Always The Same

Molecular mass of an element or a compound is calculated from the formula of the substance. So, molecular mass is numerically equal to the formula mass. However, the two terms are not always the same.

The term molecular mass represents the relative mass of a molecule of a substance while formula mass represents the relative mass of 1 formula unit of the substance.

For substances which exist as molecules, molecular mass and formula mass are the same. However, there are compounds which do not exist as molecules. For such compounds, the term molecular mass is not applicable.

For example, CO2 molecule actually exists. Hence, the term molecular mass is applicable for CO2 and both molecular mass and formula mass of CO2 is 44. However, NaCI molecule has no separate existence.

Its formula mass is 58.5 but the term molecular mass is not applicable for NaCI because it constitutes of Na+ and Cl ions.

Question 21. What is the relation between gram- molecular mass or gram-molecule and molar volume of a gas at STP?

Answer:

Relation Between Gram- Molecular Mass Or Gram-Molecule And Molar Volume Of A Gas At STP

From Avogadro’s law it can be proved that 1 mole of any gas at STP occupies 22.4 L. On the other hand, mass of 1 mol of molecules of any substance expressed in gram is called the gram- molecular mass of that substance.

Again, 1 gram- molecule of a substance is the amount of the substance expressed in gram, which is numerically equal to its molecular mass.

Thus, it can be concluded from these three relations that, at STP, volume of 1 gram-molecule of any gaseous substance is 22.4 L.

For example, 1 gram-molecule of oxygen = 32 g.

Hence, it can be said that 32 g of oxygen at STP occupies 22.4 L.

Similarly, the gram-molecular mass of carbon dioxide is 44 g. Hence, the volume of 44 g of CO2 at STP is 22.4 L.

Chapter 4 Matter Concept Of Mole Topic B Gram Atom Atomic Mass Unit Gram Mole Molar Volume Of Gases And Formula Mass Very Short Answer Type Questions Choose The Correct Answer

Question 1. 1 amu is equal to

  1. 1.66 X 10-24 kg
  2. 1.66 x 10-25 g
  3. 0.166 x 10-23 g
  4. 1.66 x 10-20 g

Answer: 3. 0.166 x 10-23 g

Question 2. The mass of 1 atom of 17CI35 is

  1. 17 u
  2. 34 u
  3. 35 u
  4. 70 u

Answer: 3. 35 u

Question 3. 1 u is equal to

Mass of one H-atom

Mass of one 12C-atom

Mass of one 16O-atom

1/12 th mass of one 12C-atom

Answer: 4. 1/12 th mass of one 12C-atom

Question 4. Molar volume of a gas at STP is

  1. 1 L
  2. 11.1 L
  3. 22.4 L
  4. 5.6 L

Answer: 3. 22.4 L

Question 5. Molar mass of sulphuric acid is

  1. 98 kg •mol-1
  2. 98 g •mol-1
  3. 49 u
  4. 49 mg•mol-1

Answer: 2. 98 g-mol-1

Question 6. For which of the following gases, the mass of 22.4 L of the gas at STP is 44 g?

  1. NH3
  2. H2S
  3. CO2
  4. CH4

Answer: 3. CO2

Question 7. The formula mass of MgCI2 (Mg = 24) is

  1. 95 u
  2. 65 u
  3. 85 u
  4. 75 u

Answer: 1. 95 u

Question 8. 117g NaCI is equal to how many gram- formula mass of NaCI?

  1. 1
  2. 2
  3. 3
  4. 4

Answer: 2. 2

Question 9. The compound whose formula mass is equal to 106 is

  1. CaCO3
  2. CaCI2
  3. NaCI
  4. Na2CO3

Answer: 4. Na2CO3

Question 10. The statement ‘molecular mass of sodium chloride is 58.5’ is wrong because

  1. It is a radioactive substance
  2. It is a solid
  3. NaCI molecule has no separate existence
  4. None of these

Answer: 3. NaCI molecule has no separate existence

Question 11. If atomic mass of an element is ‘x’ in hydrogen scale, its atomic mass in carbon scale will be

  1. x
  2. 1/12 x x
  3. 12x
  4. 12/x

Answer: 1. x

Question 12. Atomic mass of nitrogen in carbon scale is 14, its atomic mass in hydrogen scale will be

  1. 14 x 1/3
  2. 14 x 1/2
  3. 14
  4. 14 x 2

Answer: 3. 14

Question 13. If mass and gram-molecular mass of a compound are ‘m’ g and ‘M’ g respectively, number of gram-mole of the compound is

  1. m x M
  2. m/M
  3. M/m
  4. \(\frac{(m)^2}{M}\)

Answer: 2. m/M

Chapter 4 Matter Concept Of Mole Topic B Gram Atom Atomic Mass Unit Gram Mole Molar Volume Of Gases And Formula Mass Answer In Brief

Question 1. Which element is considered as the standard for determining the relative atomic mass of an element?

Answer: To determine the relative atomic mass of a 12C element, the C isotope of carbon (\({ }_6^{12} C\)) is considered as the standard.

Question 2. Find the value of 1/12 th part of the mass of  one \({ }_6^{12} C\)-atom.

Answer: The value of 1/12th part of mass of one \({ }_6^{12} C\) atom = 1.6605 x 10-24 g.

Question 3. How many grams of H2O does 1 gram-mole of H2O indicate?

Answer: 1 gram-mole of H2O indicates 18 g of water.

Question 4. 1u = how many grams?

Answer: 1u = 1.6605 x 10-24 g.

Question 5. How are mass, molar mass and number of moles related?

Answer: Given mass (m) = number of moles (n) x molar mass (M)

Question 6. What is the mass of one molecule of water (H2O) in amu?

Answer: 18 u.

Question 7. Find the actual mass of one atom of \({ }_8^{16} O\).

Answer: The actual mass of one atom of \({ }_8^{16} O\) = 16u

= 16 x 1.6605 x 10-24 g = 26.568 X 10-24 g

Question 8. What is meant by the statement ‘molecular mass of ammonia is 17 u’?

Answer: Molecular mass of ammonia is 17u means that the actual mass of one molecule of ammonia = 17 x 1.6605 x 10-24 g = 28.2285 x 10-24 g

Question 9. For which compound between CaCI2 and H2O, the use of formula mass is appropriate?

Answer: The use of formula mass is appropriate for CaCI2 as it is an ionic compound.

Question 10. Find the percentage of carbon (by mass) in CO2.

Answer: Molar mass of CO2 = 12 + 16 x 2 = 44 So, percentage of carbon (by mass) in CO2 = 12/44 x 100 =27.27%

Question 11. Find the actual mass of a hydrogen atom. (1u =1.6605 x 10-24 g)

Answer: Actual mass of a hydrogen atom = 1.008u = 1.008 x 1.6605 x 10-24 g

Question 12. Atomic mass of sodium is 23—explain.

Answer: Atomic mass of sodium is 23 means that the mass of a sodium atom is 23 times to that of 1/12 th part of the mass of a C12 isotope.

Question 13. What is the mass of a carbon atom unified atomic mass unit?

Answer: 12 u.

Question 14. Express the value of 1 amu in kg.

Answer: Value of 1 amu in kg is 1.6605 x 10-27 kg

Question 15. What is Avogram?

Answer: 1/12th part of the mass of a C12 -atom i.e.; 12 1.66 x 10-24  g is termed as 1 Avogram. 1 amu = 1 Avogram.

Question 16. Name a gaseous substance whose atomic and molecular masses are equal to each other.

Answer: Helium (He).

Question 17. In which case gram-atomic mass and gram-molecular mass are of same meaning?

Answer: For monoatomic elements;

Example: noble gases.

Question 18. If atomic mass of sodium is 23, what will be the mass of a sodium atom in ‘amu’?

Answer: 23 u or 23 amu.

Question 19. Name two substances for which both molar mass and atomic mass are same.

Answer: Neon and potassium.

Chapter 4 Matter Concept Of Mole Topic B Gram Atom Atomic Mass Unit Gram Mole Molar Volume Of Gases And Formula Mass Fill In The Blanks

Question 1. The actual mass of one ammonia molecule is _______ u.

Answer: 17

Question 2. The actual mass of an atom of an element = atomic mass of the element x ________ g.

Answer: 1.6605 x 10-24

Question 3. Actual mass of a \({ }_7^{14} \mathrm{~N}\)-atom is ______ u.

Answer: 14

Question 4. Actual mass of an \({ }_8^{16} \mathrm{~O}\)-atom is _____ g.

Answer: 16 x 1.6605 x 10-24

Question 5. At STP, the volume of 17 g NH3 is _______ L.

Answer: 22.4

Question 6. At STP, the mass of 44.8 L of CO4 (g) is ________ g.

Answer: 88

Question 7. Formula mass of CaCI2 is _______ u.

Answer: 111

Question 8. ______ part of the mass of an atom of a C12 isotope is termed as atomic mass unit.

Answer: 1/12

Chapter 4 Matter Concept Of Mole Topic B Gram Atom Atomic Mass Unit Gram Mole Molar Volume Of Gases And Formula Mass State Whether True Or False

Question 1. The relative atomic mass of an element can be defined with respect to hydrogen (1H) and carbon (12C) scales.

Answer: True

Question 2. The actual mass of one molecule of CO2 is 7.31 x 10-23 g.

Answer: True

Question 3. Formula mass is applicable for oxygen molecule.

Answer: False

Question 4. The formula mass of NaCI is 58.5.

Answer: True

Question 5. Atomic mass unit is denoted by ‘u’.

Answer: True

Question 6. 1u =6.022 x 1023 g.

Answer: False

Question 7. Atomic mass of an elemernt is generally fraction.

Answer: True

Chapter 4 Matter Concept Of Mole Topic B Gram Atom Atomic Mass Unit Gram Mole Molar Volume Of Gases And Formula Mass Numerical Examples

Question 1. Express in gram-atom

  1. 46 g sodium,
  2. 3 g carbon.

Answer:

1. Atomic mass of sodium = 23

∴ 23 g sodium = 1 gram-atom sodium

∴ 46 g sodium = 46/23 = 2 gram-atom sodium

2. Atomic mass of carbon = 12

∴ 12 g carbon = 1 gram-atom carbon

∴ 3g carbon = 3/12 = 0.25 gram-atom carbon

Question 2. Express in gram

  1. 0.25 mol nitrogen atom,
  2. 1.5 mol sodium atom.

Answer:

1. Atomic mass of nitrogen = 14

∴ 1 mol nitrogen atom = 14 g nitrogen .

∴ 0.25 mol nitrogen atom = 0.25 x 14g = 3.5 g nitrogen

2. Atomic mass of sodium = 23.

∴ 1 mol sodium atom = 23 g sodium

∴ 1.5 mol sodium atom = 1.5 x 23g = 34.5g sodium

Question 3. Express in gram-mole

  1. 64 g oxygen,
  2. 2.2 g carbon dioxide.

Answer:

1. Molecular mass of oxygen = 32

∴ 32 g oxygen = 1 gram-mole oxygen

∴ 64 g oxygen = 64/32 = 2 gram-mole oxygen

2. Molecular mass of carbon dioxide (CO2) = 12 + 2 x 16 = 44

∴ 44 g carbon dioxide = 1 gram-mole carbon dioxide

∴ 2.2 g carbon dioxide = 2.2/44 = 0.05 gram- mole carbon dioxide

Question 4. What is the mass of l gram-atom of oxygen? How many atoms are there in this quantity of oxygen?

Answer:

Atomic mass of oxygen = 16

∴ Mass of 1 gram-atom of oxygen = 16 g

Again, 16 g oxygen = 1 mol oxygen atom = 6.022 x 1023 number of oxygen atoms

Chapter 4 Matter Concept Of Mole Topic C Chemical Calculation Using Molar Mass Molar Volume And Formula Mass Synopsis

If Avogadro’s number = NA and gram-molecular mass = M, then

1. Mass of 1 atom of an element = \(\frac{\text { gram-atomic mass }}{N_A}\)

2. Mass of 1 molecule of any substance = \(\frac{M}{N_A}\)

3. Number of molecules in W gram of any substance = \(\frac{W \times N_A}{M}\)

4. Number of molecules in VI of gas at STP = \(=\frac{V \times N_A}{22.4}\)

5.

WBBSE Solutions For Class 9 Physical Science Matter Concept Of Mole Topic C Chemical Calculation Using Molar Mass Molar Volume And Formula Mass

 

6. Number of molecules in n mol of any substance = n x NA.

Chapter 4 Matter Concept Of Mole Topic C Chemical Calculation Using Molar Mass Molar Volume And Formula Mass Short And Long Answer Type Questions

Question 1. How will you determine the number of moles and number of molecules In a given mass of a substance?

Answer:

Number of moles in a given mass of a substance

(n) = \(\frac{\text { given mass of the substance in gram }}{\text { gram-molecular mass of the substance }}\)

= \(\frac{m g}{M g \cdot \mathrm{mol}^{-1}}=\frac{m}{M} \mathrm{~mol}\)

Number of molecules in a given mass of substance (N) = Number of moles x Avogadro’s number = n x NA = m/M x NA.

Question 2. What is the actual mass of 1 molecule of NH3?

Answer:

Molecular mass of NH3 = 14 + 1 x 3 = 17

Hence, actual mass of one NH3 molecule =17u

= 17 X 1.6605 X 10-24 g = 28.2285 x 10-24 g

Question 3. Explain whether 0.1 g of diamond will contain the same number of carbon atoms as present in 0.1 g of graphite.

Answer:

Both graphite and diamond are the allotropes of carbon.

Hence, equal amount (by mass) of both substances will contain same number of carbon atoms.

Thus, 0.1 g of diamond and 0.1 g of graphite contain equal number of carbon atoms.

Question 4. How many gram-atoms and gram-moles of oxygen are present in 32 g oxygen?

Answer:

Atomic mass of oxygen = 16.

∴ 16 g oxygen = 1 gram-atom oxygen.

∴ 32 g oxygen = 32/16 = 2 gram-atom oxygen.

Now, molecular mass of oxygen = 32.

∴ 32 g oxygen = 1 gram-mole oxygen.

Question 5. Among equal volumes of Cl2  and O2 at STP, which one has more mass?

Answer:

We know, the volume of 1 g-mole of any gaseous substance at STP is 22.4 L.

Now 1 g-mole Cl2 = 2 X 35.5 g = 71 g Cl2

1 g-mole O2 = 2 x 16 g = 32 g O2

∴ At STP, mass of 22.4 L of Cl2 gas = 71 g and mass of 22.4 L of O2 gas = 32 g

∴ Cl2 gas will have more mass than O2 at given conditions.

Question 6. An isotope of chlorine is 35 times heavier than 1/12 th of the mass of a C12 isotope. What is the actual mass of the Cl-leading isotope?

Answer:

1/12th of the mass of a C12 isotope = 1u = 1.6605 x 10-24  g

∴ The actual mass of the chlorine isotope = 35 x 1.6605 x 10-24 g

= 58.1175 x 10-24 g = 5.81175 X 10-23 g

Question 7. Prove that 1 gram-molecule of any substance contains the same number of molecules.

Answer:

Let, the molecular mass of a substance be M.

∴ Its molecule will be M times heavier than that of a hydrogen atom.

If the actual mass of a hydrogen atom be xg, then the actual mass of a molecule of that substance = M x xg.

Now, the mass of 1 gram-molecule of the substance is Mg.

∴ Number of molecules constituting 1 gram- molecule of the substance = \(\frac{M}{M \times x}=\frac{1}{x}\)

Since value of x is definite, so value of 1/x is also definite.

For example, number of molecules in 1 molecule oxygen gas

= \(\frac{32}{32 x}=\frac{1}{x} \quad\left[because M_{0_2}=32\right]\)

Similarly number of molecules in1 gram-molecule nitrogen gas.

= \(\frac{28}{28 x}=\frac{1}{x} \quad\left[because M_{N_2}=28\right]\)

Similarly, number of molecules in W g of B \(\frac{W}{M_B} \times 6.022 \times 10^{23}\)

∴ Ratio of number of molecules in Wgof A and

B = \(\frac{W}{M_A} \times 6.022 \times 10^{23}: \frac{W}{M_B} \times 6.022 \times 10^{23}\)

= \(M_B: M_A\)

Thus, the ratio of the number of molecules present in equal masses of two substances having different molecular masses is equal to the reciprocal of the ratio of their molecular masses.

Question 9. Molecular mass and atomic element are x and y respectively. Express the mass of1 molecule and 1 atom of that element In gram.

Answer:

Since atomic mass of the element is y, hence mass of 1 gram-atom = yg

∴ Total mass of NA number of atoms = yg [NA = Avogadro’s number]

∴ Actual mass of 1 atom = \(\frac{y}{N_A}\) g

Again, molecular mass of the element =x

∴ Mass of 1 gram-molecule of the element = xg

∴ Total mass of NA number of molecules = xg

∴ Mass of a molecule of the element = \(\frac{x}{N_A}\)g

Chapter 4 Matter Concept Of Mole Topic C Chemical Calculation Using Molar Mass Molar Volume And Formula Mass Very Short Answer Type Questions Choose The Correct Answer

Question 1. Number of molecules in 36 g water is

  1. 6.022 X 1023
  2. 12.044 X 1023
  3. 3.011 x 1023
  4. 9.0345 X 1023

Answer: 2. 12.044 X 1023

Question 2. Which of the following contains maximum number of moles?

  1. 18 g water,
  2. 34 g ammonia,
  3. 71 g chlorine,
  4. 8 g hydrogen.

Answer: 4. 8 g hydrogen.

Question 3. Which of the following contains maximum number of atoms?

  1. 28g N2,
  2. 18g H2O,
  3. 17g NH3,
  4. 16g CH4.

Answer: 4. 16gCH4.

Question 4. Number of H-atoms in 2 mol water is

  1. 1 mol
  2. 2 mol
  3. 3 mol
  4. 4 mol

Answer: 4. 4 mol

Question 5. Number of moles of H+ ions obtained from 98 g H2SO4 is

  1. 1 mol
  2. 2 mol
  3. 3 mol
  4. 4 mol

Answer: 2. 2 mol

Question 6. Amount of Ca(OH)2 required to produce 1 mol of OH ions is

  1. 74 g
  2. 148 g
  3. 37 g
  4. 60 g

Answer: 3. 37 g

Question 7. Which of the following has a volume of 22.4 L at STP?

  1. 36 g water
  2. 64 g oxygen
  3. 71 g chlorine
  4. 4 g hydrogen

Answer: 3. 71 g chlorine

Question 8. Which of the following contains maximum number of molecules?

  1. 44 g CO2
  2. 48 g O3
  3. 8 g H2
  4. 64 g SO2

Answer: 3. 8 g H2

Question 9. At STP, the mass of 44.8 L of a gas is 88 g. The molecular formula of the gas is

  1. N2
  2. CO2
  3. NH3
  4. O2

Answer: 2. CO2

Question 10. Which of the following contains maximum number of molecules?

  1. 10 g O2 (g)
  2. 15 L H2 (g) at STP
  3. 5 L N2 (g) at STP
  4. 0.5 g H2 (g)

Answer: 2. 15 L H2 (g) at STP

Question 11. Number of atoms in 4.25 g of NH3 is

  1. 6.022 x 1023
  2. 4 x 6.022 x 1023
  3. 1.7 x 1024
  4. 4.25 X 6.022 x 1023

Answer: 1. 6.022 x 1023

Question 12. How many grams of CO2 will contain the same number of molecules as that in 44.8 L of NH3 at STP?

  1. 44 g
  2. 66 g
  3. 88 g
  4. 132 g

Answer: 3. 88 g

Question 13. Which of the following contains maximum number of molecules?

  1. 1 g CO2
  2. 1 g H2
  3. 1 g O2
  4. 1 g CH4

Answer: 2. 1 g H2

Question 14. Mass of 44.8 L of a gas at STP is 92 g. Formula of the gas is

  1. N2
  2. NO2
  3. NH2
  4. O2

Answer: 2. NO2

Question 15. Mass of 2 mol of ammonia is

  1. 12 g
  2. 51 g
  3. 34 g
  4. 17 g

Answer: 3. 34 g

Question 16. The volume of 34 g ammonia at STP is

  1. 11.2 L
  2. 22.4 L
  3. 44.8 L
  4. 42.2 L

Answer: 3. 44.8 L

Question 17. If the mass of 6.022 x 1020 atoms of an element is 0.012 g, its atomic mass will be

  1. 11
  2. 12
  3. 13
  4. 24

Answer: 2. 12

Question 18. The volume of 2 g-mol CO2 at STP is

  1. 22.4 L
  2. 44.8 L
  3. 11.2 L
  4. 2.24L

Answer: 2. 44.8 L

Question 19. Number of carbon atoms in 1.71 g of sugar is

  1. 3.6 x 1022
  2. 7.2 x 1022
  3. 6.6 x 1022
  4. None of the above

Answer: 1. 3.6 x 1022

Question 20. Amount of carbon present in 0.5 mol potassium ferrocyanide (K4[Fe(CN)6]) is ®

  1. 1.5 mol
  2. 36 g
  3. 18 g
  4. 3.6 g

Answer: 2. 36 g

Question 21. Which of the following contains least number of molecules?

  1. 1 g H2
  2. 1 g N2
  3. 1.1 g O2
  4. 1.5 g O2

Answer: 3. 1.1 g O2

Question 22. Number of moles of oxygen atoms present in 6.022 x 1024 molecules of CO is

  1. 10
  2. 5
  3. 1
  4. 0.5

Answer: 1. 10

Chapter 4 Matter Concept Of Mole Topic C Chemical Calculation Using Molar Mass Molar Volume And Formula Mass Answer In Brief

Question 1. Which of the given substances has maximum number of molecules 28 g N2, 32 g O2,18 g H2O, and 100 g CaCO2?

Answer: 28g N2 = 1 mol N2;

32g O2 = 1 mol O2;

18g H2O = 1 mol H2O;

100 g CaCO3 = 1 mol CaCO3

As all the substances contain the same number of moles, the number of molecules in each of the substances will be equal i.e., 6.22x 1023.

Question 2. What is the mass of 112 mL hydrogen gas at NTP?

Answer: 0.01 g.

Question 3. A gas jar contains 17 g NH3 and another contains 44.8 L NH3 at STP. Which gas jar contains more molecules?

Answer: 17g NH3 = 17/17 mol = 1 mol NH

Again, 44.8L NH3 at STP = \(\frac{44.8}{22.4}\) mol NH3 = 2 mol NH3

Hence, the second gas jar contains more number of molecules.

Question 4. What is the volume of 7g of nitrogen at NTP?

Answer: At NTP, volume of 28 g of nitrogen = 22.4 L

Hence, at NTP, volume of 7g of nitrogen = \(\frac{22.4}{28}\) X 7 L = 5.6 L.

Question 5. What is the volume of 88 g CO2 (g) at STP?

Answer: 88gCO2 = \(\frac{88}{44}\)mol CO2 = 2mol CO2

Hence, volume of 88g of CO2 at STP = 2 X 22.4L = 44.8 L.

Question 6. Between 42g nitrogen and 64 g oxygen, which has more gram-atoms?

Answer: 42 g nitrogen = \(\frac{42}{14}\)gram-atoms of nitrogen = 3 gram-atoms of nitrogen.

64g oxygen = \(\frac{64}{16}\) gram-atoms of oxygen = 4 gram-atoms of oxygen.

Hence, 64g of oxygen contains more number of gram-atoms.

Question 7. Which of the following has maximum number of molecules at STP? 100m3 CO2, 200cm3 NH3, 150cm3 O2.

Answer: At STP, 200 cm3 of NH3 will contain the maximum number of molecules.

Question 8. How many atoms are there in 67.2L of a diatomic gas at STP?

Answer: 67.2 L of a gas at STP = \(\frac{67.2}{22.4}\) mol of the gas = 3 mol of the gas

So, number of molecules of the gas = 3 x 6.022 x 1023

As the gas is diatomic, the number of atoms = 2 x 3 X 6.022 x 1023 = 36.132 x 1023

Question 9. How many molecules are present in 109.5 g HCI?

Answer: Number of molecules in 109.5 g HCI

= \(\frac{109.5}{36.5}\) x 6.022 x 1023 = 18.066 x 1023 = 1.8066 X 1024

Question 10. Are the masses of 1 mol sodium and 1 mol oxygen are same?

Answer: The masses of 1 mol sodium and 1 mol oxygen are not same.

Question 11. How many atoms are there in 11.5 g sodium?

Answer: Number of atoms in 11.5 g sodium

= \(\frac{11.5}{23}\) X 6.022 x 1023 = 3.011 x 1023

Question 12. What will be the volume of hydrogen gas at STP produced from the reaction of 23 g of Na with water?

Answer: 2Na + 2H2O → 2NaOH + H2 ↑

∴ Volume of hydrogen gas at STP = 11.2 L

Question 13. What is the total number of electrons present in 1 mol of water?

Answer: Number of electrons in 1 mol of water = 10 mol

= 10 x 6.022 x 1023 = 6.022 x 1024

Question 14. What is the mass of 0.5 mol CO2 gas?

Answer: Mass of 0.5 mol of CO2 = 44 X 0.5 g = 22 g

Chapter 4 Matter Concept Of Mole Topic C Chemical Calculation Using Molar Mass Molar Volume And Formula Mass Fill In The Blanks

Question 1. Number of CO2 molecules in 88g CO2 is _______

Answer: 2 x 6.022 x 1023

Question 2. 16mol protons will be obtained from ______ mol oxygen.

Answer: 1

Question 3. Number of electrons present in 14 g nitrogen is _______

Answer: 7 x 6.022 X 1023

Question 4. Number of protons in 18 g water is _______

Answer: 10 X 6.022 X 1023

Question 5. 1 mol of OH ions is obtained from ______ g NaOH.

Answer: 40

Question 6. At STP, the volume of __________ mol of any gas (elemental or compound) is 22.4 L.

Answer: 1

Question 7. The number of H-atoms in 10 mol water is _________ mol.

Answer: 20

Question 8. 98 g H2SO4 = _______ mol H2SO4.

Answer: 1

Question 9. Total mass of 3.011 x 1023 number of oxygen atoms is ________ g.

Answer: 8

Question 10. At STP, the volume of a gas (in L) = number of moles of the gas molecules x _______ L.

Answer: 22.4

Chapter 4 Matter Concept Of Mole Topic C Chemical Calculation Using Molar Mass Molar Volume And Formula Mass State Whether True Or False

Question 1. The molar volume of 256 g of sulphur dioxide gas at STP is 89.6 L.

Answer: True

Question 2. The percentage of calcium in CaCO3 is 60%.

Answer: False

Question 3. 2mol OH ions can be obtained from 110 g KOH.

Answer: False

Question 4. The number of atoms present in 3.2 g CH4 is 6.022 x 1023.

Answer: True

Question 5. The number of chlorine atoms in 71 g HCI is 0.5 x 6.022 x 1023.

Answer: False

Question 6. 022 x 1023 number of molecules are present in 36 g of water.

Answer: False

Question 7. Volume of 32 g SO2 at STP is 11.2 L.

Answer: True

Question 8. 4 mol CO2 = 88 g CO2.

Answer: False

Question 9. Number of molecules present in 2 millimol of chlorine gas is 2 x 6.022 x 1020.

Answer: True

Question 10. The volume of 64 g oxygen at STP is 10 L.

Answer: False

Question 11. 44 g of CO2 contains 6.022 x 1023 number of molecules.

Answer: True

Chapter 4 Matter Concept Of Mole Topic C Chemical Calculation Using Molar Mass Molar Volume And Formula Mass Numerical Examples

Question 1. Find the mass of 1 molecule of CO2.

Answer:

Molar mass of carbon dioxide = 12 + 2 x 16 = 44

Now, 1 mol of CO2 contains 6.022 x 1023 molecules.

Hence, mass of 6.022 x 1023 CO2 molecules = 44 g

∴ Mass of one CO2 molecule = \(\frac{44}{6.022 \times 10^{23}} g\) = 7.3 x 1023 g

Question 2. Between 100 g calcium and 100 g iron, which one will contain more number of atoms? (Given: atomic masses of calcium and iron are 40 u and 56 u respectively)

Answer:

Number of moles in 100 g calcium = \(\frac{\text { given mass }}{\text { gram-atomic mass }}=\frac{100 \mathrm{~g}}{40 \mathrm{~g}}=2.5\)

∴ Number of atoms in 100 g calcium = 2.5 x 6.022 x 1023 = 1.5055 X 1024

Again, number of moles in 100 g iron

= \(\frac{\text { given mass }}{\text { gram-atomic mass }}=\frac{100 \mathrm{~g}}{56 \mathrm{~g}}=1.78\)

∴ Number of atoms in 100 g iron = 1.78 X 6.022 X 1023 = 1.07 x 1024

Therefore, 100 g calcium contains more number of atoms than 100 g iron.

Question 3. How many grams of H2SO4 are required to produce 1 gram-ion of H+?

Answer:

\(\mathrm{H}_2 \mathrm{SO}_4(a q) \rightleftharpoons 2 \mathrm{H}^{+}(a q)+\mathrm{SO}_4^{2-}(a q)\)

 

1 molecule of H2S04 gives two H+ ions in aqueous solution.

Hence, 1 gram-mole of H2SO gives 2 gram-ions of H+ ion.

∴ 1 gram-ion of H+ ion will be produced from 1/2 gram-mole of H2SO4= 98/2 = 49g H2SO4 [as 1 gram-mole of H2SO4 = 98 g]

Thus, 49 g H2SO4 is required to produce 1 gram- ion of H+ ions.

Question 4. Which of the following has maximum number of molecules? 0.5 mol H2(g), 10g O2(g) and 15 L Cl2 at STP.

Answer:

Number of molecules present in a substance is proportional to its number of moles.

Amount of H2 = 0.5 mol

\(10 \mathrm{~g} \mathrm{O}_2 =\frac{\text { given mass }}{\text { gram-molecular mass }}=\frac{10 \mathrm{~g}}{32 \mathrm{~g}}=0.31 \mathrm{~mol} \mathrm{O}_2\)

 

22.4 L Cl2 at STP = 1 mol Cl2

∴ 15 L Cl2 = \(\frac{15L}{22.4L}\) = 0.67 mol Cl2

Hence, 15 L Cl2 at STP will have maximum number of molecules.

Question 5. Arrange the gases in increasing number of atoms present in given amounts

  1. 18gH2O,
  2. 18gCO2
  3. 18gCO2,
  4. 18gCH4.

Answer:

1. 18gH2O = 18/18 = 1 mol H2O = 6.022 x 1023 molecules of H2O

As 1 molecule of H2O have 3 atoms, number of atoms = 3 x 6.022 x 1023

2. 18 g O2= 18/32 =0.5625 mol O2 = 0.5625 x 6.022 x 1023 molecules of O2.

As 1 molecule of O2 have 2 atoms, number of atoms = 2 X 0.5625 X 6.022 X 1023 = 1.125 x 6.022 x 1023

3. 18 g CO2 = 18/44 = 0.409 mol CO2 = 0.409 x 6.022 x 1023 molecules of CO2

As 1 molecule of CO2 have 3 atoms, number of atoms = 3 x 0.409 x 6.022 x 1023 = 1.227 x 6.022 x 1023

4. 18 g CH4 = 18/16 = 1.125 mol CH4 = 1.125 x 6.022 x 1023 molecules of CH4

As 1 molecule of CH4 has 5 atoms, number of atoms = 5 X 1.125 x 6.022 x 1023 = 5.625 X 6.022 X 1023

Hence, correct order is 18g O2 < 18g CO2 < 18g HO2 < 18g CH4

Question 6. 3.42 g sucrose is dissolved in 18 g water. Find the number of oxygen atoms in the formed solution.

Answer:

Molecular mass of sucrose (C12H22O11) = 12 x 12 + 1 x 22 + 16 x 11 = 342

∴ 3.42 g sucrose = 3\(\frac{3.42}{342}\) = 0.01 mol sucrose = 0.01 x 6.022 x 1023 molecules of sucrose

As 1 molecule of sucrose has 11 oxygen atoms, number of O-atoms in 3.42 g sucrose = 11 X 0.01 x 6.022 x 1023

Molecular mass of water (H2O) = 1 x 2 + 16 =18

∴ 18 g water = 1 mol water = 6.022 x 1023 molecules of water.

As each molecule of water contains 1 O-atom, number of O-atoms in 18 g water = 6.022 x 1023

∴ Total number of O-atoms in formed solution = 11 x 0.01 x 6.022 X 1023 + 1 x 6.022 x 1023 = 1.11 X 6.022 x 1023 = 6.68 X 1023

Question 7. Calculate the number of H-atoms present in 9 g water.

Answer:

Molecular mass of H2O = 1 x 2 + 16 = 18

Hence, 9 g water = 9/18 = 0.5 mol water = 0.5 x 6.022 x 1023 molecules of water

Each molecule of water contains two H-atoms.

Hence, number of H-atoms in 9 g water = 2 x 0.5 x 6.022 x 1023 = 6.022 x 1023

Question 8. State whether number of atoms in

  1. 1 g oxygen atom,
  2. 1 g oxygen molecule and
  3. 1 g ozone molecule will be same or different? Justify your answer.

Answer:

1. 16 g O-atom = NA no. of O-atoms

∴ 1 g O-atom = \(\frac{N_A}{16}\) no. of O-atoms

2. 32 g O2 molecule = NA no. of O2 molecules

∴ 1 g O2 molecule = \(\frac{N_A}{32}\)no. of O2 molecules

= \(\frac{N_A}{32}\) x 2 no. of O-atoms = \(\frac{N_A}{16}\)no. of O-atoms.

3. 48 g O3 molecule = NA no. of O3 molecules

∴ 1 g O3 molecule = \(\frac{N_A}{48}\) no. of O3 molecules

= \(\frac{N_A}{48}\) x 3 no. of O-atoms = \(\frac{N_A}{16}\) no. of O-atoms.

Hence, number of atoms in each of the substances will be equal.

Question 9. Can you drink Avogadro’s number of water molecules? (Given: Density of water = 1g • m-1)

Answer:

Molecular mass of water = 2 x 1 + 16 = 18 Gram-molecular mass of water = 18 g and molar mass of water = 18 g • mol-1

Thus, the mass of Avogadro’s number (6.022 x 1023) of water molecules = 18 g

Density of water = 1g • mL-1.

Hence, volume of 18 g water = 18 x 1 = 18mL

Thus, one can easily drink 18 mL water (approximately one test tube of water) or Avogadro’s number (6.022 x 1023) of water molecules.

Question 10. What is the volume of 4.4 g CO2 at STP?

Answer:

Gram-molecular mass of CO2 = 12 + 16 x 2 = 44 g

Hence, at STP, volume of 44 g of CO2 = 22.4 L

∴ Volume of 4.4 g of CO2 at STP = \(\frac{22.4}{44}\) x 4.4L = 2.24 L

Question 11. At STP, the volume of 0.44 g of a gas is 224 cm3. Find the molecular mass of the gas.

Answer:

At STP, mass of 224 cm3 of the gas = 0.44 g

∴ At STP, the mass of 22.4L or 22400 cm3 of the gas = \(\frac{0.44}{224}\) x 22400g = 44g

Hence, the molecular mass of the gas = 44

Question 12. What is the volume of 8 g of oxygen at standard temperature and pressure?

Answer:

Molecular mass of oxygen = 32

∴ 1 gram-mole oxygen = 32 g oxygen

Hence, at STP, volume of 32 g oxygen is 22.4 L.

∴ At STP, volume of 8 g oxygen = \(\frac{22.4}{32}\) x 8L = 5.6 L

Question 13. Find the mass (in gram) of 5.6 L ammonia at STP.

Answer:

Gram-molecular mass of ammonia (NH3) = 14+ 1×3 = 17 g

Hence, at STP, the mass of 22.4 L of ammonia = 17 g

∴ At STP, the mass of 5.6 L of ammonia = \(\frac{17}{22.4}\) x 5.6 g = 4.25 g

Question 14. Find the volume of 4 g SO2 at standard temperature and pressure.

Answer:

Molecular mass of SO2 = 32 + 16 x 2 = 64

Hence, at STP, the volume of 64 g SO2 = 22.4L

∴ At STP, the volume of 4 g SO2 = \(\frac{22.4}{64}\) x 4L= 1.4 L

Question 15. What is the mass of 4 gram-moles of oxygen? What will be the volume of this quantity of oxygen at STP?

Answer:

Molecular mass of oxygen = 32

Hence, 1 gram-mole of oxygen = 32 g oxygen

∴ 4 gram-moles of oxygen = 4 x 32 g = 128 g oxygen

Volume of 1 gram-mole of oxygen at STP = 22.4 L

Therefore, the volume of 4 gram-moles of oxygen at STP = 4 X 22.4L = 89.6L

Question 16. State whether, at STP, both 22.4 L NH3 and 22.4 L CO2 will contain

  1. same number of molecules?
  2. same number of atoms?

Answer:

1. At STP, 22.4 L NH3 = 1 mol NH3 Now, 1 mol NH3 = 6.022 x 1023 molecules NH3

At STP, 22.4 L carbon dioxide = 1 mol carbon dioxide = 6.022 x 1023 molecules of CO2

Hence, both NH3 and carbon dioxide contain same number of molecules.

2. 1 molecule of NH3 contains 4 atoms.

Hence, number of atoms in 22.4 L NH3 at STP = 4 X 6.022 X 1023 = 24.088 X 1023

Again, 1 molecule of CO2 contains 3 atoms.

Hence, number of atoms in 22.4 L CO2 at STP = 3 X 6.022 X 1023 = 18.066 X 1023

Therefore, 22.4 L NH3 will contain more atoms than that in 22.4 L CO2.

Question 17. Find the number of constituent ions in 11.1 g calcium chloride.

Answer:

Gram-formula mass of calcium chloride (CaCI2) = 40 + 35.5 X 2 = 111 g

∴ Number of formula units in 111 g CaCl2 = 6.022 x 1023

∴ Number of formula units in 11.1 g CaCI2

= \(\frac{6.022 \times 10^{23}}{111}\) x 11.1 = 6.022 x 1022

Each formula unit of calcium chloride contains Ca2+ ions and 2CI ions.

Hence, number of Ca2+ ions in 11.1 g CaCI2 = 1 x 6.022 X 1022 = 6.022 x 1022

Number of Cl ions in 11.1 g CaCI2 = 2 X 6.022 X 1022 = 12.044 x 1022

Question 18. What is the gram-formula mass of NaCI? Calculate the number of gram-formula units present in 234 g of NaCI.

Answer:

Atomic masses of sodium and chlorine are 23 and 35.5 respectively.

Hence, formula mass of NaCI = 23 + 35.5 = 58.5

∴ Gram-formula mass of NaCI = 58.5g

Number of gram-formula units in 234 g NaCI = \(\frac{234 \mathrm{~g}}{58.5 \mathrm{~g}}\) = 4

Question 19. A mixture of hydrogen and oxygen contains 20% hydrogen by weight. Calculate total number of molecules present per gram of the mixture.

Answer:

Mass of hydrogen in 1 g of the mixture = \(\frac{1 \times 20}{100} \mathrm{~g}\) 0.2 g

∴ Mass of oxygen in 1 g of mixture = (1 – 0.2) g = 0.8 g

Now, number of molecules of hydrogen in 0.2 g of hydrogen gas

= \(\frac{0.2}{2}\) X 6.022 X 1023 = 6.022 x 1023

Number of molecules of oxygen in 0.8 g of oxygen gas

= \(\frac{0.8}{32}\) x 6.022 x 1023 = 1.5055 x 1022

∴ Total number of molecules per gram of the mixture

= (6.022 + 1.5055) x 1022 = 7.5275 x 1022

Question 20. Calculate the total charge in 6 g of \(\mathrm{CO}_3^{2-}\) ion.

Answer:

Formula mass of \(\mathrm{CO}_3^{2-}\) ion = (12 + 3 x 16) = 60

∴ Number of moles of \(\mathrm{CO}_3^{2-}\) ion = \({6}{60}\) = 0.1

Now, total negative charge of a \(\mathrm{CO}_3^{2-}\) ion is equal to the negative charge of 2 electrons.

Again, total charge of 1 mol electron = 96500 C

Total charge of 6 g or 0.1 mol of \(\mathrm{CO}_3^{2-}\) ion = 2 X 0.1 X 96500 C = 19300 C

Question 21. Calculate the number of Na and ClΘ ions in 117 g of sodium chloride. [Na = 23, Cl = 35.5]

Answer:

Formula mass of sodium chloride = (23+ 35.5) = 58.5 117

∴ 117 g of NaCI = \({117}{58.5}\)= 2gram-formula mass of NaCI.

Now in NaCI, number of Naions = number of ClΘ ions

∴ Number of Na® ions in 2 gram-formula mass of NaCI = number of ClΘ ions in 2 gram-formula mass of NaCI

= 2 x 6.022 x 1023 = 12.044 x 1023 = 1.2044 X 1024

Question 22. You have been given a glass of water mixed with 10 g glucose (C6H12O6). If you drink the entire solution, calculate the number of glucose molecules you have consumed.

Answer:

Molecular mass of glucose = 6 x 12 + 12 x 1 + 6 x 16 = 180

∴ Number of moles of glucose in 10 g = \(\frac{10}{180}\)= 0.0555

∴ Number of molecules of glucose in 10 g = 0.0555 X 6.022 X 1023 = 3.345 x 1022

∴ 3.345 x 1022 number of glucose molecules will be consumed.

Question 23. Calculate the number of electrons, protons and neutrons in 1 mol O2- ion.

Answer:

Number of electrons in an O2- ion = (8 + 2) = 10

Number of protons in an O2- ion = 8

∴ Number of neutrons in an O2- ion = (16 – 8) = 8 (assuming 8O1616 isotope)

Number of electrons in 1 mol O2- ion = 1 x 10 x 6.022 x 1023 = 6.022 x 1024

∴ Number of protons in 1 mol O2- ion = 1 X 8 x 6.022 X 1023 = 4.817 X 1024

∴ Number of neutrons in 1 mol O2- ion = 1 x 8 x 6.022 x 1023 = 4.817 x 1024

Question 24. Calculate the number of atoms and molecules in 124 g phosphorus.

Answer:

Atomic mass of phosphorus = 31

∴ 1 g-atom phosphorus =31 g phosphorus

Molecular mass of phosphorus =31 x 4 = 124

∴ 1 g-molecule phosphorus = 124 g phosphorus

∴ 124 g phosphorus =  \(\frac{124}{31}\)g-atom phosphorus

= 4 g-atom phosphorus

= 4 x 6.022 x 1023 atoms of phosphorus

= 2.4088 x 1024 atoms of phosphorus 124 g phosphorus

= \(\frac{124}{124}\) g-molecule phosphorus

= 1 x 6.022 x 1023 molecules of phosphorus = 6.022 x 1023 molecules of phosphorus

Question 25. Calculate the number of hydrogen atoms in 1 L of water (at 4°C temperature).

Answer:

Density of water at 4°C = 1 g • mL-1

∴ Mass of 1 L or 1000 mL of water at 4°C = 1000 g

Now molecular mass of water = 1 x 2 + 16 = 18

∴ Number of molecules in 1000 g of water = \(\frac{6.022 \times 10^{23}}{18} \times 1000\) = 3.345 x 1025

Again, number of H-atoms in 1 molecule of water = 2

∴ Number of H-atoms in 1 L water = 2 X 3.345 X 1025 = 6.69 X 1025

Question 26. Calculate the number of hydrogen atoms present in 51 g ammonia gas. What will be the volume of that amount of gas at STP?

Answer:

Gram-molecular mass of NH3 = 17 g

∴ 51 g ammonia = \(\frac{51}{17}\) mol = 3 mol ammonia

Now, the number of H-atoms per molecule of ammonia= 3

∴ Number of H-atoms in 51 g or 3 mol ammonia gas = 3 x 3 x 6.022 x 1023 = 5.4198 x 1024

The volume of 51 g or 3 mol of ammonia gas at STP = 3 X 22.4 L = 67.2 L.

Question 27. Which one is heavier between 1 mol NO molecule and 0.5 mol NO2 molecule?

Answer:

Molecular mass of NO = 14 + 16 = 30

∴ Mass of 1 mol NO molecules = 30 g

Again, molecular mass of NO2 = 14 + 16 X 2 = 46

∴ Mass of 0.5 mol NO2 molecules = 46 X 0.5 = 23 g

∴ 1 mol NO molecules will be heavier than that of 0.5 mol NO2 molecules.

Question 28. Calculate the total number of atoms in 80 u helium.

Answer:

Mass of a helium atom = 4 u

∴ Number of atoms in 80 u helium = \(\frac{80u}{4u}\) = 20.

Question 29. Calculate the number of molecules left when 1021 molecules of CO2are removed from 200 mg of CO2

Answer:

200 mg of CO2 = 0.2 g of CO2

Now, 44g of CO23 contains 6.022 x 1023 molecules

∴ 0.2 g of CO2 contains

⇒ \(\frac{6.022 \times 10^{23} \times 0.2}{44}=2.7372 \times 10^{21}\) molecules

∴ On removing 1021 molecules, number of CO2 molecules left

= 2.7372 X 1021 – 1021 = 1.7372 X 1021

Chapter 4 Matter Concept Of Mole Topic C Chemical Calculation Using Molar Mass Molar Volume And Formula Mass Miscellaneous Type Questions

Match The Columns

1.

WBBSE Solutions For Class 9 Physical Science Matter Concept Of Mole Topic C Chemical Calculation Using Molar Mass Molar Volume And Formula Mass Match The Column 1

Answer: 1. B, 2. A, 3. D, 4. C

2.

WBBSE Solutions For Class 9 Physical Science Matter Concept Of Mole Topic C Chemical Calculation Using Molar Mass Molar Volume And Formula Mass Match The Column 2

Answer: 1. D, 2. A, 3. B, 4. C

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Solution

Chapter 4 Matter Solution Topic A True Solution Colloid And Suspension Dissolution Of Smaller Ion Or Molecules And Larger Molecules In Water Synopsis

  1. A solution is a homogeneous mixture of two or more substances in which the amount of each constituent can be varied within a certain limit.
  2. The component of a solution which is present in greater amount is called the solvent. The solution has the same physical state as that of the solvent.
  3. On the other hand, the component of the solution which is present in lesser amount is called the solute.
  4. A colloidal solution is a stable heterogeneous system of two immiscible phases in which one phase (solid, liquid or gas) is dispersed as particles with diameter ranging from 10-7-10-5 cm, in another phase (solid, liquid or gas).
  5. Colloid is not a special type of substance, rather it is a state of the substance.
  6. The medium in which the colloidal particles remain uniformly dispersed is known as the dispersion medium and the substance whose particles remain dispersed in a colloidal solution is called the dispersed phase.
  7. When light rays are passed through a colloidal solution, the rays get scattered by the colloidal particles and as a result tye path of light becomes clearly visible. This phenomenon is known as Tyndai. effect.
  8. Tyndall effect helps to distinuguish between true solutions and colloidal solutions.
  9. On the basis of the physics states of the re dispersion medium and dispersed phase, colloids areclassified into eight groups.
  10. Colloids are classified into eight groups. They are sol, solid, sol, gel, emulsion, solid aerosol, liquid aerosol, solid foam and foam.
  11. An emulsion is a colloidal solution in which both the dispersed phase and dispersion medium are liquid.
  12. Emulsions are of two types: oil-in-water type and water-in-oil type emulsion.
  13. The chemical substances used to enhanse the stability of an emulsion are called emulsifiers. Soaps and detergents are well-known emulsifiers.
  14. A suspension is a heterogeneous and unstable system in which particles having diameter greater than 10-5 cm remain dispersed in the solvent. On standing, the particles slowly separate out from the mixture and settle at the bottom.

Chapter 4 Matter Solution Topic A True Solution Colloid And Suspension Dissolution Of Smaller Ion Or Molecules And Larger Molecules In Water Short And Long Answer Type Questions

Question 1. What is a solution?

Read and Learn More WBBSE Solutions for Class 9 Physical Science and Environment

Answer:

Solution:-

If a homogeneous mixture of two or more substances (solid, liquid or gas) have uniform properties (in terms of constituents and structure) throughout the mixture and the amounts of the constituents can be varied within certain limits, then the homogeneous mixture is said to be a solution.

Question 2. What do you mean by solvent and solute?

Answer:

Solvent And Solute:-

Solvent: The component of a solution which is generally present in greater amount and whose physical state is the same as that of the solution is known as the solvent.

Solute: The component of a solution which is present in lesser amount and which remains dissolved in another solid, liquid or gaseous substance to form a homogeneous mixture (solution) is called a solute. For example, sugar dissolves in water to form a homogeneous mixture. Here, sugar is the solute.

In some solutions, where both solvent and solute are in the same phase, the terms solvent and solute is defined with respect to their relative quantities in the solution.

For example, when 70 parts of alcohol mix with 30 parts of water, alcohol is considered as the solvent and water as the solute. On the other hand, in a mixture of 70 parts of water and 30 parts of alcohol, water is the solvent and alcohol is the solute.

Question 3. All solutions are mixtures but all mixtures are not solutions. Justify.

Answer:

All Solutions Are Mixtures But All Mixtures Are Not Solutions:-

If a mixture of two or more substances is homogeneous in nature and their composition can be varied only within a certain limit, then the mixture is called a solution. For example, common salt dissolved in water forms a solution.

However, a mixture may also be heterogeneous in nature, in which the composition is not uniform throughout and can be changed in any proportion. This type of mixtures cannot be termed as solution. For example, a mixture of sand and sugar cannot be called a solution.

Thus, all solutions are mixtures but all mixtures are not solutions.

Question 4. Define homogeneous and heterogeneous mixtures.

Answer:

Homogeneous And Heterogeneous Mixtures:-

Homogeneous Mixtures: A homogeneous mixture is a solid, liquid or gaseous mixture that has the same proportions of its components throughout any given sample.

Example: air or sugar solution.

Heterogeneous Mixture: A heterogeneous mixture is simply a mixture that is not uniform in composition. It has components in which proportions vary throughout the sample.

Example: mixture of sand and sulphur powder.

Question 5. Why is an aqueous solution of sugar called a true solution?

Answer:

An Aqueous Solution Of Sugar Called A True Solution:-

When sugar is added to water, the sugar molecules occupy the intermolecular spaces between the water molecules and dissolve in it. As a result, the sugar molecules form a homogeneous mixture with water and cannot be separated easily. This is why an aqueous solution of sugar is called a true solution.

Question 6. Discuss the properties of a true solution.

Answer:

The Properties Of A True Solution:-

  1. A true solution possesses the following properties
  2. A true solution is a homogeneous mixture of solvent and solute. It has uniform properties (in terms of components, physical properties and structural features) throughout the solution.
  3. The solute particles present in a true solution are so small (particle diameter is less than or equal to 10-8 cm) that they cannot be seen even under an ultramicroscope.
  4. In a true solution, the solvent and solute may exhibit some changes in their physical properties, but their chemical properties remain unchanged.
  5. The relative amount of the components in a solution can be increased or decreased within a certain limit.
  6. The solute and solvent in a true solution cannot be separated by filtration or by gravitational separation. Even if the solution is kept undisturbed for a long period of time, the solute does not settle down.
  7. The components of a true solution can be separated by physical methods such as, evaporation, distillation or crystallisation.
  8. Heat may or may not be evolved or absorbed during formation of a true solution.

Question 7. What is a colloidal solution?

Answer:

Colloidal Solution:-

A colloidal solution is said to be a stable heterogeneous system of two immiscible phases in which one phase (solid, liquid or gas) comprises of the particles with diameter ranging from 10-7 – 10-5 cm, is dispersed into another phase (solid, liquid or gas).

For example, freshly precipitated ferric hydroxide when shaken with water and small amount of ferric chloride, it forms a colloidal solution.

Question 8. What is meant by dispersion medium and dispersed phase of a colloidal solution?

Answer:

Dispersion Medium: The medium in which the colloidal particles remain uniformly dispersed is called the dispersion medium.

Dispersed Phase: The component of a colloidal solution which remain uniformly dispersed in the dispersion medium and consists of particles with diameter ranging from 10~7-10-5 cm is called the dispersed phase.

Example: In a gold sol, water is the dispersion medium while gold particles form the dispersed phase.

Question 9. State whether colloidal particles can be separated from a solution by filtration using a filter paper.

Answer:

  1. The pores of the filter paper are generally larger than 10-5 cm. So, the colloidal particles having diameter of 10-7 to 10-5 cm can easily pass through a filter paper. So, colloidal particles cannot be separated from a solution using a filter paper.
  2. On the other hand, colloidal particles cannot pass through a parchment paper. So, parchment paper is effective in separating colloidal particles from a solution.

Question 10. State the imporatnt properties of a colloidal solution.

Answer:

The Important Properties Of A Colloidal Solution:-

A colloidal solution has the following properties

1. Heterogeneity:

Colloidal solutions are heterogeneous in nature. The dispersed phase neither dissolves completely in the dispersion medium nor does it separate out from the dispersion medium. The colloidal particles remain dispersed in the dispersion medium.

2. Tyndall Effect:

When a beam of light is allowed to pass through a colloidal solution, scattering of light by the colloidal particles occurs and the path of the beam through the solution gets illuminated and clearly visible. This phenomenon is called Tyndall effect.

3. Brownian Motion:

The colloidal particles in the dispersion medium are in continuous random motion moving in zigzag paths. This motion is known as Brownian motion.

4. Electrophoresis:

The colloidal particles are charged particles. If electricity is passed through a colloidal solution, then the colloidal particles move towards the oppositely charged electrodes. This movement of colloidal particles towards a specific electrode, under the influence of an electric field is called electrophoresis.

5. Passage Through Filter Paper And Parchment Paper:

Colloidal particles can pass through filter paper but cannot pass through parchment paper.

Question 11. What are the diameters of the solute particles in true solutions, colloidal solutions and suspensions?

Answer:

In true solutions, diameter of solute particles is less than or equal to 10-8 cm or 0.1 nm.

In colloidal solutions, the diameter of solute particles ranges from 10-7 -10-5 cm or 1-100 nm. In suspensions, the diameter of solute particles is greater than 10-5 cm or 100 nm.

Lyophilic And Lyophobic Sol:

If the dispersed phase of a sol has high affinity for the molecules of dispersion medium, then it is called a lyophilic sol. Sols of starch, gum, gelatin, glue etc. are examples of such sol. If the dispersion medium of such sols is water, then these are called hydrophilic sols.

If the dispersed phase of a sol has a very little or almost no affinity for the molecules of the dispersion medium, then it is called a lyophobic sol. Some common examples are ferric hydroxide [Fe(OH)3] sol, arsenius sulphide [As2S3] sol, etc. If water is the dispersion medium, then such sols are called hydrophobic sols.

Question 12. Discuss the significance of Brownian motion.

Answer:

The Significance Of Brownian Motion Are As Follows:-

  1. Brownian motion is an inherent evidence of the incessant motion of particles in a solution.
  2. The value of Avogadro’s number can be determined by measuring Brownian motion.
  3. Due to Brownian motion, the colloidal particles are in continuous motion. This motion prevents the particles to settle down even under gravitational force. Thus, Brownian motion plays a very significant role in stabilising a colloidal solution.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Solution Topic A Brownian Motion

Question 13. What is a suspension?

Answer:

Suspension:-

A heterogeneous and unstable system in which particles of a substance (usually a solid) of diameter greater than 10-5 cm remain suspended in another substance (usually a liquid) is called a suspension. On standing, the particles slowly separate out from the mixture and settle down at the bottom of the vessel.

Example:

When finely powdered barium sulphate (BaSO4) is shaken vigorously with water, it is observed that the particles get evenly distributed throughout the solvent and remain suspended in water to form a suspension. If the beaker is kept undisturbed for some time, then barium sulphate particles begin to settle down at the bottom of the beaker.

Question 14. State the properties of a suspension.

Answer:

Properties Of A Suspension:-

A suspension has the following properties

  1. A suspension is a heterogeneous mixture.
  2. When kept undisturbed for some time, the suspended particles gradually settle down under gravity.
  3. The solution is turbid and hence light cannot pass through a suspension. However, a suspension may sometimes exhibit the Tyndall effect.
  4. The suspended particles are visible under ordinary microscopes and sometimes even to the naked eyes.
  5. The suspended particles cannot pass through a parchment paper or a filter paper.

Question 15. Colloidal particles do not settle down at the bottom of the beaker, but particles of suspensions do. Explain with reason.

Answer:

Due to Brownian motion of the colloidal particles, they are always in a continuous random motion. The possibility of aggregation of colloidal particles is thus very low. So, colloidal particles do not settle down at the bottom of the beaker.

On the other hand, particles of a suspension do not exhibit Brownian motion. Hence, on standing, the particles come closer to each other and aggregate to form larger particles which eventually settle down under gravity.

Question 16. Explain why turbid water collected from ponds or rivers do not become clear on prolong standing.

Answer:

Mud particles act as colloid particles in the turbid water of ponds or rivers. These colloid particles can not be precipitated out naturally. That is why those water do not become clear on prolong standing.

Question 17. Compare the properties of a true solution, a colloidal solution and a suspension.

Answer:

The properties of a true solution, a colloidal solution and a suspension are compared in the following table

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Solution Topic A Properties Of True Solution A Colloidal Solution And Suspension

Question 18. What is the Tyndall effect?

Answer:

Tyndall Effect:-

When a beam of light is allowed to pass through a colloidal solution, scattering of light by the colloidal particles occurs and the illuminated path of light can be visualised from a direction at right angles to the incident light. This phenomenon is called Tyndall effect and it has been named after the Irish physicist John Tyndall.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Solution Topic A Tyndall Effect

Question 19. Mention some practical applications of Tyndall effect.

Answer:

Some Practical Applications Of Tyndall Effect:-

  1. Ultramicroscope was developed on the basis of Tyndall effect. It is used to determine the size of particles in aerosols by using the principle of Tyndall effect.
  2. Tyndall effect is used to determine the presence of particulate matter (which destroys the efficiency of catalyst) in air during the reaction of sulphur dioxide and air in the preparation of sulphuric acid by contact process.

Question 20. Cigarette smoke sometimes light blue in colour. Why?

Answer:

Cigarette Smoke Sometimes Light Blue In Colour Because:-

Cigarette smoke is a colloidal system which belongs to the category of solid aerosol. It consists of fine particles of carbon dispersed in air (dispersion medium).

Due to its colloidal nature, it can scatter light beam passing through it, which is known as Tyndall effect. As blue light is scattered to the maximum extent, cigarette smoke sometimes appears light blue in color.

Question 21. Classify colloids on the basis of the physical states of dispersed phase and dispersion medium and give examples for each category.

Answer:

On the basis of the physical states of the dispersed phase and dispersion medium, colloids can be classified into eight categories as described below

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Solution Topic A Colloids On the Basis of Physical States Of Dispersed Phase And Medium And Examples

Question 22. Why do colloidal solutions formed by different methods exhibit different colours?

Answer:

The colour of a colloidal solution depends on the wavelength of visible light scattered by the colloid particles. The scattering of light, on the other hand, depends on the size of colloidal particles in the solution.

The size of colloidal particles formed is different for different methods. So, a colloidal solution exhibits different colours when formed by different methods.

Question 23. What is a sol? Give example.

Answer:

Sol:-

Colloidal solutions in which the dispersed phase is a solid while the dispersion medium is a liquid is called a sol.

Example: Gold sol, sulphur sol, arsenius sulphide sol etc. In each of these sols, gold, sulphur and arsenius sulphide particles respectively form the dispersed phase while water acts as the dispersibn medium.

Question 24. What is an aerosol? How are they categorised?

Answer:

Aerosol:-

An aerosol is a colloidal system in which the dispersion medium is a gas or air. p If the dispersed phase in an aerosol is solid, then it is called solid aerosol such as smoke, dust particles floating in air etc. On the other hand, if the dispersed phase of an aerosol is liquid, then it is called liquid aerosol such as fog, cloud etc.

Question 25. What are alcohol and hydrosol?

Answer:

Alcohol And Hydrosol:-

Colloids with alcohol as dispersion medium and having solids as dispersed phase are termed as alcohols.

Again, colloids whose dispersed phases and dispersion medium are solid and water respectively are termed as hydrosol.

Question 26. State some important applications of emulsions.

Answer:

Some Important Applications Of Emulsions Are Given Below:-

  1. Milk, butter, margarine, vanishing cream, and cold cream used in our everyday life are all emulsions.
  2. Cleansing action of soap is due to the emulsification of grease with water. Water and soap together forms a colloidal solution and removes grease along with dirt and dust from clothes by forming emulsion.
  3. A wide variety of medicines like cod liver oil, and vitamin B-complex are oil-in-water type emulsions. These are easily absorbed by our digestive system and provide quick relief.

Question 27. Classify emulsions and define each of them with examples.

Answer:

Emulsions Are Of Two Types:-

  1. Oil-in-water type emulsion and
  2. Water-in-oil type emulsion.

Here, the term oil represents any liquid that is immiscible with water.

1. Oil-In-Water Type Emulsion:

When small quantity of oil is agitated with an excess amount of water, an oil-in-water type emulsion is formed.

Here, water is the dispersion medium and oil is the dispersed phase.

Example: Milk, vanishing cream etc. When small amount of nitrobenzene is agitated with water, this type of emulsion is produced.

2. Water-In-Oil Type Emulsion:

When small quantity of water is agitated with an excess amount of oil, a water-in-oil type emulsion is formed. Here, oil is the dispersion medium and water is the dispersed phase.

Examples: Butter, cod liver oil, cold cream etc.

Question 28. Write the differences between oil-in-water type emulsions and water-in-oil type emulsions.

Answer:

The major differences between oil-in-water type emulsions and water-in-oil type emulsions are given below

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Solution Topic A Differences Between Oil In Water And Water In Oil Type Emulsion

Question 29. What is Brownian motion? How does this motion arise?

Answer:

Brownian Motion:-

  1. The continuous, random motion of particles of the dispersed phase in a colloidal solution in zigzag paths is known as Brownian motion.
  2. Brownian motion arises due to random collisions of the particles of the dispersion medium with the colloidal particles in dispersed phase.

Question 30. What is an emulsifier? Give example.

Answer:

Emulsifier:-

While preparing an emulsion of two immiscible liquids, a third substance is generally added to the mixture. The mixture is then vigorously agitated. The third substance which is added in small quantity to stabilise the emulsion is known as an emulsifier or emulsifying agent.

Example: In case of milk which is a colloid (emulsion), a protein named casein acts as the emulsifier.

Question 31. How will you prepare colloidal sulphur?

Answer:

Preparation Of Colloidal Sulphur:-

Dilute hydrochloric acid reacts with sodium thiosulphate to produce sulphur. When the concentration of sulphur in the solution reaches a certain level, the sulphur particles instead of getting precipitated, form colloidal particles and remain in the colloidal solution as the dispersed phase.

\(\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3+2 \mathrm{HCl} \rightarrow 2 \mathrm{NaCl}+\mathrm{S}+\mathrm{SO}_2+\mathrm{H}_2 \mathrm{O}\)

This reaction is called clock reaction.

Question 32. Name a substance that is used as an emulsifying agent. Mention its role as an emulsifying agent in our daily life.

Answer:

  1. Soaps, detergents etc. act as emulsifying agents.
  2. Soaps and detergents help in the formation of emulsion between water and grease or oil. When soap or sodium stearate is dissolved in water, it dissociates to form a stearate ion.

The ion has two different parts—the long hydrocarbon part (C17H35 —) which is hydrophobic and the ionic part ( — COO) which is hydrophilic. Thus, the hydrocarbon chain dissolves in oil or grease of the clothes while the ionic part dissolves in water.

In this way, an emulsion is formed and oil or grease, along with dirt and dust is easily removed from clothes by the action of soap.

Question 33. Water from a tube well sometimes becomes turbid and brownish on standing. Explain with reason.

Answer:

Water From A Tube Well Sometimes Becomes Turbid And Brownish On Standing:-

Water collected from tube wells sometimes contains ferrous bicarbonate [Fe(HCO3)2] salt dissolved in it. This compound is oxidised by aerial oxygen to produce brown ferric hydroxide [Fe(OH)3]. This ferric hydroxide forms colloidal solution with water. As a result, the water becomes turbid and brownish on standing.

4Fe(HCO3)2 + 2H2O + O2 → 4Fe(OH)↓+ 8CO2

Question 34. Proteins, DNA, starch etc. form colloidal solution with water. Why?

Answer:

Proteins, starch, DNA etc. are polymeric biomolecules. These are large molecules known as macromolecules. The diameter of the molecules of these substances is larger than that of the solute particles of a true solution (diameter ≤ 10-8 cm).

Hence, these molecules cannot occupy the intermolecular spaces between water molecules. So, these substances cannot form true solutions with water.

However, macromolecules have great affinity towards water molecules and when these substances are added to water, they do not separate out from the mixture. The dimension of the particles is in the range of ≤10-7 – ≤10-5 cm. Therefore, they form a colloidal solution with water.

Chapter 4 Matter Solution Topic A True Solution Colloid And Suspension Dissolution Of Smaller Ion Or Molecules And Larger Molecules In Water Very Short Answer Type Questions Choose The Correct Answer

Question 1. Which of the following is a true solution?

  1. Aqueous solution of sugar
  2. Cod liver oil
  3. Mixture of sand and water
  4. Milk

Answer: 1. Aqueous solution of sugar

Question 2. Which of the following is a colloidal solution?

  1. Aqueous solution of sugar
  2. Mixture of BaS04 and water
  3. Mixture of sand and water
  4. Milk

Answer: 4. Milk

Question 3. Which of the following is an emulsion?

  1. Curd
  2. Milk
  3. Milk of magnesia
  4. Soda water

Answer: 2. Milk

Question 4. The solution in which scattering of light can be observed is

  1. Blood
  2. Mixture of sand and water
  3. Aqueous solution of copper sulphate
  4. Aqueous solution of sodium chloride

Answer: 1. Blood

Question 5. The correct order of stability is

  1. Suspension < colloidal solution < true solution
  2. Colloidal solution < true solution < suspension
  3. True solution < colloidal solution < suspension
  4. Colloidal solution < suspension < true solution

Answer: 1. Suspension < colloidal solution < true solution

Question 6. For which of the following solutions, the solute particles can pass through parchment paper?

  1. Milk
  2. Cod liver oil
  3. Aqueous solution of sugar
  4. Sand mixed with water

Answer: 3. Aqueous solution of sugar

Question 7. Which of the following solutions is homogeneous in nature?

  1. Milk
  2. Cod liver oil
  3. Aqueous solution of sugar
  4. Mixture of sand and water

Answer: 3. Aqueous solution of sugar

Question 8. Which of the following acts as dialyzer in human body?

  1. Lungs
  2. Kidneys
  3. Liver
  4. Stomach

Answer: 2. Kidneys

Question 9. In which of the following, the dispersion medium is not a liquid?

  1. Fog
  2. Foam
  3. Sulphur sol
  4. Cream

Answer: 1. Fog

Question 10. Milk is a type of

  1. Gel
  2. Foam
  3. Oil-in-water type emulsion
  4. Water-in-oil type emulsion

Answer: 3. Oil-in-water type emulsion

Question 11. An example of a solid sol is

  1. Metal alloy
  2. Gold sol
  3. Paneer
  4. Pumice stone

Answer: 1. Metal alloy

Question 12. An example of gel is

  1. Milk of magnesia
  2. Paneer
  3. Cream
  4. Lather of soap

Answer: 2. Paneer

Question 13. A colloidal solution of a liquid dispersed in another liquid is known as

  1. Sol
  2. Gel
  3. Foam
  4. Emulsion

Answer: 4. Sol

Question 14. The easiest way to identify a colloidal solution is by

  1. Observing Tyndall effect
  2. Observing Brownian motion
  3. Electrodialysis

Answer: 1. Observing Tyndall effect

Question 15. The molecules of which of the following substances will not occupy the intermolecular spaces between water molecules when it is dissolved in water?

  1. Ethanol
  2. Sugar
  3. Glycerin
  4. Protein

Answer: 4. Protein

Question 16. In presence of colloidal particles, light rays get

  1. Reflected
  2. Refracted
  3. Scattered
  4. Deviated

Answer: 3. Scattered

Question 17. For which of the following dispersion medium is not liquid?

  1. Fog
  2. Lather
  3. Sulphur sol
  4. Cream

Answer: 1. Fog

Question 18. Descending order of the size of particles of true solution (A), Colloidal sol (B) and suspension (C) is

  1. A >B > C
  2. B > A > C
  3. C >B > A
  4. A < B > C

Answer: 3. C >B > A

Question 19. Particle with which diameter, can fit in the intermolecular space of water

  1. 10-9 cm
  2. 10-6 cm
  3. 10-5 cm
  4. 10-4 cm

Answer: 1. 10-9 cm

Question 20. Example of liquid aerosol

  1. Cloud
  2. Smoke
  3. Milk
  4. Cream

Answer: 1. Cloud

Question 21. For which of the following dispersed phase and dispersion medium are liquid and gas respectively?

  1. Emulsion
  2. Sol
  3. Gel
  4. Liquid Aerosol

Answer: 4. Liquid Aerosol

Question 22. Example of emulsifier

  1. Arrowroot
  2. Gelatine
  3. Dettol
  4. Cod liver oil

Answer: 2. Gelatine

Question 23. Which one is suspension?

  1. Milk
  2. Milk of magnesia
  3. Cake
  4. Shaving cream

Answer: 4. Shaving cream

Question 24. Which one is not an example of emulsion?

  1. Butter
  2. Shampoo
  3. Cake
  4. Shaving cream

Answer: 4. Shaving cream

Chapter 4 Matter Solution Topic A True Solution Colloid And Suspension Dissolution Of Smaller Ion Or Molecules And Larger Molecules In Water Answer In Brief

Question 1. The diameter of a particle ‘O’ present in water is 80 nm. What is the nature of the solution?

Answer: Q will form a colloidal solution.

Question 2. Name a colloidal solution which is used in our daily life.

Answer: Milk.

Question 3. Mention the type of solutions formed when sulphur is separately dissolved in water and alcohol.

Answer: Sulphur remains dispersed in water to form a colloidal solution. On the other hand, it completely dissolves in alcohol to form a true solution.

Question 4. Give some examples of solid sol.

Answer: Coloured glass, ruby glass (Au/glass), gemstones, metal alloys etc.

Question 5. The particulate matter in air belongs to which type of colloid?

Answer: Solid aerosol.

Question 6. What is meant by gel?

Answer: Gel is a colloidal solution in which a liquid is dispersed in a solid. So, the dispersion medium is solid and dispersed phase is liquid. Jelly, gelatin etc. are some examples of gel.

Question 7. Give some examples of liquid aerosol.

Answer: Fog and cloud are some examples of liquid aerosol.

Question 8. Give an example of a colloid in which the dispersion medium is solid and the dispersed phase is gas.

Answer: The colloids in which the dispersion medium is solid and the dispersed phase is gas are known as solid foams. Some common examples are, cake, pumice stone (air dispersed in silicate compounds) etc.

Question 9. Give some examples of hydrophilic colloids.

Answer: Starch, gelatin, protein, cellulose, soap etc. are some examples of hydrophilic colloids.

Question 10. Give some examples of hydrophobic colloids.

Answer: Silver sol, gold sol, arsenius sulphide sol, ferric hydroxide sol etc. are some examples of hydrophobic colloids.

Question 11. Which substance acts as the emulsifying agent in milk?

Answer: In milk, casein (which is a protein) acts as the emulsifying agent.

Question 12. What is the type of colloid in which both the dispersion medium and dispersed phase are liquid known as?

Answer: The colloids in which both the dispersion medium and dispersed phase are liquid are known as emulsions.

Question 13. Name some covalent compounds which are soluble in water.

Answer: Some covalent compounds which are soluble in water are sugar, alcohol, HCl etc.

Question 14. Name a water soluble organic compound.

Answer: Sugar is a water-soluble organic compound.

Question 15. Name an electrovalent compound which is insoluble in water.

Answer: An electrovalent compound which is insoluble in water is barium sulphate (BaSO4).

Question 16. Give an example of a colloidal solution in which the colloidal particles are large molecules.

Answer: Starch remains dispersed in water to form a colloidal solution. Here, large starch molecules are present as colloidal particles.

Question 17. What is the diameter of solute particles in a true solution?

Answer: ≤ 10-8 cm.

Question 18. What is the diameter of solute particles in a colloidal solution?

Answer: 10-7-10-5 cm.

Question 19. What is the diameter of solute particles in a suspension?

Answer: 10-5 cm.

Question 20. Which method is used for the separation of crystalloids from colloids?

Answer: electrodialysis.

Question 21. Name the random, continuous motion exhibited by colloidal particles in a colloidal solution

Answer: Brownian motion.

Question 22. Why gas mixtures cannot form colloidal systems?

Answer: Gas mixtures always form homogeneous mixtures irrespective of the ratio in which they are mixed. Thus, they cannot form colloidal systems.

Question 23. The suspended particulate matters belong to which type of colloids?

Answer: SPM belong to the solid aerosol class of colloid.

Question 24. By which property colloids and True solutions can be distinguished?

Answer: Tyndall Effect.

Question 25. Name an emulsifier used in daily life.

Answer: Soap or detergent.

Question 26. Name the process by which crystalloid can be separated from colloid using diffusion through a semi-permeable membrane.

Answer: Dialysis.

Question 27. Which solution is used with HSO to prepare colloidal Sulphur?

Answer: Hypo solution or the dilute solution of Sodium thiosulphate.

Question 28. Name two diseases caused due to solid aerosol.

Answer: Asthma and silicasis.

Question 29. Which protective colloid is used in icecream?

Answer: Gelatine.

Question 30. Given an example of a solution where salute and solvent both are solid.

Answer: Brass.

Question 31. Give an example of a colloid which is an essential component of our body.

Answer: Blood (colloidal solution of Albumin).

Question 32. Give an example of a colloidal system where finer particles of water are dispersed in fat.

Answer: Butter.

Question 33. Mention the dispersion phase in O/W type Emulsion.

Answer: Oil.

Question 34. Give an example of colloid causing air pollution.

Answer: Solid aerosol.

Question 35. Why Gelatine is used to prepare ice cream?

Answer: Gelatine is used to produce ice cream to give the stability of the colloidal system (ice cream).

Question 36. Give an example of a colloidal system where the dispersed phase and the dispersion medium are gas and liquid respectively.

Answer: Soda-water.

Question 37. Which type of colloid is the soap-lather? Mention the nature of dispersed phase and dispersion medium of soap lather.

Answer:

  1. Soap-lather belong to the ‘Foam’ family of colloid.
  2. The dispersed phase and dispersion medium here are of gaseous and liquid respectively.

Question 38. Which type of solution will formed by dissolving starch in water?

Answer: Colloidal solution.

Question 39. Which type of solution is formed by dissolving barley in hot water.

Answer: Colloidal solution.

Question 40. Name the property of scattering of light by colloid particles.

Answer: Tyndall effect.

Question 41. Give an example of a colloid whose dispersed phase and dispersion medium, both are solid.

Answer: Curd.

Question 42. Give an example of a colloid which is used as disinfectant.

Answer: Dettol.

Question 43. How solute and solvent are termed in colloidal system?

Answer: Solute and solvent are termed as dispersed phase and dispersion medium respectively in colloidal system.

Question 44. Name the solvent and solutes of air.

Answer: Solvent of air is nitrogen and oxygen, carbon dioxide and other gases are the solutes of air.

Question 45. Name the solutes and solvents in—brass and sodium amulgum.

Answer: In brass the solvent is copper and the solute is zinc.ln sodium amulgum solute and solvents are mercury and sodium respectively.

Chapter 4 Matter Solution Topic A True Solution Colloid And Suspension Dissolution Of Smaller Ion Or Molecules And Larger Molecules In Water Fill In The Blanks

Question 1. True solutions are _______ in nature.

Answer: Homogeneous

Question 2. A Colloidal solutions are ____ in nature.

Answer: Heterogeneous

Question 3. Milk is a colloid in which ______ is the dispersed phase and ______ is the dispersion medium.

Answer: Far, water

Question 4. ________ solutions show Tyndall effect.

Answer: Colloidal

Question 5. The colloidal particles in a colloidal solution can be observed by using ______

Answer: Ultramicroscope

Question 6. When sodium chloride is dissolved in water, it forms a ________ solution.

Answer: True

Question 7. When starch is dissolved in water, it forms a _________ solution.

Answer: Colloidal

Question 8. A ________ is a is a colloidal solution in which the dispersed phase is solid.

Answer: Sol

Question 9. A _________ is a colloidal system in which the dispersed phase is liquid and the dispersion medium is solid.

Answer: Gel

Question 10. An example of solid foam is ________

Answer: Cake

Question 11. A colloidal system is known as _______ if the dispersed phase is a gas and the dispersion medium is solid.

Answer: Solid foam

Question 12. Dispersed phase and dispersion medium, in case of pumice stone, are _______ and _______ respectively.

Answer: Gas, solid

Question 13. The solutes in case of colloidal solution are termed as _______

Answer: Dispersed phase

Question 14. Dispersed phase and dispersion medium for emulsions are _______ and _______ respectively.

Answer: Liquid, liquid

Chapter 4 Matter Solution Topic A True Solution Colloid And Suspension Dissolution Of Smaller Ion Or Molecules And Larger Molecules In Water State Whether True Or False

Question 1. The component of a solution, which is present in a lesser amount is known as the solute.

Answer: True

Question 2. The amount of CaCI2 present in 500 mL of 25% CaCI2 solution is 125 g.

Answer: True

Question 3. The solute particles of a colloidal solution can pass through parchment paper.

Answer: False

Question 4. The solute particles of a true solution are visible under an ultramicroscope.

Answer: False

Question 5. A true solution does not exhibit Tyndall effect.

Answer: True

Question 6. If the diameter of solute particles is 10-4 cm, the corresponding solution is a true solution.

Answer: False

Question 7. Colloidal system whose dispersed phase and dispersion medium both are gaseous, does not exist.

Answer: True

Question 8. Particles of a true solution can not be observed even with the help of a powerful microscope.

Answer: True

Question 9. Example of water-in-oil type emulsion is butter.

Answer: True

Question 10. Colloidal solution are homogeneous.

Answer: False

Question 11. Colloidal solution are transparent.

Answer: False

Chapter 4 Matter Solution Topic A True Solution Colloid And Suspension Dissolution Of Smaller Ion Or Molecules And Larger Molecules In Water Numerical Examples

Question 1. 8 g of NaOH is dissolved in 2 L of its aqueous solution, Calculate the strength of the solution in mole/litre unit.

Answer:

Molecular mass of NaOH = (23 + 16 + 1) = 40

mass of 1 mol NaOH = 40 g

∴ 8g NaOH = 8/40 mol =0.2 mol

∴ Amount of NaOH in 2 L solution =0.2 mol

∴ Amount of NaOH in 1 L solution = 0.2/2 mol = 0.1 mol

∴ Strength of the solution =0.1 mol/L

Question 2. 300 g saturated solution of KNO3 at 100°C is cooled down to 10°C. Calculate the amount of salt that should be crystallised out from the solution (Solubility of KNO3 at 100°C and 10°C are 250 and 20 respectively).

Answer:

At 100°C, amount of KNO3 dissolved in 100 g H2O = 250 g

∴ (100 + 250)g saturated solution contains 100 g solvent

∴ 300 g saturated solution contains \(\frac{100 \times 300}{350}\)g solvent = 85.71 g solvent

∴ Amount of solute in 300 g solution (300 – 85.71)g = 214.29g

Now at 10°C,

Amount of KNO3 that can be dissolved in 100 g solvent = 20 g

∴ Amount of KNO3 that can be dissolved in 85.71 g solvent = \(\frac{20 \times 85.71}{100} \mathrm{~g}\) g = 17.142 g

∴ Amount of KNO3 that should be crystallised out = (214.29 – 17.142)g = 197.148g

Question 3. 4 g of NaOH (molar mass 40) is dissolved in 500 ml of a solution. Calculate the strength of the solution in

  1. g • L-1,
  2. % strength (w/v) and
  3. mol • L-1.

Answer:

1. Amount of NaOH in 500 ml solution = 4 g

∴ Amount of NaOH in 1000 ml or 1 L

= \(\frac{4 \times 1000}{500} \mathrm{~g}\) = 8g

∴ Strength of solution = 8 g • L-1

2. Amount of NaOH in 500 ml solution = 4 g

∴ Amount of NaOH in 100 ml solution

= \(\frac{4 \times 100}{500} g\) = 0.8 g

∴ Strength of the solution is 0.8% (w/v)

3. Again, 4 g NaOH = 4/40 mol = 0.1 mol

∴ Strength of the solution = 0.1/500 x 1000 = 0.2 mol L-1

Question 4. Strength of a glucose solution is 0.8 mol • L-1. Express the strength in % (w/v).

Answer:

Molar mass of glucose is = 180.

1 L or 1000 ml. solution contains 0.8 mol glucose = 0.8 x 180 g = 144 g of glucose

∴ 100 mol of solution contains = \(\frac{144 \times 100}{1000}\) g of glucose = 14.4 g glucose.

∴ % strength (w/v) of the solution is 14.4%.

Question 5. Calculate the amount of salt required to prepare 200 ml 2% sodium bicarbonate solution. Express the strength in gram per litre.

Answer:

100 ml of 2% solution contains 2 g of sodium bicarbonate.

∴ Amount of salt in 200 ml solution = 2/100 x 200 g = 4g

∴ Amount of salt in 1000 ml solution = 4/100 X 1000 = 20 g

Strength of the solution is 20 g • L-1.

Question 6. Among 120 ml 10 % NaCI solution and 120 ml, 1 mol • L-1 NaCI solution, which contains a greater amount of NaCI?

Answer:

Amount of NaCI present in 120 ml of 10% NaCI solution = 10/100 x 120 g = 12 g

Amount NaCI present in 120 ml of 1 mol • L-1 solution = \(\frac{(23+35.5) \times 120}{1000}\) g = 7.02 g

∴ 120 ml 18% NaCI solution contains greater amount of sodium chloride.

Chapter 4 Matter Solution Topic B Solubility Strength Of Solution And Its Unit Synopsis

  1. Solubility of a substance at a given temperature is the maximum number of gram units of the substance that can be dissolved in 100g of a solvent at that temperature.
  2. The curve obtained by plotting the solubility of a substance in a given solvent along the y-axis against temperature along x-axis is called the solubility curve of that solute.
  3. Solubility of solids in liquids generaily increases with increase in temperature.
  4. Example: KNO3, NaCI, NH4CI. sugar, blue vitriol etc.
  5. Solubility of CaSO4, Ca(OH)2, etc. in water decreases with incease in temperature.
  6. Solubility of gases in liquid decreases with increase in temperature and increases with decrease in temperature.
  7. At constant temperature solubility of gaseous solutes in liquid solvents increases with increase in pressure and decreases with decreases in pressure.

Chapter 4 Matter Solution Topic B Solubility Strength Of Solution And Its Unit Short And Long Answer Type Questions

Question 1. Why It necessary to mention temperature in the definition of solubility?

Answer:

The solubility of a substance in a given solvent varies with temperature. Hence, the amount of a particular solute required to saturate 100 g of a solvent will be different at different temperatures. Thus, it is necessary to mention temperature in the definition of solubility.

Question 2. Why is it necessary to mention a definite amont of solvent in the definition of solubility?

Answer:

At a particular temperature, the amount of solute required to make a saturated solution depends on the amount of solvent. If greater amount of solvent is taken, then it can dissolve more amount of solute whereas, lesser amount of solvent dissolves lesser amount of solute.

Hence, we cannot determine the amount of solute required to make a saturated solution, if the amount of solvent is not specified. Thus, it is necessary to specify the amount of solvent in the definition of solubility.

Question 3. What are the units used to express the strength of a solution? Comment on their dependence on temperature.

Answer:

Different units used to express the concentration or strength of a solution are volume percentage (% W/V), moles per litre or molarity (mol • L-1) and gram per litre (g • L-1).

Although mass or number of moles of a substance is independent of temperature, volume of a solution changes with temperature. As all the units are dependent on volume of the solution, these are also dependent on temperature.

Question 4. What is meant by molarity or moles per litre strength of a solution? Give example.

Answer:

At a given temperature, the number of gram-moles of a solute dissolved in 1 litre (or 1000mL) of a solution is known as the molarity or moles per litre strength of the solution.

Example: 5 gram-moles of HN03 is dissolved in 1 litre aqueous solution. Hence, the molarity of the solution will be 5. The molar strength is expressed as mol • L-1 or (M).

Question 5. What is meant by gram per litre strength of a solution? Give example.

Answer:

The amount of solute in gram dissolved per litre of the solution is expressed in gram per litre unit.

Example: If 5.0g of sodium chloride is dissolved in 1 L of solution, then the strength of the sodium chloride solution in gram per litre unit will be 5. The unit is expressed as g • L-1.

Question 6. What changes will be observed in the (W/V) concentration of a solution if the temperature of the solution is raised?

Answer:

The following changes will be observed in the (W/V) concentration of a solution if the temperature of the solution is raised

  1. With the increase in temperature, the volume of the solution increases. Hence, the ratio of mass of solute to the volume of solution decreases. As a result, concentration of the solution also decreases.
  2. If some amount of solvent vapourises due to increase in temperature, then the volume of solution also decreases. Hence, the concentration of the solution also increases.

Question 7. Why molarity depends on temperature?

Answer:

We know molarity of solution = \(\frac{\text { number of } \mathrm{g} \cdot \text { mol of solute }}{\text { volume in } \mathrm{L} \text { of the solution }}\)

Now volume of the solution changes with temperature. Thus molarity of the solution also changes with temperature, although the number of g • mol of the solute remains unaltered. Hence molarity depends on temperature.

Question 8. If the amount of the dissolved solute in a gram of saturated solution is b gram, calculate the solubility of the solute at that temperature.

Answer:

We know, solubility of a solute at t°C = \(\frac{\text { mass of the solute in saturated solution }}{\text { mass of the solvent in-the solution }} \times 100\)

= \(\frac{b \mathrm{~g}}{(a-b) g} \times 100=\frac{100 b}{(a-b)}\)

Question 9. What are the factors on which the solubility of a substance depends?

Answer:

The solubility of a substance in a given solvent depends on the following factors

1. Nature of solute:

Different solutes dissolve to various extents in a given solvent at a particular temperature.

2. Temperature:

For most solid substances, solubility in a given liquid increases with the rise in temperature. However, there are some gases whose solubility in a given liquid decreases with the rise in temperature.

3. Atmospheric pressure:

At a particular temperature, the solubility of a gas in a liquid generally increases with the increase in pressure above the liquid and vice-versa. This is why C02 dissolved in water under very high pressure comes out as bubbles when a sodawater bottle is opened.

Question 10. What is a solubility curve? Draw the solubility curve of Glauber’s salt in water and explain its solubility with the help of the curve.

Answer:

The curve obtained by plotting the solubility of a solute in a given solvent along the Y-axis against temperature along X-axis, is called the solubility curve of that solute. This curve easily demonstrates the relation between temperature and solubility of a substance.

The solubility curve of Glauber’s salt in water (hydrated sodium sulphate) rapidly rises up with the rise in temperature upto 32.38°C. After 32.38°C, if the temperature is further increased, the solubility gradually starts decreasing.

Above 32.38°C, hydrated crystals of sodium sulphate (Na2SO4 – 10H2O) give up ten water molecules to form anhydrous sodium sulphate (Na2SO4). So, above 32.38°C, the solubility curve actually represents the solubility of anhydrous sodium sulphate.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Solution Topic B Solubility Curve Of Glauber's Salt

Question 11. Draw the solubility curves of KNO3 and NaCI in water and mention the effect of temperature on solubility of these salts.

Answer:

The solubility curve of KNO3 rises steeply with rise in temperature. This means that the solubility of KNO3 increases rapidly with the increase in temperature.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Solution Topic B Solubility Curve Of KNO3 And Nacl

The solubility curve of NaCI is almost parallel to the X-axis. This indicates that the solubility of NaCI in water remains almost unchanged over a wide range of temperature.

Question 12. How does the solubility of Ca(OH)2 in water changes with the rise in temperature? Explain with the help of its solubility curve.

Answer:

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Solution Topic B Solubility Curve Of Ca(OH)2

From the solubility curve, it can be said that the solubility of Ca(OH)2 in water decreases with the rise in temperature.

Question 13. State the significance of solubility curve.

Answer:

  1. Without actually performing an experiment, the solubility of a substance at different temperatures can be determined with the help of the solubility curve.
  2. It helps to compare the solubilities of different substances at a given temperature.
  3. The solubility curve helps us to determine which substance will preferentially crystallise out if a solution of a mixture of substances is subjected to cooling or evaporation.

Question 14. Discuss the effect of temperature on the solubility of a gas in a liquid.

Answer:

In general, the solubility of a gas in a liquid decreases with the rise in temperature. For example, CO2 is soluble in water. At ordinary temperature, some amount of C02 gas dissolves in water. However on heating, the dissolved CO2 fizzles out from the solution.

Question 15. Discuss the effect of pressure on the solubility of a gas in a liquid.

Answer:

At a particular temperature, the solubility of a gas in a liquid generally increases with the increase in pressure and decreases with the decrease in pressure. So, at a higher pressure relatively large amount of gas dissolves in a liquid. This principle is used in the preparation of soda water where excess CO2 is dissolved in water by applying pressure.

Question 16. Why does soda water fizz when the cap of the bottle is removed?

Answer:

At a particular temperature, the solubility of a gas in a liquid increases with increase in pressure. In soda water bottles, carbon dioxide is dissolved in water applying high pressure.

When the cap of the bottle is removed, the pressure suddenly decreases and as a result the solubility of CO2 in water decreases. Consequently, excess carbon dioxide comes out from water and eventually causes the fizz.

Question 17. A bottle containing liquor ammonia must be cooled before opening. Why?

Answer:

A saturated aqueous solution of ammonia is called liquor ammonia. With rise in temperature, solubility of NH3 in water decreases and hence some NH3 gas comes out of the solution and causes excess pressure inside the bottle.

If the bottle is opened under this condition, then excess NH3 gas along with some dissolved NH3 gas may spurt out from the bottle causing serious accident. However, if the bottle is cooled down, the excess NH3 again dissolves in water and no such accident occurs when the bottle is opened.

Question 18. Why bubbles are formed when water is heated?

Answer:

Some amount of air generally remains dissolved in water at room temperature. Now solubility of gases in liquid decreases with increase in temperature. That is why solubility of air decreases when we heat the water and hence the excess air comes out of water forming bubbles.

Chapter 4 Matter Solution Topic B Solubility Strength Of Solution And Its Unit Very Short Answer Type Questions Choose The Correct Answer

Question 1. HCI (g) dissolves in water, because HCI

  1. Is a gaseous substance
  2. Reacts with water
  3. Is an ionic compound
  4. Is a polar covalent compound

Answer: 4. Is a polar covalent compound

Question 2. Water can dissolve ionic compounds, because water is a/an

  1. Covalent compound
  2. Ionic compound
  3. Polar covalent compound
  4. Liquid

Answer: 3. Polar covalent compound

Question 3. The temperature upto which the solubility curve for the solubility of Glauber’s salt in water rises is

  1. 25.5°C
  2. 32.4°C
  3. 35.2°C
  4. 37.24°C

Answer: 2. 32.4°C

Question 4. At a constant temperature, the solubility of a gas in a given volume of liquid is directly proportional to the pressure of the gas. This statement is known as

  1. Boyle’s law
  2. Charles’ law
  3. Raoult’s law
  4. Henry’s law

Answer: 4. Henry’s law

Question 5. 01 Unit of solubility is

  1. g • cm-3
  2. kg • m-3
  3. g
  4. Unitless

Answer: 4. Unitless

Question 6. With rise in temperature, the solubility of Glauber’s salt (Na2SO4 • 10H2O) in water

  1. Decreases
  2. Increases
  3. Remains almost unchanged
  4. Initially increases and then decreases

Answer: 4. Initially increases and then decreases

Question 7. With rise in temperature, the solubility of potassium nitrate (KNO3) in water

  1. Decreases
  2. Increases
  3. Remains almost unchanged
  4. Initially increases and then decreases

Answer: 1. Decreases

Question 8. Which of the following units of concentration is not affected by any change in temperature?

  1. Mass-volume percentage (% W/V)
  2. Mass percentage (% W/W)
  3. Moles per litre
  4. Gram per litre

Answer: 2. Mass percentage (% W/W)

Question 9. With rise in temperature, the solubility of a gas in a liquid

  1. Increases
  2. Decreases
  3. Remains almost unchanged
  4. Initially increases and then decreases.

Answer: 2. Decreases

Question 10. 10% Na2CO3 solution means

  1. 10g Na2CO3 is dissolved in 100 mL solution
  2. 100g Na2CO3 is dissolved in 10 mL solution
  3. 10g Na2CO3 is dissolved in lOmL solution
  4. 10g Na2CO3 is dissolved in 100g solution

Answer: 1. 10g Na2CO3 is dissolved in 100 mL solution

Question 11. The amount of NaCI present in 200 mL of 50% NaCI solution is

  1. 50 g
  2. 75 g
  3. 100 g
  4. 150 g

Answer: 3. 100 g

Question 12. The concentration of a solution prepared by dissolving 40 g NaOH in 1000 mL water will be

  1. 1 mol • L-1
  2. 2 mol • L-1
  3. 3 mol • L-1
  4. 4 mol • L-1

Answer: 1. 1 mol • L-1

Question 13. If xg of a solute is required to make yg saturated solution at f°C temperature; solubility at that temperature will be

  1. x/y x 100
  2. y
  3. \(\frac{x}{y-x} \times\) x 100
  4. \(\frac{y}{x-y} \times\) x 100

Answer: 3. \(\frac{x}{y-x} \times\) x 100

Question 14. With increase in temperature, the solubility of slaked lime in water

  1. Increases
  2. Decreases
  3. Remains same
  4. Increases firstly then decreases

Answer: 2. Decreases

Question 15. Solubility does not depend upon

  1. Temperature
  2. Pressure
  3. Nature of solvent
  4. Gravitational force of the earth

Answer:  4. Gravitational force of the earth

Question 16. The gas which comes out as effervescence from a bottle of soda water when we open it is

  1. NH3
  2. H2
  3. He
  4. CO2

Answer: 4. CO2

Question 17. The solubility of which one decreases with increase in temperature?

  1. KNO3
  2. Ca(OH)2
  3. NaCI
  4. NH4CI

Answer: 2. Ca(OH)2

Question 18. Solubility of NaCI in water at 273K is 37. Amount of water required to make a saturated solution containing 100g of NaCI at that temperature will be

  1. 270.27g
  2. 245 g
  3. 222.22g
  4. 145 g

Answer: 1. 270.27 g

Question 19. Solubility of which one in water increases reading with increase in temperature?

  1. Ca(OH)2
  2. NaNO3
  3. NaCI
  4. NH4CI

Answer: 2. NaNO3

Question 20. Solubility of which one increases slightly with increase in temperature?

  1. ZnSO4
  2. CaSO4
  3. NaCI
  4. Na2SO4-10H2O

Answer: 3. NaCI

Question 21. With increase in pressure, the solubility of a solid in a liquid

  1. Increases
  2. Decreases
  3. Firstly increases and then decreases
  4. Remains same

Answer: 4. Remains same

Question 22. Amount of Na2CO3 in 100 ml of 25% Na2CO3 solution

  1. 25 g
  2. 2.5 g
  3. 225 g
  4. 250g

Answer: 4. 250g

Question 23. If we express the strength of 10% Na2CO3 solution as mol/litre, it will be

  1. 2.00
  2. 1.00
  3. 0.74
  4. 0.94

Answer: 4. 0.94

Chapter 4 Matter Solution Topic B Solubility Strength Of Solution And Its Unit Answer In Brief

Question 1. What is the unit of solubility?

Answer: Solubility is a unitless physical quantity. It has no unit.

Question 2. Name some compounds whose solubility in water decreases with the increase in temperature.

Answer: The compounds whose solubility in water decreases with the increase in temperature are calcium sulphate (CaSO4), calcium nitrate [Ca(NO3)2], slaked lime or calcium hydroxide [Ca(OH)2], cerium sulphate [Ce2(SO4)3] etc.

Question 3. Name a solid whose solubility in water remains almost constant with change in temperature.

Answer: Solubility of common salt or sodium chloride (NaCI) in water remains almost constant with change in temperature

Question 4. Name a solid whose solubility in water initially increases with the rise in temperature but starts to decrease when the temperature is further increased.

Answer: The solubility of Glauber’s salt (Na2SO4-10H2O) in water initially increases with the rise in temperature up to 32.4°C and then the solubility begins to decrease when the temperature is further increased.

Question 5. How is the solubility of a solid solute in a liquid solvent affected by pressure?

Answer: There is no effect of pressure on the solubility of a solid solute in a liquid solvent.

Question 6. State Henry’s law.

Answer: At constant temperature, the solubility of a given gas in a given volume of liquid solvent is directly proportional to the pressure of that gas.

Question 7. What will happen if a saturated solution of sugar prepared at room temperature is cooled down to 5°C?

Answer: When a saturated solution of sugar prepared at room temperature is cooled down to 5°C, the solubility of sugar decreases and some amount of sugar settles down at the bottom.

Question 8. How much NaCI is present in 50 mL of 20% aqueous solution of NaCI?

Answer: 50 mL of 20% aqueous solution of NaCI will contain 10 g of NaCI.

Question 9. What change in W/V concentration of a • solution will be observed with the rise in temperature?

Answer: If the solvent does not evaporate on heating, then the W/V concentration of a solution generally decreases if the temperature is increased.

Question 10. Why does molarity of a solution depend on temperature?

Answer: Molarity of a solution is the number of moles, of solute present per litre of a solution. As the volume of solution depends on temperature (although mass of solute is independent of temperature), the molarity of a solution also depends on temperature.

Question 11. Why the statement—“Solubility of NaCO3 in water is 20“—is erroneous?

Answer: The temperature is not mentioned here which makes the statement erroneous.

Question 12. What do you mean by the statement— Solubility curve of NaCI is almost parallel to x-axis?

Answer: It means that the solubility of NaCI in water remains almost unaltered with change in temperature.

Question 13. State whether the solubility of gas in water increases or decreases with decrease in pressure.

Answer: Solubility of gas in water decreases with decrease in pressure.

Question 14. How the solubility of Ca(OH)2 changes with increase in temperature.

Answer: Solubility of Ca(OH)2 in water decreases with increase in temperature.

Question 15. Which of the following solutions of higher concentration: 100 g/L NaOH and 10 molar NaOH.

Answer: Concentration of NaOH in 10 molar NaOH solution is higher.

Question 16. What do you mean by 15% Na2CO3 solution.

Answer: 15% Na2CO3 solution means that 15 g of Na2CO3 is dissolved in 100 ml solution.

Chapter 4 Matter Solution Topic B Solubility Strength Of Solution And Its Unit Fill In The Blanks

Question 1. If 5 g of a salt is dissolved in 50 mL water, then the strength of the solution will be _______ % (W/V).

Answer: 10

Question 2. The solubility of those substances which absorb heat during dissolution in water, _______ with the rise in temperature.

Answer: Increase

Question 3. The chemical substance added to a mixture of two immiscible substances to increase the stability of the emulsion is called a/an _______ agent.

Answer: Emulsifying

Question 4. The solubility-curve for the solubility of NaCI in water is almost ________ to the temperature-axis.

Answer: Parallel

Question 5. The solubility curve of ________ rises up with the rise in temperature and then starts decreasing after reaching a maximum.

Answer: Glauber’s salt

Question 6. The solubility of calcium hydroxide in water ________ with rise in temperature.

Answer: Decreases

Question 7. The solubility of Glauber’s salt at 32.4°C as obtained from its solubility curve actually represents the solubility of ________

Answer: Anhydrous Na2SO4

Question 8. CO2 is dissolved in water at ________ pressure to produce soda water.

Answer: High

Question 9. With rise in temperature, the %(W/V) strength of a solution _________

Answer: Decreases

Question 10. To express the solubility of a substance, mentioning of ________ is necessary.

Answer: Temperature

Question 11. If 10 g of NaOH is dissolved in 100 ml of NaOH solution, strength of the solution is ________

Answer: 10%

Question 12. Solubility of CO2 in water _______ with increase in temperature.

Answer: Decreases

Question 13. Solubility of solids in liquid depends on _______

Answer: Temperature

Chapter 4 Matter Solution Topic B Solubility Strength Of Solution And Its Unit State Whether True Or False

Question 1. The amount of CaCI2 present in 500 mL of 25% CaCI2 solution is 125 g.

Answer: True

Question 2. The unit of solubility is gram per litre.

Answer: False

Question 3. The solubility of cerium sulphate decreases with the rise in temperature.

Answer: True

Question 4. Molarity of a solution is independent of temperature.

Answer: False

Question 5. The solubility of a gas in a liquid usually decreases with the increase in pressure above the liquid.

Answer: False

Question 6. W/V concentration of a solution increases with the rise in temperature.

Answer: False

Chapter 4 Matter Solution Topic B Solubility Strength Of Solution And Its Unit Numerical Examples

Question 1. At 50°C, 100g of a saturated solution on complete evaporation gives 5O g of residue. Find the solubility of the substance at that temperature.

Answer: Amount of solvent = Amount of solution – Amount of solute = (100 – 50) g = 50 g

∴ At 50°C, 50 g of solvent dissolves 50 g of solute

∴ 1 g of solvent dissolves 50/50 g of solute

∴ 100 g of solvent dissolves 50/50 x 100 g of solute = 100 g of solute.

Hence, solubility of the substance at 50°C is 100.

Question 2. At 30°C, the solubility of a salt in water is 40. Find the amount of water required to make a saturated solution of 30 g of the salt at that temperature.

Answer:

At 30°C, the solubility of the salt in water is 40.

∴ 40 g of the solute dissolves in 100 g of water.

∴ 30 g of the solute dissolves in x 30 g of water = 75 g of water.

Thus, 75 g of water will be required to make a saturated solution of 30 g of the salt at 30°C.

Question 3. The solubility of a salt in water at 60 30°C are 140 and 75 respectively. 50 g of the saturated solution of that salt at 60°C is cooled down to 30°C. How much of the salt will separate out from the solution?

Answer: 

At 60° C, 100g of water disisolves 140g salt.

∴ 50g of water dissolves (\(\frac{140}{100} \times 50\))g =70g salt.

At 30°C, 100g of water dissolves 75 g salt.

∴ 50g of water dissolves (\(\frac{75}{100} \times 50\)) g = 37.5g salt.

∴ When the solution is cooled down to 30°C, the amount of solute that will separate out from the solution = (70 – 37.5)g = 32.5g

Question 4. At 40°C, 20g of an unsaturated solution on evaporation gives 4g residue. What amount of salt has to be dissolved in 100 g of solution to make a saturated solution? [Solubility of the salt in water at 40°C is 40]

Answer:

20 g of solution has 4g of salt dissolved in it.

∴ 100g of solution will contain 4/20 x 100 g of the salt = 20g of the salt

∴ 100g of the solution contains = (100 – 20)g = 80 g water

∴ At 40°C, 100 g of water is saturated by \(\frac{40 \times 80}{100}\) g salt = 32 g of the salt

Hence, amount of salt required to make the solution saturated = (32 – 20) g = 12g

Question 5. Concentration of a solution containing NaOH is 100 g • L-1. Express the concentration in mass-volume percentage (% W/V).

Answer:

1000 mL of the solution contains 100g NaOH.

∴ 100 mL of the solution contains \(\frac{100 \times 100}{1000}\) g of NaOH = 10 g of NaOH.

Hence, concentration in mass-volume percentage (% W/V) = 10%

Question 6. 1.575 g oxalic acid is dissolved in 250 mL water. Find the cone, of the solution in g-L-1 & %W/V.

Answer:

In 250 mL water, the amount of oxalic acid dissolved is 1.575 g.

∴ In 1000 mL water, the amount of oxalic acid dissolved will be = \(\frac{1.575 \times 1000}{250} \mathrm{~g}\) = 6.3 g

Hence, concentration of the solution is 6.3g • L-1.

Again, 1000 mL of water dissolves 6.3 g oxalic acid.

∴ 100 mL of water dissolves \(\frac{6.3 \times 100}{1000}\) g = 0.63 g

oxalic acid. Hence, concentration of the solution in % W/V = 0.63%

Question 7. 60g NaOH is dissolved in 500 mL water. Express the cone, of the solution in mol • L-1.

Answer:

In 500 mL water, mass of NaOH dissolved = 60g

∴ In 1000 mL water, mass of NaOH dissolved = 60/500 x 1000 = 120g

Now, molecular mass of NaOH = 23 + 16 + 1 = 40.

∴ 1 mol NaOH = 40 g NaOH.

∴ 120 g NaOH = 120/40 mol Na0H = 3 mo1 NaOH

Hence, 1 L of the solution contains 3 mol of NaOH. So, the concentration of the solution is 3mol • L-1.

Question 8. How many grams of 50% (W/W) H2SO4 solution will be required to prepare 500 mL 2 (M) H2SO4 solution?

Answer:

Molecular mass of H2SO4 = 98.

∴ Mass of H2SO4 in 1000mL 2 (M) H2SO4 = 2 x 98 = 196g

Mass of H2SO4 dissolved in 500 mL 2 (M) H2SO4= 2 x 98/1000 x 5008 = 98g

Now, in 50% (W/W) H2SO4 solution, 100 g of the solution has 50g of H2SO4 dissolved in it.

∴ 98g H2SO4is dissolved in 100/50 x 98g = 196 g solution.

Hence, 196 g of H2SO4 solution will be required.

Question 9. Mass percentage & molar concentration of an HNO3 solution are 31.5 and 4 (M) respectively. Express the density in g • mL-1.

Answer:

Mass percentage of HNO3 solution is 31.5. Hence, 100g of solution has 31.5 g HNO3.

Let, the density of the solution be d g • mL-1.

∴ Volume of 100 g of the solution = 100/d mL

Now, 31.5g HN03 = 31.5/63 mol = 0.5 mol HNO3

Hence, the molar concentration of the solution = \(\frac{0.5}{100 / d} \times 1000\)  = 5  x d(M)

Now, 5 x d = 4; Hence, d = 0.8

Thus, density of the solution is 0.8 g • mL-1.

Chapter 4 Matter Solution Topic C Saturated Unsaturated Supersaturated Solution Synopsis

  1. The solution in which the maximum amount of solute remains dissolved in the solvent at a given temperature is called a saturated solution.
  2. An unsaturated solution is one in which, the solvent has the capacity to dissolve more amount of solute at a particular temperature.
  3. A supersaturated solution at a particular temperature is defined as the solution in which the solvent, under special conditions, contains more amount of solute than that required to form a saturated solution.
  4. A supersaturated solution is highly unstable. It readily precipitates the excess solute, when the solution is disturbed or shaken.
  5. The strength of a solution is generally expressed in mass percentage (% W/W), mass-volume percentage (% W/V), volume percentage (% V/V), mol • L-1 or g • L-1 units.
  6. x% W/V aqueous solution of a substance means that, at a given temperature x g of the substance is dissolved in 100mL water.

Chapter 4 Matter Solution Topic C Saturated Unsaturated Supersaturated Solution Short And Long Answer Type Questions

Question 1. What is a saturated solution?

Answer:

A solution in which, the maximum amount of solute remains dissolved in the solvent at a given temperature, is called a saturated solution. If more amount of solute is further added to the solution at that temperature, the excess solute settles at the bottom, but the concentration of the solution remains unchanged

Question 2. What is an unsaturated solution?

Answer:

An unsaturated solution is one in which, at a particular temperature, the solvent has the capacity to dissolve more amount of solute. On adding more solute, the concentration of the solution increases and after some time the solution becomes saturated.

Question 3. Differentiate between saturated and unsaturated solutions.

Answer:

The differences between saturated and unsaturated solutions are as follows

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Solution Topic C Differences Between Saturated Solution And Unsaturated Solution

Question 4. How will you convert a saturated solution into an unsaturated solution?

Answer:

A saturated solution can be converted to an unsaturated solution by the following methods

  1. At a particular temperature, a saturated solution can be made unsaturated by adding some more amount of solvent.
  2. In most cases, on heating, a saturated solution becomes unsaturated at a higher temperature.

Question 5. How will you convert an unsaturated solution into a saturated solution?

Answer:

An unsaturated solution can be converted to a saturated solution by the following methods

  1. At a particular temperature, an unsaturated solution can be made saturated by adding some more amount of solute.
  2. In most cases, on cooling, an unsaturated solution becomes saturated at a lower temperature.
  3. Evaporating some amount of solvent from a solution by heating and then subsequently cooling the solution to its initial temperature will decrease the amount of solvent in the solution. The solution thus obtained is a saturated solution at that temperature.

Question 6. Why the saturated solution of glauber salt cannot be unsaturated by heating the solution?

Answer:

The solubility of glauber salt (Na2SO4 – 10H2O) increases rapidly with increase in temperature upto 32.4°C. At a higher temperature than 32.4°C, the salt looses its water of crystallisation and transform to anhydrous sodium sulphate.

On increasing the temperature above 32.4°C solubility of anhydrous sodium sulphate decreases with increase in temperature. As a consequence, the excess salt gets precipitated out from the saturated solution. Hence the above solution cannot be unsaturated by heating.

Question 7. What is a supersaturated solution?

Answer:

A supersaturated solution at a particular temperature is defined as the solution in which the solvent contains more solute at a specific condition than that required to form a saturated solution at that temperature.

Question 8. State two important characteristics of a supersaturated solution.

Answer:

Two important characteristics of a supersaturated solution are

A supersaturated solution is highly unstable. Excess solute from the solution settles down at the bottom of the beaker on slightly shaking the solution, by adding a small crystal of the solute or in presence of dust particles.

On adding more solute to a supersaturated solution at a particular temperature, the concentration of the solution decreases.

Question 9. How will you prepare the super saturated solution of sodium thiosulphate?

Answer:

A small amount of hypo or sodium thiosulphate crystal is taken in a test tube and closed at top with cotton. The test tube is then immersed partially in boiling water bath. Crystals lose their water of crystallisation and get dissolved in that water forming a clear solution.

The test tube is then cooled, without stirring, to room temperature. The solution thus obtained is the super saturated solution of sodium thiosulphate.

Question 10. Differentiate between saturated and supersaturated solutions.

Answer:

The differences between saturated and supersaturated solutions are as follows

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Solution Topic C Differences Between Saturated Solution And Supersaturated Solution

Question 11. With the help of a simple experiment, how will you determine whether a solution is saturated, unsaturated or supersaturated?

Answer:

A solution of a given solute in a given solvent is taken. Some more amount of the solute is added to the solution. The following observations will help us to identify whether this solution is saturated, unsaturated or supersaturated

  1. If the solute completely or partially dissolves in the solution on stirring at the same temperature, then the solution is an unsaturated solution.
  2. If the added solute deposits at the bottom of the beaker instead of dissolving and its concentration remains unchanged, then it is a saturated solution.
  3. If on addition of excess solute, the added solute separates out from the solution and settles at the bottom of the beaker instead of dissolving and the amount of deposited solute is more than the added amount of solute, then the solution is a supersaturated solution.

Question 12. You have been given a saturated solution of common salt at room temperature. If the temperature of the solution has been reducedto5°Cby keeping it in an ice block, will the solution remain saturated? What will bethe probable observations?

Answer:

  1. The solution will remain saturated.
  2. Solubility of common salt changes slightly with temperature. So by cooling the solution only a little amount of salt will be observed to get precipitated out.

Question 13. what is the reason behind the damage of health of the paint workers?

Answer:

Many volatile substances like methylated spirit, tarpene oil, naptha, acetone etc. are used as solvent of paints during painting. These solvents evaporated at room temperature and enter easily to the body of the paint workers through inhalation. These can easily be in contact with the skin of the workers also.

These solvents cause allergy and several bronchiole or lung diseases which causes harm to the health of the workers involved in painting.

Chapter 4 Matter Solution Topic C Saturated Unsaturated Supersaturated Solution Very Short Answer Type Questions Choose The Correct Answer

Question 1. A supersaturated solution in water can be prepared by

  1. CuSO4 • 5H2O
  2. Na2SO4 • 10H2O
  3. FeSO4 • 7H2O
  4. Na2S2O3 • 5H2O

Answer: 4. Na2S2O3 • 5H2O

Question 2. The amount of dissolved solute present in a supersaturated solution compared to the maximum amount of solute that can be present in a saturated solution at a given temperature is

  1. Less
  2. More
  3. Equal
  4. None of these

Answer: 2. More

Question 3. Formula of hypo

  1. Na2SO4  • 10H2O
  2. Na2S2O3
  3. Na2S2O3  • 5H2O
  4. NaHSO4

Answer: 3. Na2S2O3  • 5H2O

Question 4. What happens when the temperature of the KNO3 solution saturated at 80°C reduces to 40°C?

  1. No change
  2. Solution will become unsaturated
  3. Some amount of KNO3 will settle down
  4. Solution will become supersaturated

Answer: 3. Some amount of KNO3 will settle down

Question 5. The solution in which more solute can be dissolved, is

  1. Saturated solution
  2. Unsaturated solution
  3. Supersaturated solution
  4. Colloid

Answer: 2. Unsaturated solution

Question 6. If some amount of solute is added to a saturated solution at a definite temperature, density of the solution will

  1. Increase
  2. Reduce
  3. Remain same
  4. None of the above

Answer: 3. Remain same

Chapter 4 Matter Solution Topic C Saturated Unsaturated Supersaturated Solution Answer In Brief

Question 1. Classify solutions on the basis of their concentration.

Answer: On the basis of their concentration, solutions can be classified into three types—saturated solution, unsaturated solution and supersaturated solution.

Question 2. Among saturated, unsaturated and supersaturated solutions, which one is the least stable?

Answer: A supersaturated solution is the least stable.

Question 3. Name a solid which can form a supersaturated solution.

Answer: Sodium thiosulphate pentahydrate (Na2S2O3 5H2O), commonly known as hypo, can form a supersaturated solution.

Question 4. State whether the solubility of common salt dissolved in a cup of water will be equal in Kolkata and Sikkim.

Answer: The solubility of a substance salt depends on temperature. As the temperature of the two places are different, the solubility of common salt in a cup of water will also be different.

Question 5. What amount of of acid (in g) is present per litre of 2 (M) H2SO4 solution? (Molecular mass of H2SO4 is 98)

Answer: Per litre of 2 (M) H2SO4 solution will contain 98 x 2 = 196 g of acid.

Question 6. Which type of solutions cannot form a saturated solution?

Answer: Solutions consisting of two miscible liquids can never form a saturated solution. For example, a solution of water and ethyl alcohol cannot form a saturated solution.

Question 7. On cooling a saturated solution, whether it remains saturated or not?

Answer: If we cool a saturated solution, some amount of solute gets precipitated out. But the remaining solution remains saturated.

Chapter 4 Matter Solution Topic C Saturated Unsaturated Supersaturated Solution Fill In The Blanks

Question 1. A _______  solution is one, whose concentration increases if some more solute is added to the solution.

Answer: Unsaturated

Question 2. In a ________ solution, the concentration remains unchanged even if some amount of solute is added to it.

Answer: Saturated

Question 3. A saturated solution dissolves more solute under a specific condition to produce _______ solution.

Answer: Supersaturated

Question 4. At a particular temperature, the dissolved solute particles in a saturated solution is in _______ with the precipitated particles.

Answer: Equilibrium

Question 5. At a given temperature, if some more solute is added to a supersaturated solution, its solubility ________

Answer: Decreases

Question 6. Unsaturated solution can absorb more _______

Answer: Solute

Chapter 4 Matter Solution Topic C Saturated Unsaturated Supersaturated Solution State Whether True Or False

Question 1. An unsaturated aqueous solution of ammonia is called liquid ammonia.

Answer: False

Question 2. Supersaturated solutions are stable in nature.

Answer: False

Question 3. A saturated solution remains saturated at any temperature.

Answer: False

Question 4. When more solute is added to a saturated solution, the excess solute will be sedimented.

Answer: True

Question 5. Supersaturated solution cannot be prepared by heating copper sulphate crystal in water- bath.

Answer: True

Chapter 4 Matter Solution Topic D Crystallisation Motion Of ParticlesIn Solution And Different Solvents Other Than Water Synopsis

  1. The continuous random motion of colloidal particles in a colloidal solution is known as Brownian motion.
  2. Colloidal particles exhibit Brownian motion in a colloidal solution. However, Brownian motion is not observed in case of particles of a suspension.
  3. The process by which crystals of a substance are formed from its saturated solution or its molten state either by cooling the solution or by sublimation is known as crystallisation.
  4. The phenomenon of spontaneous mixing of two or more substances irrespective of their nature without any external help to form a homogeneous mixture is called diffusion.
  5. Apart from water, compounds such as ethyl alcohol, methyl alcohol, acetone, chloroform, kerosene etc, are also used as solvents. These are known as aprotic solvents.
  6. Use of volatile solvents may cause serious health hazards.

Chapter 4 Matter Solution Topic D Crystallisation Motion Of ParticlesIn Solution And Different Solvents Other Than Water Short And Long Answer Type Questions

Question 1. What are crystals? On which factors does the shape of a crystal depend?

Answer:

Crystals:-

  1. Homogeneous solid substances having definite geometrical shape and bounded by symmetrically arranged plane surfaces that meet up at sharp edges are known as crystals. The substances formed of crystalline particles are known as crystalline substances. Some common examples are alum, diamond, sugar, copper sulphate etc.
  2. The shape of a crystal entirely depends on the number of particles (atoms, molecules or ions) present in the crystal and their geometrical arrangement.

Question 2. What do you mean by crystallisation and residue?

Answer:

Crystallisation And Residue:-

Crystallisation: The process by which crystals of a substance are formed from its saturated solution or molten state either by cooling the solution or by sublimation is known as crystallisation.

Residue: The remaining solution that is left behind in the beaker after a solid is separated out as crystals from its saturated solution on cooling is called residue.

Question 3. What is the primary condition for the purification of a solid by crystallisation?

Answer:

The Primary Condition For The Purification Of A Solid By Crystallisation:-

To purify a solid by crystallisation method, the solid must be soluble in a specific solvent while the impurities present in the solid must be insoluble or have different solubilities in that particular solvent.

Question 4. Mention two important applications of crystallisation.

Answer:

Two Important Applications Of Crystallisation:-

  1. Sea water contains different impurities along with dissolved salts. The method of crystallisation is applied for the removal of those impurities and extraction of salt from sea water.
  2. Impure samples of potash alum [K2SO4 – AI2(SO4)3– 24H2O] and nitre (KNO3) are purified by crystallisation.

Question 5. Crystallisation is a more effective method for purifying a solid than evaporation. Explain.

Answer:

Crystallisation Is A More Effective Method For Purifying A Solid Than Evaporation:-

  1. Crystallisation is a more effective method for purifying a solid than evaporation because of the following reasons
  2. During evaporation, the solvent is evaporated from the solution by applying heat. However on heating, some solids may decompose to produce new substances or substances like sugar may also get burnt to form a black mass of carbon particles.
  3. During evaporation, the solid impurities present in the solution do not separate out from the solution. So, the residue may contain large amount of impurities along with the desired substance. Thus, the solid obtained by evaporation may not be pure. However, during crystallisation, the impurities remain in the solution, while the solid crystallises out in pure form.

Question 6. What is overgrowth? What is seed crystal? Give examples.

Answer:

Overgrowth:-

  1. When a small crystal of the solute is dipped into its saturated solution with the help of a thread, the crystal grows bigger in size due to deposition of more solute crystals on its surface. This phenomenon is known as overgrowth of crystal.
  2. Seed crystal is a small crystal of the solute, dipped into its saturated solution with the help of a thread. The crystal becomes larger due to deposition of more solute crystals on its surface.
  3. If a small crystal of potash alum is dipped into a saturated solution of alum, alum starts depositing on the surface of the crystal and it grows bigger in size. Here, the small crystal is the seed crystal and the phenomenon is called overgrowth.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Solution Topic D Overgrowth What Is Seed Crystal

Question 7. Mention some uses of seed crystals in different industries.

Answer:

Uses Of Seed Crystals In Different Industries:-

  1. Seed crystals find important applications in many industries, such as
  2. During production of cane sugar (sucrose) from sugarcane juice, pure crystals of sugar are used as seed crystals.
  3. Seed crystals are used to accelerate the crystallisation of precious stones (like diamond), gemstones (like sapphire) and semiconductors (like silicone, germanium etc.)
  4. In Baeyer’s process of extraction of alumina, hydrated crystals of Al2O3 are used as seed crystals to accelerate precipitation of alumina.

Question 8. What is diffusion?

Answer:

Diffusion:-

The phenomenon of spontaneous mixing of two or more substances irrespective of their nature and. molar mass, without any external help to form a homogeneous mixture is called diffusion. In a solution, the solute particles move from a region of higher concentration to a region of lower concentration due to diffusion until a homogeneous mixture is formed.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Solution Topic D Diffusion

Question 9. How does diffusion take place?

Answer:

Diffusion Take Place As Follows:-

In a solution, both solvent and solute molecules move incessantly due to their high kinetic energy.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Solution Topic D Diffusion Process

Consequently, the solute molecules collide among themselves as well as with the solvent molecules leading to continuous and random motion (Brownian motion). As a result, the solute particles move from a region of higher concentration to a region of lower concentration by the process of diffusion until an equilibrium is reached.

Question 10. “Like dissolves like”— explain the statement with examples.

Answer:

“Like Dissolves Like”:-

When a solute is added to a solvent, three different types of interactions take place in the solution

  1. Solvent-solvent interaction,
  2. Solute-solute interaction and
  3. Solvent-solute interaction.

When the nature and extent of these three interactions are similar, then the solute dissolves in the solvent. For example, polar solutes like sodium chloride dissolve easily in polar solvents like water.

In this solution, the interaction between two water molecules is almost identical to the interaction between sodium chloride molecules as well as that between water and sodium chloride molecules. Similarly, a non-polar solute is found to be highly soluble in non-polar organic solvents.

Question 11. Write some uses of solvent.

Answer:

Some Uses Of Solvent:-

In different industries, methyl alcohol is extensively used as a solvent for colours, varnishes, celluloid substances, cements, fats etc.

Question 12. Write some uses of ethyl alcohol as a solvent.

Answer:

Uses Of Ethyl Alcohol As A Solvent:-

Ethyl alcohol is a very important solvent. It is used in different industries to dissolve resins, soaps, varnishes, colours, rayons, lax, synthetic rubbers, synthetic fibres, dyes etc. Ethyl alcohol is also used extensively as a solvent in pharmaceutical industries.

Question 13. Write some uses of acetone as a solvent.

Answer:

Uses Of Acetone As A Solvent:-

Acetone is a widely used organic solvent in chemical industries. It is used as a solvent for acetylene and semi-synthetic polymers like nitrocellulose, cellulose acetate etc. It is also used to dissolve varnishes and polishing materials.

Question 14. Write some uses of kerosene as a solvent.

Answer:

Uses Of Kerosene As A Solvent:-

Kerosene is an important organic solvent. It is used to dissolve different organic substances as well as colours. It can also remove colour stains from clothes.

Question 15. Write some uses of chloroform as a solvent.

Answer:

Uses Of Chloroform As A Solvent:-

Chloroform is an excellent organic solvent. In different industries, it is used to dissolve substances like fat, rubber, wax, resin etc. It is also widely used in the extraction of oils, gums and alkaloids.

Question 16. Why are volatile solvents used to dissolve colours and varnishes?

Answer:

Volatile Solvents Used To Dissolve Colours And Varnishes:-

Colours and varnishes are generally dissolved in volatile solvents because of the following reasons

  1. When the colour or varnish is applied on a surface, the volatile solvent evaporates easily. As a result, a smooth layer of colour or varnish is obtained.
  2. When a colour is sprayed by dissolving it in a volatile solvent, less amount of colour is required thus, making the process more cost-effective.

Question 17. Discuss some harmful effects of using volatile solvents.

Answer:

Harmful Effects Of Using Volatile Solvents:-

In our everyday life, we come across a number of volatile solvents. Volatile organic solvents like methyl alcohol, methylated spirit, turpentine or tarpin oil, naphtha, acetone etc. are widely used as solvents for oil paints, paint thinners, spray paints, varnishes etc.

Being volatile in nature, these solvents evaporate very fast even at ordinary temperature and can enter our body through respiratory system and get absorbed in the blood through our lungs.

In fact, these solvents may directly come in contact with our skin. Thus exposure to these solvent may cause serious damage to our body. The harmful effects caused by volatile solvents can be categorised into two groups

  1. Harmful effects due to short-term exposure and
  2. Harmful effects due to long-term exposure.

Harmful effects due to short term exposure to volatile solvents include

  1. Allergies,
  2. Headache,
  3. Asthma,
  4. Nausea and vomiting tendency,
  5. Drowsiness,
  6. Burning sensation in respiratory tract eyes, nose etc.

Harmful effects due to long term exposure to volatile solvents include

  1. Increased risk of cancer,
  2. Damage of kidneys and liver,
  3. Delayed childhood development and
  4. Damage of central nervous system.

Question 18. Mention the harmful effects of ethyl alcohol, methyl alcohol and acetone when they are used as solvents.

Answer:

Harmful Effects Of Ethyl Alcohol:

It is used as a solvent for different medicines. On entering the human body, it affects the nervous system. Thus, transmission of nervous impulse across the body gets disrupted and the person suffers from dizziness, Ethyl alcohol also reacts with some medicines and cause adverse side effects.

Harmful Effects Of Methyl Alcohol:

Methyl alcohol (CH3OH) is a volatile toxic compound and it decomposes to form formaldehyde (HCHO). If 10 mL CH3OH enters into the body, it may cause blindness due to the effect of HCHO, if 30 mL enters, the person may become unconscious and if 100 mL enters the body, the person may even die.

Harmful Effects Of Acetone:

If excess acetone enters into the body through respiration, then it may lead to temporary nervous breakdown.

Question 19. Mention the harmful effects of chloroform and kerosfene when they are used as solvents.

Answer:

Harmful Effects Of Chloroform:

Excess chloroform on entering the body through respiration, paralyses the central nervous system. It can even cause cardiac problems leading to death. In presence of light and air, chloroform gets oxidised to form poisonous carbonyl chloride or phosgene (COCI2) gas.

Harmful Effects Of Kerosene:

Kerosene has a suffocating and choking odour and it adversely affects living organisms.

Chapter 4 Matter Solution Topic D Crystallisation Motion Of ParticlesIn Solution And Different Solvents Other Than Water Very Short Answer Type Questions Choose The Correct Answers

Question 1. Brownian motion of colloidal particles is caused due to

  1. Convection
  2. Attraction between the particles of dispersed phase and dispersion medium
  3. Unequal collision between the particles of dispersed phase and dispersion medium
  4. Change in temperature

Answer: 3. Unequal collision between the particles of dispersed phase and dispersion medium

Question 2. The correct order for diffusion is

  1. Solid > liquid > gas
  2. Liquid > solid > gas
  3. Gas > solid > liquid
  4. Gas > liquid > solid

Answer: 4. Gas > liquid > solid

Question 3. The Brownian motion of colloidal particles indicates

  1. Linear motion
  2. Circular motion
  3. Spiral motion
  4. Random motion

Answer: 4. Random motion

Question 4. Which of the following is not an organic solvent?

  1. Liquid ammonia
  2. Benzene
  3. Chloroform
  4. Acetone

Answer: 1. Liquid ammonia

Question 5. Which of the following substances cannot form crystals by sublimation?

  1. Iodine
  2. Ammonium chloride
  3. Sodium chloride
  4. Benzoic acid

Answer: 3. Sodium chloride

Question 6. Just before crystallisation, the solution somewhat becomes

  1. Saturated
  2. Unsaturated
  3. Supersaturated
  4. Precipitated

Answer: 3. Supersaturated

Question 7. An example of a crystalline solid is

  1. Glass
  2. Wax
  3. Pitch
  4. Common salt

Answer: 4. Common salt

Question 8. An example of an amorphous solid is

  1. Sugar
  2. Salt
  3. Glass
  4. Blue vitriol

Answer: 3. Glass

Question 9. Which of the following is a method of crystallisation?

  1. Sublimation
  2. Filtration
  3. Fractional distillation
  4. Decantation

Answer: 1. Sublimation

Question 10. Mixing of the solute and solvent particles due to continuous, random motion of the particles is known as

  1. Osmosis
  2. Diffusion
  3. Effusion
  4. Distillation

Answer: 2. Diffusion

Question 11. An example of an organic solvent is

  1. Water
  2. Liquid ammonia
  3. Acetone
  4. Liquid SO2

Answer: 3. Acetone

Question 12. An inorganic solvent except water is

  1. Chloroform
  2. Kerosene
  3. Liquid ammonia
  4. Ethyl alcohol

Answer: 3. Liquid ammonia

Question 13. An example of lyophilic sol is

  1. Gold sol
  2. Silver sol
  3. Sulphur sol
  4. Starch sol

Answer: 4. Sulphur sol

Question 14. An example of a lyophobic sol is

  1. Ferric hydroxide sol
  2. Starch sol
  3. Gelatin
  4. Soap

Answer: 1. Ferric hydroxide sol

Question 15. In which of the following diffusion does not take place?

  1. Hydrogen and nitrogen
  2. Oxygen and water
  3. NaCI and Na2SO4
  4. Sugar and water

Answer: 3. NaCI and Na2SO4

Question 16. In which of the following, there is no water of crystallisation?

  1. Blue vitriol
  2. Borax
  3. Washing soda
  4. Sugar

Answer: 4. Sugar

Question 17. Anhydrous copper sulphate, by nature, is

  1. Deliquescent
  2. Hygroscopic
  3. Supersaturated
  4. None of these

Answer: 2. Hygroscopic

Question 18. The residual solution after separating the crystal by crystallisation is

  1. Saturated
  2. Unsaturated
  3. Supersaturated
  4. Colloid

Answer: 1. Saturated

Question 19. Inorganic, non-aqueous solvent is

  1. Chloroform
  2. Liquid CO2
  3. Ethanol
  4. Kerosene

Answer: 2. Liquid CO2

Question 20. Universal solvent is

  1. Alcohol
  2. Water
  3. Ether
  4. All of the above

Answer: 2. Water

Question 21. Which solvent is used to remove nail polish?

  1. Ethyl alcohol
  2. Methyl alcohol
  3. Acetone
  4. Tarpene oil

Answer: 3. Acetone

Question 21. Crystal without water of crystallisation is

  1. Common salt
  2. Blue vitriol
  3. Alum
  4. Epsom salt

Answer: 1. Common salt

Chapter 4 Matter Solution Topic D Crystallisation Motion Of ParticlesIn Solution And Different Solvents Other Than Water Answer In Brief

Question 1. Define water of crystallisation.

Answer: One or more than one water molecule attached to the crystals of certain compounds on which the colour and shape of the crystal largely depend is known as water of crystallisation.

Question 2. Write the formula of common alum.

Answer: The formula of common alum is K2SO4 – AI2(SO4)3– 24H2O.

Question 3. Why does diffusion take place in a solution?

Answer: Diffusion takes place due to continuous random motion of solute and solvent particles in a solution.

Question 4. What is the effect of temperature on diffusion of a liquid?

Answer: With the rise in temperature, diffusion in a liquid occurs at a faster rate.

Question 5. Give some examples of organic solvents.

Answer: Ethyl alcohol, methyl alcohol, acetone, chloroform, kerosene, carbon tetrachloride etc., are some examples of organic solvents.

Question 6. Give some examples of inorganic solvents other than water.

Answer: Some examples of inorganic solvents other than water are liquid ammonia, liquid nitrogen dioxide, liquid sulphur dioxide etc.

Question 7. Name a poisonous organic solvent.

Answer: Methyl alcohol

Question 8. Name an organic solvent which is actually a mixture.

Answer: Kerosene is an organic solvent which is actually a mixture of hydrocarbons.

Question 9. Which volatile solvent acts as a anaesthetic?

Answer: Chloroform is used as anaesthetic.

Question 10. Give an example of a solvent which can dissolve both of sulphur and phosphorous.

Answer: Carbon disulphide (CS2).

Question 11. Name two non-aqueous solvent.

Answer: Ethyl alcohol and chloroform.

Question 12. Name a crystal containing water of crystallisation.

Answer: Hypo or sodium thiosulphate (Na2S2O3 – 5H2O).

Question 13. Name the solvent used in paint and varnish.

Answer: Methylated spirit, tarpene oil, naptha, acetone etc.

Question 14. Give an example of efflorescent crystal.

Answer: Blue vitriol or blue stone i.e. CuSO4, 5H2O is an example of efflorescent crystal.

Question 15. Give an example of a crystal which control water of crystallisation but is not an efflorescent one.

Answer: Common alum [K2SO4 – AI2(SO4)3– 24H2O].

Question 16. What happens when blue vitriol is heated over 250°C?

Answer: Water of crystallisation of blue vitriol crystal evaporated upon heating over 250°C to produce white powder of amorphous copper sulphate.

Question 17. Mention the number of water of crystallisation in common alum.

Answer: 24

Chapter 4 Matter Solution Topic D Crystallisation Motion Of ParticlesIn Solution And Different Solvents Other Than Water Fill In the Blanks

Question 1. The size of a crystal depends upon the _______ and _______ of the constituent atoms or particles present in the crystal.

Answer: Size, arrangement

Question 2. Glass is a/an _______ solid.

Answer: Amorphous

Question 3. Acetone is an example of ________

Answer: Organic

Question 4. Number of water of crystallisation in glauber Salt is ______

Answer: 10

Question 5. Example of a crystalline compound without water of crystallisation is _________

Answer: Sodium chloride

Chapter 4 Matter Solution Topic D Crystallisation Motion Of ParticlesIn Solution And Different Solvents Other Than Water State Whether True Or False

Question 1. Brownian motion of colloidal particles is continuous and random in nature.

Answer: True

Question 2. In diffusion, the solute particles move from a region of higher concentration to lower concentration.

Answer: True

Question 3. Methyl alcohol on entering the body decomposes to form acetaldehyde.

Answer:  False

Question 4. Chloroform is a volatile solvent which is oxidised to form the toxic compound, phosgene.

Answer: True

Question 5. Seed crystals can accelerate the process of crystallisation of gemstones.

Answer: True

Question 6. Sublimation is a method of crystallisation.

Answer: True

Chapter 4 Matter Solution Topic D Crystallisation Motion Of ParticlesIn Solution And Different Solvents Other Than Water Miscellaneous Type Questions

Match The Column

1.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Solution Topic D Match The Column 1

Answer: 1.D, 2. C, 3. A, 4. B

2.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Solution Topic D Match The Column 2

Answer: 1. B, 2. D, 3. A, 4. C

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Acids Bases And Salts

Chapter 4 Matter Acids Bases And Salts Topic A Concept Of Acid And Bases Synopsis

  1. According to Arrhenius theory of electrolytic dissociation, an acid is a hydrogen-containing compound which dissociates in aqueous solution to produce H+ ion (or H3O+ ion) as the only cation. On the other hand, compounds that yield OH ions as the only anion in their aqueous solutions are called bases.
  2. Acids like H2SO4, HCI, HNO3 and bases like NaOH, KOH find important applications in different industries.
  3. Metals do not produce hydrogen on reacting with cold and very dilute HNO3. Only Mg and Mn can displace hydrogen from cold and very dilute nitric acid.
  4. Hydrochloric acid is identified by using aqueous solution of silver nitrate. Aqueous solution of barium chloride is used to identify sulphuric acid. Nitric acid is identified by the ring test.
  5. Strong acids and strong bases are highly corrosive in nature. So, proper precautionary measures are adopted while using them.
  6. When concentrated nitric acid is refluxed With small amount of starch, a yellow-coloured solution of nitric acid containing different oxides of nitrogen is obtained. It is called fuming nitric acid.

Chapter 4 Matter Acids Bases And Salts Topic A Concept Of Acid And Bases Short And Long Answer Type Questions

Question 1. Define acids in the light of Arrhenius’s theory of electrolytic dissociation.

Read and Learn More WBBSE Solutions for Class 9 Physical Science and Environment

Answer:

Arrhenius’s Theory Of Electrolytic Dissociation :-

According to Arrhenius theory of electrolytic dissociation, an acid is defined as a hydrogen-containing compound that dissociates in; water to produce H+ ions (or H3O+ ions) as the only cation.

For example, HCI ionises in water to produce H+ (or H3O+) ions as the only cation. Hence, HCI is an acid.

\(\mathrm{HCl} \rightarrow \mathrm{H}^{+}+\mathrm{Cl}^{-} ; \mathrm{H}^{+}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_3 \mathrm{O}^{+}\)


Question 2. Mention the general properties of acids.

Answer:

General Properties Of Acids Are

  1. Acids are ionised in aqueous solution forming H ion (H3O ion actually) and thus can conduct electricity.
  2. Acids are sour in taste.
  3. Acids turn blue litmus paper red.
  4. Acids generally produce hydrogen gas in reaction with metal.
  5. Acids generally produce carbon dioxide in reaction with carbonates and bicarbonates.
  6. Acids produce salt and water in reaction with alkali.

Question 3. Why do free H+ Ions not exist in water or aqueous solution?

Answer:

H+ ions combine with electronegative oxygen atoms in water to produce hydroxonium or hydronium ions (H3O+). So, free H+ ions do not exist in water or aqueous solution.

Question 4. What are inorganic or mineral acids and give some examples.

Answer:

Inorganic Acids :

Acids which are obtained from minerals or produced from inorganic substances are known as mineral acids. Hydrochloric acid (HCI), sulphuric acid (H2SO4), nitric acid (HNO3) are some examples of mineral acids.

Question 5. What are organic acids and give some examples.

Answer:

Organic Acids :

Acids (containing carbon atoms) which are obtained from plants or animals are called organic acids. For example, formic acid is found in ant venom, lemon contains citric acid, lactic acid is present in curd etc.

Question 6. What is meant by ionisation of an acid in aqueous solution and give example.

Answer:

Ionisation Of An Acid In Aqueous Solution

In aqueous solution, an acid dissociates to form H+ ions which are unstable in nature. So these ions combine with water molecules to form hydronium ions (H3O+). This decomposition of an acid in its aqueous solution is known as the ionisation of an acid.

\(\mathrm{HCl}(a q)+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{Cl}^{-}(a q)\)


Question 7. What are strong acids and give examples.

Answer:

Strong Acids :

Acids which almost completely dissociate in aqueous solution to produce large amount of H3O+ ions are called strong acids. Some examples are hydrochloric acid (HCI), sulphuric acid (H2SO4), nitric acid (HNO3), perchloric acid (HCIO4) etc.

Question 8. What are weak acids and give examples.

Answer:

Weak Acids :

Acids which partially dissociate in aqueous solution to produce small amount of H3O+ ions and most of the molecules remain undissociated in the solution are called weak acids. Some examples are acetic acid (CH3COOH), carbonic acid (H2CO3), citric acid etc.

Question 9. Give definition of a base according to Arrhenius’s theory.

Answer:

Arrhenius’s Theory :

According to Arrhenius theory, a base can be defined as a compound (mainly the metal oxides and metal hydroxides) which on dissociating in water produces hydroxyl ions (OH) as the only anion.

Na2O, NaOH, CaO are some examples of bases.

Question 10. Mention the general properties of bases.

Answer:

General Properties Of Bases Are :

  1. Bases form OHe ion as anion when they are dissolved in water.
  2. Bases taste bitter and are generally soapy in nature.
  3. Bases turn red litmus paper blue.
  4. Bases form salt and water in reaction with acids.

Question 11. NaOH is a strong base but NH4OH is a weak base. Explain.

Answer:

Difference Between Strong And Weak Base :

NaOH dissociates completely into Na and OHΘ ions in aqueous solution.

⇒ \(\mathrm{NaOH} \rightleftharpoons \mathrm{Na}^{\oplus}+\mathrm{OH}^{\ominus}\)

But NH4OH dissociates partially into NH and OHΘ ion in aqueous solution.

⇒ \(\mathrm{NH}_4 \mathrm{OH} \rightleftharpoons \mathrm{NH}_4^{\oplus}+\mathrm{OH}^{\ominus}\)

Depending of the nature of ionic dissociation in their respective aqueous solutions, NaOH acts as a strong base and NH4OH acts as a weak base.

Question 12. Mention the limitations of Arrhenius acid-base concept.

Answer:

Limitations Of Arrhenius Acid-Base Concept :

The limitations of Arrhenius acid-base concept are as follows

1. According to this concept, an acid or a base is defined on the basis of their dissociation in aqueous solutions. Thus, the concept is limited to aqueous medium only. It cannot explain the acidic or basic property of compounds in solvents other than water (non-aqueous solvents) like liquid ammonia, benzene etc.

2. According to this concept, a base must yield OH ions in aqueous medium. Thus, the theory fails to explain the basic nature of compounds like ammonia (NH3), methyl amine (CH3NH2), aniline (C6H5NH2) etc. Similarly acidic nature of compounds like BF3 or AICI3 cannot be explained by this theory as these compounds do not produce H+ ions in aqueous medium.

Question 13. All acids are hydrogen-containing compounds, but all hydrogen-containing compounds are not acids—justify the statement with example.

Answer:

All Acids Are Hydrogen-Containing Compounds, But All Hydrogen-Containing Compounds Are Not Acids :

According to Arrhenius theory, an acid must contain H-atom. However, a compound containing hydrogen will be called an acid only if it has the following properties

  1. It will dissociate in aqueous solution to produce H3O+ (hydroxonium) ions.
  2. It will produce hydrogen gas on reacting with metals.
  3. It will react with a base or an alkali to produce salt and water.
  4. Its aqueous solution will turn blue litmus red. However there are compounds containing hydrogen, which do not exhibit these properties.

For Example

  1. Compounds like PH3, NH3, CH4, sugar (C12H22O11) do not H3O+ ions in aqueous solution. Neither do they react with metals to produce hydrogen gas nor do they react with bases to form salt and water.
  2. Sodium reacts with H2O to produce hydrogen gas but the reaction produces NaOH instead of a salt.
  3. Al, Zn etc. react with compounds like sodium hydroxide, potassium hydroxide etc. to produce hydrogen gas but these compounds do not yield H3O+ ions in their aqueous solution. So, these compounds are not acids.

Question 14. Why are CaH2 and CH4 not acids?

Answer:

CaH2 and CH4 Are Cannot Be Termed As An Acids Because :

CaH2 produces H ion or hydride ion. That is why it cannot be termed as an acid.

\(\mathrm{CaH}_2 \rightarrow \mathrm{Ca}^{2+}+2 \mathrm{H}^{\ominus}\)

CH4 is a covalent compound and cannot form H ion in aqueous solution. Hence CH4 cannot be termed as acid.

Question 15. All alkalis are bases, but all bases are not alkalis—explain.

Answer:

All Alkalis Are Bases, But All Bases Are Not Alkalis :

A base dissolves in water to form an alkali.

However, all bases are not soluble in water and these bases do not form alkalis. Thus, all alkalis are bases, but bases insoluble in water are not alkalis.

For example, Na2O is a base. It dissolves in water to form NaOH. Sodium hydroxide (NaOH), potassium hydroxide (KOH), calcium hydroxide [Ca(OH)2] dissolve in water to give OH ions. So these are alkalis.

⇒ \(\mathrm{NaOH} \rightarrow \mathrm{Na}^{+}+\mathrm{OH}^{-} ; \mathrm{KOH} \rightarrow \mathrm{K}^{+}+\mathrm{OH}^{-}\)

⇒ \(\mathrm{Ca}(\mathrm{OH})_2 \rightarrow \mathrm{Ca}^{2+}+2 \mathrm{OH}^{-}\)

On the other hand, aluminium hydroxide [AI(OH)3], zinc hydroxide [Zn(OH)2], ferric hydroxide [Fe(OH)3] etc. react with acids to form salt and water. So they are bases. However, as they are insoluble in water, they cannot be called alkalis.

Question 16. Sodium hydroxide (NaOH) is a base, but  CIOH is an acid. Why?

Answer:

Sodium Hydroxide (NaOH) Is A Base, But  CIOH Is An Acid Because :

NaOH dissociates in water to produce Na+ and OH ions. As it yields OH ions in water, it is called a base. On the other hand, CIOH dissociates in water to give CIO and H+ ions.

H+ ions further combine with water molecules to form H3O+ ions. As it produces H3O+ ions in aqueous solution, it is an acid.

Question 17. State whether AI(OH)3 is a base or an alkali. Justify your choice.

Answer:

Aluminium Hydroxide AI(OH)3 Is Not An Alkali.

Water soluble metal hydroxides are known as alkalis. Aluminium hydroxide [AI(OH)3] is a base as it reacts with acid to produce salt and water but it is insoluble in water. So, aluminium hydroxide is not an alkali.

Question 18. Give an example of a reaction where two gases react to form a solid.

Answer:

Example Of A Reaction Where Two Gases React To Form A Solid :

Ammonia gas reacts with hydrogen chloride gas to form fine particles of solid ammonium chloride which floats in air forming white fumes.

Equation: \(\mathrm{NH}_3+\mathrm{HCl} \rightarrow \mathrm{NH}_4 \mathrm{Cl}\)

Question 19. Write down equations for the reaction of HCI with NaOH, Mg(OH)2 and AI(OH)3 respectively.

Answer:

Equations For The Reaction Of HCI With NaOH, Mg(OH)2 And AI(OH)3 Respectively : 

HCI forms corresponding salts and water in reaction with the mentioned bases:

⇒ \(\mathrm{NaOH}+\mathrm{HCl} \rightarrow \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}\)

⇒ \(\mathrm{Mg}(\mathrm{OH})_2+2 \mathrm{HCl} \rightarrow \mathrm{MgCl}_2+2 \mathrm{H}_2 \mathrm{O}\)

∴ \(\mathrm{Al}(\mathrm{OH})_3+3 \mathrm{HCl} \rightarrow \mathrm{AlCl}_3+3 \mathrm{H}_2 \mathrm{O}\)

Question 20. Give examples of chemical reactions of hydrochloric acid with carbonates and bicarbonate compounds.

Answer:

Chemical Reactions Of Hydrochloric Acid With Carbonates And Bicarbonate Compounds :-

Hydrochloric acid reacts with carbonate and bicarbonate compounds to liberate carbon dioxide and produce corresponding chloride salts.

⇒ \(\mathrm{Na}_2 \mathrm{CO}_3+2 \mathrm{HCl} \rightarrow 2 \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \uparrow\)

⇒ \(\mathrm{NaHCO}_3+\mathrm{HCl} \rightarrow \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \uparrow\)

Question 21. Under what condition does Cu react with HCI? Give equation.

Answer:

Condition For The Reaction Of Cu With HCI

Copper slowly reacts with hot and concentrated HCI in the presence of air or oxygen O2(g) to produce cupric chloride (CuCI2) and water.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Topic A Concept Of Acid And Bases

Question 22. Give an example of the reducing property of hydrochloric acid (HCI).

Answer:

Reducing Property Of Hydrochloric Acid (HCI) :

Hydrochloric acid reduces manganese dioxide (MnO2) to manganous chloride (MnCI2) and itself gets oxidised to evolve greenish-yellow chlorine gas.

⇒ \(\mathrm{MnO}_2+4 \mathrm{HCl} \rightarrow \mathrm{MnCl}_2+\mathrm{Cl}_2 \uparrow+2 \mathrm{H}_2 \mathrm{O}\)

Question 23. How will you identify hydrochloric acid? Give corresponding equation.

Answer:

Identification Of Hydrochloric Acid:

When silver nitrate (AgNO3) solution is added to an aqueous solution of hydrochloric acid (HCI), a curdy white precipitate of silver chloride (AgCI) is obtained. The precipitate is insoluble in nitric acid but dissolves readily in excess ammonium hydroxide (NH4OH) solution.

⇒ \(\mathrm{AgNO}_3+\mathrm{HCl} \rightarrow \mathrm{AgCl} \downarrow+\mathrm{HNO}_3
(white ppt.)\)

WBBSE Solutions For Class 9 Physical Science Chapter 4 Topic A Concept Of Acid And Bases Hydrochloric Acid

Question 24. Mention some uses of hydrochloric acid in industries.

Answer:

Uses Of Hydrochloric Acid In Industries.

Some important uses of hydrochloric acid are as follows

  1. In the preparation of chlorine and different metal chlorides.
  2. Preparation of aqua regia and glucose (from starch) and in removing scales from boilers
  3. In dye industries, tanneries, pharmaceutical industries and during galvanisation and tinplating of iron.

Question 25. What happens when carbonate and bicarbonate compounds react with nitric acid (HNO3)? Give equations.

Answer:

Nitric acid reacts with carbonate and bicarbonate compounds to liberate carbon dioxide and produce corresponding nitrate salts.

\(\mathrm{Na}_2 \mathrm{CO}_3+2 \mathrm{HNO}_3 \rightarrow 2 \mathrm{NaNO}_3+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \uparrow\) \(\mathrm{NaHCO}_3+\mathrm{HNO}_3\rightarrow \mathrm{NaNO}_3+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \uparrow\)

Question 26. Write with equation what happens when copper turnings are heated with hot and concentrated nitric acid.

Answer:

When copper turnings are heated with hot and concentrated nitric acid, brown fumes of nitrogen dioxide (NO2) gas is evolved. Copper is oxidised by nitric acid to copper nitrate [Cu(NO3)2] which makes the solution bluish-green in colour.

\(\mathrm{Cu}(s)+4 \mathrm{HNO}_3(a q) \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_3\right)_2(a q)+2 \mathrm{NO}_2(g) \uparrow+2 \mathrm{H}_2 \mathrm{O}(l)\)

Question 27. Give An example of the oxidising property of nitric acid.

Answer:

Example Of The Oxidising Property Of Nitric Acid :

Hot and concentrated nitric acid oxidises metallic zinc to zinc nitrate [Zn(NO3)2] and itself gets reduced to nitrogen dioxide.

\(\mathrm{Zn}+4 \mathrm{HNO}_3 \rightarrow \mathrm{Zn}\left(\mathrm{NO}_3\right)_2+2 \mathrm{NO}_2 \uparrow+2 \mathrm{H}_2 \mathrm{O}\)

Question 28. Which acid is termed as ‘aqua fortis’ or ‘strong water’? How can hydrogen be produced form that acid?

Answer:

  1. Nitric acid is termed as ‘aqua fortis’.
  2. In the reaction with Magnesium (Mg), very dilute (1%) nitric acid produces hydrogen.

⇒ \(\mathrm{Mg}+2 \mathrm{HNO}_3 \rightarrow \mathrm{Mg}\left(\mathrm{NO}_3\right)_2+\mathrm{H}_2 \uparrow\)

Question 29. What is fuming nitric acid?

Answer:

Fuming Nitric Acid :

When concentrated nitric acid (HNO3) is refluxed with starch, a brownish-yellow solution is obtained containing different oxides of nitrogen (mainly NO2) dissolved in it. This solution is called fuming nitric acid. It has been so named because the dissolved NO2 continuously comes out from the solution as brown fumes.

Question 30. Why does concentrated nitric acid turn yellow in presence of sunlight?

Answer:

In presence of sunlight, concentrated nitric acid decomposes to form brown-coloured NO2 gas. This gas dissolves in nitric acid and turns the solution yellow.

⇒ \(4 \mathrm{HNO}_3 \rightarrow 4 \mathrm{NO}_2 \uparrow+2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2 \uparrow\)

Question 31. Prove that oxygen is present in nitric acid.

Answer:

Oxygen Is Present In Nitric Acid :

When concentrated HNO3 is added drop-wise on red-hot pumice stone, a brown gas mixture evolves. When this gas mixture in passed through a ‘Ll’ tube dipped in freezing mixture, a colourless gas is evolved through ‘U’ tube which ignites a flameless burning stick.

The colourless gas gets absorbed by the alkaline potassium pyrogallate solution and turns the colour of the solution brown. So the gas is oxygen which comes from nitric acid.

\(4 \mathrm{HNO}_3 \rightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{NO}_2+\mathrm{O}_2\)

This reaction proves the presence of oxygen in nitric acid.

Question 32. Concentrated nitric acid can not be kept in copper vessel but the same can be kept in aluminium vessel. Why?

Answer:

Nitric acid produces brown fumes of nitrogen dioxide and copper nitrate if it is kept in a copper vessel.

⇒ \(\mathrm{Cu}+4 \mathrm{HNO}_3 \rightarrow \mathrm{Cu}\left(\mathrm{NO}_3\right)_2+2 \mathrm{NO}_2+2 \mathrm{H}_2 \mathrm{O}\)

That is why concentrated HNO3 can not be kept in copper vessel.

If concentrated nitric acid is kept in an aluminium vessel, nitric acid reacts with aluminium to form aluminium oxide. This aluminium oxide forms a thin layer which is insoluble in HNO3.

So the concentrated HNO3 can not come in contact with aluminium furthermore. That is why concentrated HNO3 can be kept in aluminium vessel.

Question 33. Describe ring test performed for identification of nitric acid. Give relevant equations.

Answer:

Ring Test Performed For Identification Of Nitric Acid :

A sample of the given solution is taken in a test tube and freshly prepared ferrous sulphate (FeSO4) solution is added to it. Then the solution is cooled and concentrated H2SO4 is gradually poured into the solution along the side of the test tube.

Sulphuric add being denser moves down and a brown ring is formed at the junction of the two liquid layers (acid + solution consisting of FeSO4 and sample). This test confirms the presence of nitric acid or any other nitrate salt.

HNO3 reacts with FeSO4 in the presence of concentrated H2SO4 to produce nitric oxide (NO). The formed nitric acid then combines with FeSO4 to form a complex [Fe(H2O)5NO]SO4 [pentaaquanitrosyliron (II) sulphate] which is responsible for the formation of the brown ring.

⇒ \(2 \mathrm{HNO}_3+3 \mathrm{H}_2 \mathrm{SO}_4+6 \mathrm{FeSO}_4 \longrightarrow 3 \mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3+2 \mathrm{NO}+4 \mathrm{H}_2 \mathrm{O}\)

⇒ \(\mathrm{FeSO}_4+6 \mathrm{H}_2 \mathrm{O} \rightarrow\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{SO}_4\)

⇒ \({\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{SO}_4+\mathrm{NO} \longrightarrow}{\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{NO}_3\right] \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O}}\)

WBBSE Solutions For Class 9 Physical Science Chapter 4 Topic A Concept Of Acid And Bases Ring Test Formed For Identification Nitric Acid

Question 34. Why is freshly prepared ferrous sulphate solution used in the ring test for the identification of nitric acid?

Answer:

The Reason For Freshly Prepared Ferrous Sulphate Solution Used In The Ring Test For The Identification Of Nitric Acid Is : 

In the presence of atmospheric oxygen, ferrous sulphate solution is rapidly oxidised to ferric sulphate. Ferric sulphate cannot form any complex with NO.

Hence, ferrous sulphate solution stored in laboratory cannot form the brown ring under the given reaction conditions. Thus, freshly prepared ferrous sulphate is used in the ring test for identification of nitric acid.

Question 35. The brown ring formed in the ring test disappears when the test tube is shaken. Why?

Answer:

The Brown Ring Formed In The Ring Test Disappears When The Test Tube Is Shaken Because:

When the test tube is shaken, concentrated H2SO4 combines with water and evolves large amount of heat which decomposes the complex compound, [Fe(H2O)5NO]SO4 and liberates nitric oxide (NO) gas.

NO comes out as bubbles from the test tube. As the complex responsible for ring formation decomposes, the brown ring disappears.

Question 36. Mention some uses of nitric acid in industries.

Answer:

Uses Of Nitric Acid In Industries:

Some important uses of nitric acid in industries are as follows

  1. It is used in the manufacture of different explosives like nitroglycerine, picric acid, TNT (trinitrotoluene) etc.
  2. It is used in the production of celluloids, synthetic colours, synthetic silk etc.
  3. HNO3 is used in the production of electrical cells, nitrate salts, fertilisers, aqua regia etc. with a glass rod. As a result, the produced heat is evenly distributed, thereby reducing the chance of any accident.

Question 37. What happens when carbonate and bicarbonate compounds react with sulphuric acid (H2SO4)? Give equations.

Answer:

Carbonate and bicarbonate compounds react with sulphuric acid to liberate carbon dioxide and produce corresponding sulphate salts.

⇒ \(\mathrm{Na}_2 \mathrm{CO}_3+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{Na}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \uparrow\)

⇒ \(\mathrm{NaHCO}_3+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{NaHSO}_4+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \uparrow\)

Question 38. Give an example of the oxidising property of sulphuric acid.

Answer:

Example Of The Oxidising Property Of Sulphuric Acid:

Hot and concentrated sulphuric acid oxidises metallic zinc to zinc sulphate (ZnSO4) and itself gets reduced to sulphur dioxide.

⇒ \(\mathrm{Zn}+2 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{ZnSO}_4+\mathrm{SO}_2 \uparrow+2 \mathrm{H}_2 \mathrm{O}\)

Question 39. Give suitable conditions and equations of the reaction between copper and sulphuric acid.

Answer:

Copper reacts with hot and concentrated sulphuric acid to produce blue-coloured copper sulphate and sulphur dioxide gas which has a smell of burnt sulphur.

⇒ \(\mathrm{Cu}(s)+2 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CuSO}_4(a q)+\mathrm{SO}_2 \uparrow+2 \mathrm{H}_2 \mathrm{O}(l)\)

Question 40. Give an example of hygroscopic property of concentrated sulfuric acid.

Answer:

When cone. H2SO4 is added to blue vitriol crystals, white powder of anhydrous copper sulfate is formed. This is an example of the hygroscopic property of cone. H2SO4 .

⇒ \(\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O} \stackrel{\text { Conc. } \mathrm{H}_2 \mathrm{SO}_4}{-5 \mathrm{H}_2 \mathrm{O}} \mathrm{CuSO}_4\)

Question 41. What happens when a few drops of cone. H2SO4 are added to sugar?

Answer:

Sugar turns to black carbon when a few drops of cone. H2SO4 are added to it.

⇒ \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} \frac{\text { Conc. } \mathrm{H}_2 \mathrm{SO}_4}{-11 \mathrm{H}_2 \mathrm{O}} 12 \mathrm{C} \text { (black) }\)

 

WBBSE Solutions For Class 9 Physical Science Chapter 4 Topic A Concept Of Acid And Bases Metal Activity Series

Question 42. Fe reacts with dilute H2SO4 to liberate hydrogen, but Cu does not—explain.

Answer:

Only metals present above hydrogen in the metal activity series can replace hydrogen from a dilute acid. Iron is placed above hydrogen while copper is placed below hydrogen in the metal activity series.

So iron reacts with dilute sulphuric acid to liberate hydrogen but copper is not able to do so. It must be noted that, copper does not react with dilute sulphuric acid but it is oxidised by hot and concentrated sulphuric acid.

Question 43. How will you identify sulphuric acid? Give equation.

Answer:

When barium chloride (BaCI2) solution is added to an aqueous solution of sulphuric acid (H2SO4), white barium sulphate (BaSO4) is precipitated which is insoluble in hydrochloric acid (HCI) or nitric acid (HNO3).

⇒ \(\mathrm{H}_2 \mathrm{SO}_4+\mathrm{BaCl}_2 \rightarrow \mathrm{BaSO}_4 \downarrow+2 \mathrm{HCl}\)

Question 44. Three bottles contain equal volumes of concentrated HCI, concentrated HNO3 and concentrated H2SO4 How will you identify the bottle containing H2SOwithout conducting any chemical test?

Answer:

The specific gravity of concentrated H2SO4 is much greater than that of concentrated HCI and concentrated HNO3. Hence, the bottle which is heaviest contains concentrated H2SO4.

Question 45. What is oleum?

Answer:

Oleum:

When an excess amount of SO3 gas is passed through 98% cone. H2SO4, the gas gets absorbed by the acid and an oily liquid substance of formula H2S2O3 is formed. This substance is termed as oleum.

⇒ \(\mathrm{H}_2 \mathrm{SO}_4+\mathrm{SO}_3 \rightarrow \mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7\)

Question 46. Prove the presence of hydrogen in sulphuric acid.

Answer:

Presence Of Hydrogen In Sulphuric Acid :

At room temperature, dil. H2SO4 and granulated zinc react to produce a colourless, odourless gas in which an ignited splint is extinguished and the gas burns with blue flame in air forming a colourless liquid.

This colourless liquid turns the white anhydrous copper sulphate blue. So this colourless liquid is water which was formed by the combustion of the colourless, odourless gas.

Therefore, the said gas is hydrogen which was formed by the reaction of granulated zinc and dil. H2SO4. Hence the pressence of hydrogen in H2SO4 is proved.

⇒ \(\mathrm{Zn}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{ZnSO}_4+\mathrm{H}_2\)

Question 47. Mention some uses of sulphuric acid in industries.

Answer:

Some important uses of sulphuric acid are as follows

  1. H2SO4is used in the preparation of inorganic acids like hydrochloric acid and nitric acid.
  2. H2SO4 is used in the preparation of fertilisers like super phosphate, ammonium sulphate.
  3. Sulphuric acid finds application in the production of alum, glucose, ethers, alcohols, dyes, explosives etc.
  4. It is used for processes such as, petroleum refining and galvanisation.
  5. H2SO4 is also used in lead storage batteries, extraction of metals and manufacture of films and synthetic silk.

Question 48. Acids are not stored in metal containers. Explain.

Answer:

Metals placed above hydrogen in the activity series react with acids to produce salts and hydrogen gas. So, acids cannot be kept in contact with active metals. For example, Fe reacts with dilute H2SO4 to produce FeSO4 salt and hydrogen gas.

Question 49. Nowadays, the contept of nascent hydrogen is considered to be obsolete— explain with an example.

Answer:

When hydrogen gas is passed through an aqueous solution of ferric chloride acidified with dilute H2SO4, no change is observed. Flowever when, zinc dust is added to the solution, ferric chloride (FeCl3) is reduced to ferrous chloride (FeCI2) and the solution turns colourless.

Earlier, it was considered that nascent hydrogen (which is more reactive in nature) produced due to the reaction between zinc dust and dilute sulphuric acid was responsible for the reduction of ferric chloride to ferrous chloride.

However, it has been proved that the reduction is due to the electrons released by the metals in the acid solution. Nascent hydrogen has no role in reducing the compounds, rather the metal acts as the reductant.

Zn atoms release electrons and get converted to Zn2+ ions. Fe3+ ions accept those electrons and are reduced to Fe2+ ions. Again H+ ions present in the solution are also reduced to hydrogen gas simultaneously by accepting the electrons released by zinc atoms.

⇒ \(\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+}+2 e \text { (oxidation) }\)

⇒ \(2 \mathrm{Fe}^{3+}+2 e \rightarrow 2 \mathrm{Fe}^{2+} \text { (reduction) }\)

⇒ \(\mathrm{H}^{+}+e \rightarrow \mathrm{H} \text { (reduction); } \mathrm{H}+\mathrm{H} \rightarrow \mathrm{H}_2 \uparrow\)

It should also be noted that, nowadays the concept of nascent oxygen is also considered to be obsolete.

Question 50. Write with equation what happens when zinc dust is added to hot and concentrated caustic soda solution.

Answer:

Zinc dust reacts with hot and concentrated caustic soda solution (NaOH) to produce sodium zincate (Na2ZnO2) salt and hydrogen gas.

⇒ \(\mathrm{Zn}+2 \mathrm{NaOH} \rightarrow \mathrm{Na}_2 \mathrm{ZnO}_2+\mathrm{H}_2 \uparrow\)

Question 51. Concentrated sodium hydroxide (NaOH) solution should not be boiled in aluminium vessels. Why?

Answer:

Aluminium reacts with sodium hydroxide solution (NaOH) to form soluble sodium aluminate salt along with the evolution of hydrogen gas. As a result, the vessels get corroded.

⇒ \(2 \mathrm{Al}+2 \mathrm{NaOH}+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NaAlO}_2+3 \mathrm{H}_2 \uparrow\)

So, concentrated NaOH solution should not be boiled in aluminium vessels.

Question 52. Mention some uses of sodium hydroxide in industries.

Answer:

Sodium hydroxide is a widely used chemical in different industries. Some of its uses are as follows

  1. Sodium hydroxide is used in the manufacture of soaps, paper and synthetic fibres.
  2. It is used in the production of organic dyes.
  3. It is used in the refining of petroleum-based products.
  4. Caustic soda (NaOH) is used as a reagent in the industrial preparation of different chemicals.

Question 53. Concentrated acids are diluted by adding the acid into water and not by adding water into the acid. Why?

Answer:

Large amount of heat is evolved when water is added to acid and this causes the acid to boil. As a result, the acid may spurt out from the beaker and cause serious accident.

Apart from this, the glass beaker may break due to evolution of excess amount of heat. So, the acid is gradually added to large amount of water with continuous stirring.

Question 54. Why should one be cautious while working with concentrated acids or Concentrated alkalis?

Answer:

Concentrated acids or concentrated alkalis are highly corrosive substances. They cause blister-like wounds on human skin. Burns caused due to acids and alkalis are known as acid burns and alkali burns respectively.

Concentrated acids are extremely harmful for eyes. Apart from these, acids may cause burning sensation, blackening of skin and also damage the living tissues.

Acid vapours if inhaled may cause breathing troubles. Concentrated alkalis also produce similar problems. Thus, one should, be careful while working with concentrated acids or concentrated alkalis.

Question 55. What do you mean by acid burns and alkali bums?

Answer:

When concentrated mineral acids like HCI, H2SO4, HNO3 etc., and concentrated alkalis like NaOH, KOH etc., come in contact with our skin, these cause blisters on our skin and even burn it.

The burns caused by the corrosive action of acids are known as acid burns while the burns caused by the corrosive action of strong alkalis ate known as alkali burns.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Topic A Concept Of Acid And Bases Acid Burns And Alkali Burns

Question 56. What are the steps that should be immediately taken in case of an acid burn or an alkali burn?

Answer:

Concentrated acids or alkalis cause blisters on our skin. In such cases, the affected part should be immediately washed with plenty of water repeatedly.

In case of an acid burn, the affected part should be treated with 1% dilute sodium bicarbonate solution and in case of an alkali burn, the affected part should be washed with 1% acetic acid solution. After the preliminary treatment is done, one should consult a doctor.

Question 57. Human skin becomes yellowish when it mcomes in contact with dilute nitric acid. Why?

Answer:

Dilute nitric acid reacts with proteins present in the human skin to form yellow-coloured xanthoproteic acid. This makes the skin yellowish.

Question 58. Human skin blackens when it is exposed to concentrated sulphuric acid. Why?

Answer:

Concentrated sulphuric acid is strongly hygroscopic and corrosive in nature. When it comes in contact with human skin, it absorbs water from the skin causing the skin to blacken.

Question 59. What changes will be observed in the colour of a litmus paper when it comes in contact with dry HCI gas and an aqueous solution of hydrogen chloride gas?

Answer:

No change will be observed in the colour oflitmus paper when it comes in contact with dry HCI gas. However, the litmus paper turns red in aqueous solution of hydrogen chloride gas.

Question 60. HCI gas is not an acid, but aqueous solution of HCI is strongly acidic. Why?

Answer:

Hydrogen chloride is a covalent compound. It does not dissociate in gaseous state or in liquid state to produce H+ ion. Hence, hydrogen chloride gas is not an acid but is a neutral compound.

However, in aqueous solution, polar water molecules separate the partially positively charged H-atom from partially negatively charged Cl-atom in the polar H — Cl molecule.

Consequently, HCI molecules ionise to form H+ and Cl ions. H+ ions further combines with water molecules to form H3O+ ions and thus, aqueous solution of HCI shows acidic property.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Topic A Concept Of Acid And Bases Sodium Biocarbonate And Tartaric Acid

For example, if a dry mixture of sodium bicarbonate and tartaric acid is taken in a glass and the mouth of the glass is covered with a deflated balloon, no change is observed.

However, if some water is added to the glass and the deflated balloon is placed at the mouth of the flask, it is seen that the balloon gets inflated due to the liberation of CO2 gas.

Question 61. A mixture of tartaric acid crystals and sodium bicarbonate undergo chemical reaction only in presence of water. Explain with reason.

Answer:

Tartaric acid and sodium bicarbonate do not react with each other in dry state because they do not behave as acids or bases in dry state. However, when water is added to the mixture, tartaric acid ionises to produce H+ ions which in turn react with bicarbonate ions (HCO3) produced due to dissociation of sodium bicarbonate to liberate CO2 gas.

⇒ \(\mathrm{H}^{+}+\mathrm{HCO}_3^{-} \rightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \uparrow\)

Question 62. HCI is added to silver nitrate solution.

Answer:

When silver nitrate (AgNO3) solution is added to hydrochloric acid, insoluble silver chloride (AgCI) is produced which forms a curdy white precipitate.

⇒ \(\mathrm{AgNO}_3+\mathrm{HCl} \underset{\text { (curdy white) }}{\rightarrow \mathrm{AgCl} \downarrow}+\mathrm{HNO}_3\)

Question 63. Dilute hydrochloric acid is added to calcium carbonate.

Answer:

Calcium carbonate or marble reacts with dilute hydrochloric acid to produce calcium chloride (CaCI2) and carbon dioxide gas.

\(\mathrm{CaCO}_3+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_2+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \uparrow\)

Question 64. A mixture of concentrated HCI and Mno2 is heated.

Answer:

When a mixture of manganese dioxide (MnO2) and concentrated HCI is heated, HCI reduces MnO2 to manganous chloride (MnCl2) and itself gets oxidised to evolve greenish-yellow chlorine gas.

⇒ \(\mathrm{MnO}_2+4 \mathrm{HCl} \rightarrow \mathrm{MnCl}_2+\mathrm{Cl}_2 \uparrow+2 \mathrm{H}_2 \mathrm{O}\)

Question 65. Ammonia is allowed to react with hydrogen chloride gas.

Answer:

Ammonia gas reacts with hydrogen chloride (HCI) gas to produce dense white fumes of ammonium chloride (NH4CI).

⇒ \(\mathrm{NH}_3(g)+\mathrm{HCl}(g) \rightarrow \mathrm{NH}_4 \mathrm{Cl}(s)\)

Question 66. Dilute hydrochloric acid (HCI) is kept in an aluminium vessel.

Answer:

If dilute HCI is kept in an aluminium vessel, then aluminium reacts with the dilute acid to produce aluminium chloride (AlCI3) along with evolution of hydrogen gas. Thus, the vessel gets corroded.

⇒ \(2 \mathrm{Al}+6 \mathrm{HCl} \rightarrow 2 \mathrm{AlCl}_3+3 \mathrm{H}_2 \uparrow\)

Question 67. Aqueous sodium carbonate is treated with hydrogen chloride (HCI) gas.

Answer:

Aqueous sodium carbonate reacts with hydrogen chloride (HCI) gas to produce sodium chloride, carbon dioxide and water.

⇒ \(\mathrm{Na}_2 \mathrm{CO}_3+2 \mathrm{HCl} \rightarrow 2 \mathrm{NaCl}+\mathrm{CO}_2 \uparrow+\mathrm{H}_2 \mathrm{O}\)

Question 68. Metallic copper Is treated with concentrated HCI in presence of oxygen.

Answer:

Metallic copper slowly reacts with hot and concentrated HCI in presence of oxygen to produce cupric chloride (CuCI2) and water.

⇒ \(2 \mathrm{Cu}(s)+\mathrm{O}_2(g)+4 \mathrm{HCl}(a q) \rightarrow \stackrel{\Delta}{2 \mathrm{CuCl}_2(a q)+2 \mathrm{H}_2 \mathrm{O}(I)}\)

Question 69. Concentrated HNO3 is added dropwiseon hot charcoal.

Answer:

When concentrated nitric acid is added dropwise to hot charcoal (carbon), carbon gets oxidised to carbon dioxide (CO2) gas while nitric acid gets reduced to form reddish-brown nitrogen dioxide (NO2) gas.

⇒ \(\mathrm{C}+4 \mathrm{HNO}_3 \rightarrow \mathrm{CO}_2 \uparrow+4 \mathrm{NO}_2 \uparrow+2 \mathrm{H}_2 \mathrm{O}\)

Question 70. Copper is heated with concentrated nitric acid.

Answer:

When copper is heated with concentrated nitric acid, copper is oxidised by nitric acid to form cupric nitrate. [Cu(NO3)2]. Nitric acid itself gets reduced to nitrogen dioxide.

⇒ \(\mathrm{Cu}+4 \mathrm{HNO}_3 \rightarrow \mathrm{Cu}\left(\mathrm{NO}_3\right)_2+2 \mathrm{NO}_2 \uparrow+2 \mathrm{H}_2 \mathrm{O}\)

Question 71. Hot and concentrated nitric acid is kept in an iron vessel.

Answer:

Hot and concentrated nitric acid initially reacts with iron to produce ferric nitrate [Fe(NO3)3], NO2 and water.

⇒ \(\mathrm{Fe}+6 \mathrm{HNO}_3 \rightarrow \mathrm{Fe}\left(\mathrm{NO}_3\right)_3+3 \mathrm{NO}_2 \uparrow+3 \mathrm{H}_2 \mathrm{O}\)

Ferric nitrate then decomposes to form ferric oxide which forms a layer over the iron vessel and prevents further reaction of iron with the acid.

⇒ \(4 \mathrm{Fe}\left(\mathrm{NO}_3\right)_3 \rightarrow 2 \mathrm{Fe}_2 \mathrm{O}_3+12 \mathrm{NO}_2+3 \mathrm{O}_2 \uparrow\)

Question 72. Metallic magnesium is treated with very dilute nitric add.

Answer:

Metallic magnesium reacts with very dilute nitric acid to form magnesium nitrate [Mg(NO3)2] and hydrogen gas.

⇒ \(\mathrm{Mg}+2 \mathrm{HNO}_3 \rightarrow \mathrm{Mg}\left(\mathrm{NO}_3\right)_2+\mathrm{H}_2 \uparrow\)

Question 73. Cold and dilute nitric acid is added to copper.

Answer:

Copper reacts with cold and dilute nitric acid to form cupric nitrate [Cu(NO3)2], nitrous oxide (N2O) and water.

⇒ \(4 \mathrm{Cu}+10 \mathrm{HNO}_3 \rightarrow 4 \mathrm{Cu}\left(\mathrm{NO}_3\right)_2+\mathrm{N}_2 \mathrm{O} \uparrow+5 \mathrm{H}_2 \mathrm{O}\)

Question 74. Hot and concentrated nitric acid is added dropwise on red-hot pumice stone.

Answer:

When hot and concentrated nitric acid is added dropwise on red-hot pumice stone, the acid decomposes to form nitrogen dioxide (NO2) gas along with oxygen and water vapour.

⇒ \(4 \mathrm{HNO}_3 \rightarrow 4 \mathrm{NO}_2 \uparrow+2 \mathrm{H}_2 \mathrm{O} \uparrow+\mathrm{O}_2 \uparrow\)

Question 75. Nitric acid vapour is passed over metallic copper.

Answer:

Metallic copper reacts with nitric acid vapour to produce cupric oxide (CuO) and water along with evolution of nitrogen gas.

⇒ \(5 \mathrm{Cu}+2 \mathrm{HNO}_3 \rightarrow 5 \mathrm{CuO}+\mathrm{N}_2 \uparrow+\mathrm{H}_2 \mathrm{O}\)

Question 76. Copper is treated with 1:1 HNO3 at room temperature.

Answer:

Blue-coloured cupric nitrate, nitric oxide gas and water are formed by the reaction of cccper 1 : 1 HNO3 at room temperature.

⇒ \(3 \mathrm{CU}+8 \mathrm{HNO}_3 \rightarrow 3 \mathrm{Cu}\left(\mathrm{NO}_3\right)_2+4 \mathrm{H}_2 \mathrm{O}+2 \mathrm{NO} \uparrow\)

Question 77. A mixture of potassium nitrate and concentrated H2SO4 is heated around 200ºC.

Answer:

When a mixture of potassium nitrate and concentrated H2SO4 is heated around 200ºC, potassium hydrogen sulphate KHSO4 and nitric acid are produced.

⇒ \(\mathrm{KNO}_3+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{KHSO}_4+\mathrm{HNO}_3\)

Question 78. Dilute sulphuric acid is added to barium chloride solution.

Answer:

When dilute sulphuric acid is aaced to barium chloride (BaCI2) solution, white barium su pnate is precipitated which is insoluble in hytrochiorc acid (HCI) and nitric acid (HNO3).

⇒ \(\mathrm{H}_2 \mathrm{SO}_4+\mathrm{BaCl}_2 \underset{\text { (white) }} \rightarrow \mathrm{BaSO}_4 \downarrow+2 \mathrm{HCl}\)

Question 79. Concentrated H2SO4 is heated with copper turnings.

Answer:

When concentrated H2SO4 is heated with copper turnings, copper sulphate is produced and ‘ sulphur dioxide gas having a smell of burnt S sulphur is liberated.

⇒ \(\mathrm{Cu}(s)+2 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CuSO}_4(a q)+\mathrm{SO}_2 \uparrow+2 \mathrm{H}_2 \mathrm{O}(I)\)

Question 80. Concentrated sulphuric acid H2SO4 is added to formic acid.

Answer:

When concentrated H2SO4 is added to formic acid, sulphuric acid absorbs water molecules from formic acid and produces carbon monoxide (CO).

⇒ \(\mathrm{HCOOH}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CO} \uparrow+\left[\mathrm{H}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{SO}_4\right]\)

Chapter 4 Matter Acids Bases And Salts Topic A Concept Of Acid And Bases Very Short Answer Type Questions Choose The Answer

Question 1. A covalent compound whose aqueous solution is acidic is

  1. CH4
  2. CCI4
  3. HCI
  4. NH

Answer: 3. HCI

Question 2. A covalent compound whose aqueous solution is alkaline

  1. CH4
  2. CCI4
  3. HCI
  4. NH

Answer: 4. NH

Question 3. One of the limitations of Arrhenius acid-base theory is that

  1. It is applicable only for organic solvents
  2. It is applicable only for inorganic solvents
  3. Water is essentially required as the solvent
  4. No solvent is required

Answer: 3. Water is essentially required as the solvent

Question 4. In aqueous solution H+ ion is present as

  1. Hion
  2. H2+ ion
  3. H3O+ ion
  4. H3O ion

Answer: 3. H3O+ ion

Question 5. Which of the following is used to prepare soaps and detergents?

  1. HNO3
  2. HCI
  3. H2SO4
  4. NaOH

Answer: 4. NaOH

Question 6. Which of the following is known as ‘aqua fortis’?

  1. NaOH
  2. HNO3
  3. HCI
  4. H2SO4  

Answer: 2. HNO3

Question 7. Which of the following is called the ‘king of chemicals’?

  1. NaOH
  2. HNO3
  3. HCI
  4. H2SO4  

Answer: 4. H2SO4  

Question 8. The acid used in the preparation of starch from glucose is

  1. HCI
  2. HNO3
  3. H2SO4  
  4. H3PO4  

Answer: 1. HCI

Question 9. Aqua regia is a

  1. 3:1 mixture of concentrated HCI and concentrated HNO3
  2. 1:3 mixture of concentrated HCI and concentrated H2SO4
  3. 2:3 mixture of HNO3 and H2SO4
  4. 1:1:1 mixture of HCI, HNO3 and H2SO4

Answer: 1. 3:1 mixture of concentrated HCI and concentrated HNO3

Question 10. Which of the following does not produce hydrogen on reacting with dilute acids?

  1. Fe
  2. Zn
  3. Cu
  4. Mg

Answer: 3. Cu

Question 11. Which of the following can produce hydrogen by reacting with cold and very dilute HNO3?

  1. Fe
  2. Zn
  3. Cu
  4. Mg

Answer: 4. Mg

Question 12. The compound used to identify H2SO4 is

  1. BaSO4
  2. BaCI2
  3. Ba(OH)2
  4. BaCO3

Answer: 2. BaCI2

Question 13. Which of the following is identified by using an aqueous solution of AgNO3?

  1. HCI
  2. HNO3
  3. H2SO4
  4. NaOH

Answer: 1. HCI

Question 14. NaOH reacts with HCI in

  1. Dry condition
  2. Aqueous solution
  3. Benzene solution
  4. CCL solution

Answer: 2. Aqueous solution

Question 15. Which of the following is not a mineral acid?

  1. Hydrochloric acid
  2. Citric acid
  3. Sulphuric acid
  4. Nitric acid

Answer: 2. Citric acid

Question 16. Which of the following is not a base?

  1. NaOH
  2. KOH
  3. NH4OH
  4. C2H5OH

Answer: 4.C2H5OH

Question 17. Which of the following is a strong hygroscopic substance?

  1. Cone. H2SO4
  2. Cone. HNO3
  3. Cone. HCI
  4. CH3COOH

Answer: 1. Cone. H2SO4

Question 18. Which of the following can be identified through ring test?

  1. HCI
  2. H2SO4
  3. NaOH
  4. HNO3

Answer: 4. HNO3

Question 19. \(\mathrm{Zn} \underset{\mathrm{HNO}_3}{\stackrel{\text { dil, cold }}{\longrightarrow}} \mathrm{Zn}\left(\mathrm{NO}_3\right)_2+X+\mathrm{H}_2 \mathrm{O}\); Here ‘X’ is

  1. N2O
  2. NO
  3. NO2
  4. NH4NO3

Answer: 4. NH4NO3

Question 20. The brown ring compound formed in the ring test is

  1. \(\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5(\mathrm{NO})_2\right] \mathrm{SO}_4\)
  2. \(\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{NO}\right] \mathrm{SO}_4\)
  3. \(\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{NO}\right]\left(\mathrm{SO}_4\right)_2\)
  4. \(\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{NO}\right] \mathrm{SO}_4\)

Answer: 2. \(\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{NO}\right] \mathrm{SO}_4\)

Question 21. Brown fumes is seen to be coming out of the surface of which acid?

  1. HNO3
  2. H2CO3
  3. H2SO4
  4. HCI

Answer: 1. HNO3

Question 22. The gas formed by heating Cu-turnings with I cone. HN03 is

  1. CO2
  2. NO2
  3. NO
  4. N2O

Answer: 2. NO2

Question 23. According to Arrhenius Theory, an acid produces (in aqueous medium)

  1. H ion as only anion
  2. H ion as only cation
  3. OH ion as only anion
  4. OH ion as only cation

Answer: 2. H ion as only cation

Question 24. The metal which produces H2 in reaction with acid as well as with base is

  1. Copper
  2. Magnesium
  3. Zinc
  4. Iron

Answer: 3. Zinc

Question 25. If acid falls on any body part, which one of the following should be used to wash it off?

  1. Sodium bicarbonate solution
  2. Lemon juice
  3. Lime water
  4. Caustic soda solution

Answer: 1. Sodium bicarbonate solution

Question 26. Which one of the following is formed by the reaction of copper with cold and dil. HNO3?

  1. NO
  2. NO2
  3. N2O
  4. N2

Answer: 3. N2O

Question 27. Which one of the following is hydracid?

  1. H2SO4
  2. HNO3
  3. HCI
  4. HCOOH

Answer: 3. HCI

Question 28. The acid which is secreted inside the stomach is

  1. HNO3
  2. H2SO4
  3. HCI
  4. H2CO3

Answer: 3. HCI

Question 29. Cone. H2SO4 is a

  1. Deliquescent substance
  2. Efflorescent substance
  3. Crystalline substance
  4. Hygroscopic substance

Answer: 4. Hygroscopic substance

Question 30. The gas produced by the reaction of MnO2 with dil. hydrochloric acid is

  1. H2
  2. O2
  3. CI2
  4. N2

Answer: 3. CI2

Chapter 4 Matter Acids Bases And Salts Topic A Concept Of Acid And Bases Answer In Brief

Question 1. For which property of water, an acid ionises in water to produce H3O+ ions?

Answer: Due to the polar nature of water, an acid ionises in water to produce H3O+ ions.

Question 2. Give some examples of organic acid.

Answer: Acetic acid (CH3COOH), formic acid (HCOOH) etc., are some examples of organic acid.

Question 3. Name an inorganic acid which is synthesised in human body.

Answer: Hydrochloric acid (HCl).

Question 4. Show the reaction for dissociation of HNO3 and CH3COOH in aqueous medium.

Answer:

⇒ \(\mathrm{HNO}_3+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_3 \mathrm{O}^{\oplus}+\mathrm{NO}_3^{\ominus}\)

⇒ \(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOH}^{\ominus}+\mathrm{H}_3 \mathrm{O}^{\oplus}\)

Question 5. Give an example (name and formula) of a weak acid.

Answer: Acetic acid (CH3COOH).

Question 6. Name two acids used in daily life.

Answer: Examples of acids used in daily life are

  1. Acetic acid (vinegar is used in cooking),
  2. Hydrochloric acid (muriatic acid is used to clean bathrooms).

Question 7. Name a base which is not a metallic oxide or hydroxide.

Answer: Ammonia (NH3).

Question 8. Name two basic substances used in daily life.

Answer: Washing soda (Na2CO3) and baking soda (NaHCO3) are two basic substances used in daily life.

Question 9. Can BCI3 be considered as an acid according to Arrhenius theory?

Answer: BCI3 cannot be considered as an acid as it does notyield any H+ ion in aqueous solution.

Question 10. Name a metal which cannot produce hydrogen gas by reacting with dilute mineral acids.

Answer: Copper.

Question 11. Give example of a volatile mineral acid.

Answer: Hydrochloric Acid (HCI).

Question 12. Name a covalent gaseous compound which ionises in water and acts as a strong acid.

Answer: Hydrogen chloride (HCI) is a gaseous covalent compound which ionises in water and acts as a strong acid.

Question 13. What is muriatic acid?

Answer: Muriatic acid is hydrochloric acid (HCI).

Question 14. Which metals can displace hydrogen from cold and dilute nitric acid?

Answer: Metals such as magnesium (Mg) and manganese (Mn) can displace hydrogen from cold and dilute nitric acid.

Question 15. Which acid is used to remove impurities from gold?

Answer: Nitric acid (HNO3).

Question 16. State the major use of aqua regia.

Answer: Aqua regia is used to dissolve noble metals like gold, silver, platinum etc.

Question 17. Which acid is used in the preparation of TNT?

Answer: Nitric acid (HNO3).

Question 18. Which gas is excreted from jewellery shops?

Answer: Nitrogen dioxide (NO2).

Question 19. Which acid is used more in jewellery industry?

Answer: HNO3 (nitric acid).

Question 20. Which acid is used to prepare nitroglycerine?

Answer: Nitric acid is used to prepare nitroglycerine.

Question 21. Which acid is used in the ring test for identification of nitric acid?

Answer: Sulphuric acid.

Question 22. Write down the name and formula of the compound which forms the brown ring in the ring test for identification of nitric acid.

Answer: Pentaaquanitrosyliron (II) sulphate and its formula is [Fe(H2O)5(NO)]SO4.

Question 23. Which acid is used in the preparation of inorganic fertiliser, ammonium sulphate?

Answer: Sulphuric acid (H2SO4).

Question 24. Name the acid which is used in the preparation of HCI(g).

Answer: Sulphuric acid (H2SO4).

Question 25. Give example of a hygroscopic acid.

Answer: Cone. H2SO4

Question 26. Which salt is produced when aluminium reacts with hot and concentrated aqueous solution of NaOH?

Answer: Sodium aluminate (NaAIO2).

Question 27. What is oil of vitriol?

Answer: Sulfuric acid is termed as oil of vitriol.

Chapter 4 Matter Acids Bases And Salts Topic A Concept Of Acid And Bases Fill In The Blanks

Question 1. The formula of hydronium ion is ________

Answer: H3O+

Question 2. A base is a compound which dissociates in water to produce _______ ions.

Answer: OH  

Question 3. Ammonia is a base because it reacts with acids to produce _______

Answer: Salts

Question 4. Fuming nitric acid is a strong ______ agent.

Answer: Oxidising

Question 5. ______ is used in the preparation of picric acid.

Answer: HNO3

Question 6. ________ is used to absorb water vapour.

Answer: H2SO4

Question 7. The base used to prepare synthetic fibre or rayon is _______

Answer: NaOH

Question 8. Copper reacts with hydrochloric acid in presence of _______

Answer: O2

Question 9. The metal which is not used in the preparation of hydrogen is ________

Answer: Cu

Question 10. Freshly prepared solution of _________ is required for the ring test.

Answer: FeSO4

Question  11. The water pollutant which is produced when Cu reacts with hot and cone. HNO3 is ______

Answer: Cu(NO3)2

Question 12. The gas which is produced when Cu reacts with hot and cone. H2SO4 is _______

Answer: SO2

Question 13. A concentrated acid is diluted by gradually adding_______ into _______

Answer: Acid, water

Question 14. A mixture of tartaric acid and sodium bicarbonate undergoes chemical reaction when ________ is added to it.

Answer: Water

Question 15. Hydrochloric acid (HCI) ionises in ________ solution and exhibits its acidic property.

Answer: Aqueous

Question 16. Hydrochloric acid (HCI) is identified by using ______ solution.

Answer: Silver nitrate

Question 17. Hydrochloric acid (HCI) reacts with Zn to produce hydrogen and ________

Answer: ZnCI2

Question 18. The basicity of an acid is the number of replaceable __________ atoms present in one molecule of the acid.

Answer: Hydrogen

Question 19. An example of a tribasic acid is _________

Answer: H3PO4

Question 20. ________ solution is used to treat alkali burns.

Answer: 1% acetic acid

Question 21. Excess secretion of acid causes _________ in stomach.

Answer: Ulcer

Question 22. If oxygen atom is present in acid, the acid is termed as ________ acid.

Answer: Oxy

Question 23. Aluminium hydroxide reacts with NaOH to produce _______ and ________

Answer: Sodium aluminate, water

Chapter 4 Matter Acids Bases And Salts Topic A Concept Of Acid And Bases State Whether True Or False

Question 1. Water soluble metal hydroxides are called alkalis.

Answer: True

Question 2. The compound responsible for the formation of brown ring in the ring test is pentaaquanitrosyliron(ll) sulphate.

Answer: True

Question 3. Prolonged secretion of acid in the stomach causes ulcer.

Answer: True

Question 4. Hydrochloric acid exhibits reducing property while nitric acid exhibits oxidising property.

Answer: True

Question 5. The basicity of H2SO4 is 1.

Answer: False

Question 6. Nitric acid is extensively used in the manufacture of explosives.

Answer: True

Question 7. HCI is the organic acid produced in human body.

Answer: False

Question 8. BaCI2 solution is used to identify HNO3.

Answer: False

Chapter 4 Matter Acids Bases And Salts Topic B Qualitative Concept Of pH Synopsis

  1. The H+ ion concentration of a dilute solution is expressed in terms of pH.
  2. pH is the negative logarithm of H+ ion concentration of a solution. pH of a neutral solution is 7 at 25°C. Any acidic solution has pH less than 7 and alkaline solution has pH greater than 7.
  3. pH-paper is used as a universal indicator to get an idea about the pH value of a solution. The paper shows different colours in solutions of different pH values.
  4. The decay of tooth enamel occurs when the pH level of mouth cavity falls below 5.5.
  5. The pH level of soil or water bodies is increased by addition of quick lime.

Chapter 4 Matter Acids Bases And Salts Topic B Qualitative Concept Of pH Short And Long Answer Type Questions

Question 1. What is meant by pH of a solution?

Answer:

pH of a solution is defined as the negative logarithm (to the base 10) of the molar concentration of H3O+ ions in the solution. pH comes from the German word Potenz de Hydrogen where Potenz means power.

⇒ \(p H=-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\log _{10} \frac{1}{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]}\)

where [H3O+ ] = molar concentration of H3O+ ions in the solution.

Question 2. What idea is obtained about the nature of a solution from its pH value?

Answer:

At 25°C, if the pH of a solution is less than 7, then the solution is acidic in nature. If the pH is greater than 7, then the solution is alkaline in nature whereas if the pH is equal to 7, then the solution is a neutral solution.

Question 3. What Is a pH scale? Mention the range of a pH scale.

Answer:

The scale used to express the acidic or alkaline nature of a solution with respect to the pH of the solution is called a pH-scale.

In general, the H3O+ ion concentration or OH   ion concentration of a dilute solution does not exceed 1mol • L-1.

If the H3O+ ion concentration of a solution is 1 mol • L-1 then its pH will be = \(-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right] =-\log _{10^1} 1=0\)

On the other hand, if the OH ion concentration of a solution is 1 mol • L 1, then its H3O+ ion concentration will be 10-14mol • L-1 (at 25°C \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right] \times\left[\mathrm{OH}^{-}\right]=10^{-14}\)).

Hence its pH will be = \(-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=-\log _{10} 10^{-14}=14\)

Thus, pH of a dilute aqueous solution at 25°C ranges from 0 to 14.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Topic B Qualitative Concept Of pH

Question 4. The H3O+ ion concentration in a solution at 25°C is 10-4  mol • L-1. Find the pOH of the solution.

Answer:

The H3O+ ion concentration in the solution, [H3O+] = 10-4 mol • L-1

∴ pH of the solution = \(-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\) = \(-\log \left(10^{-4}\right)=-(-4)=4\)

At 25°C, pH + pOH = 14

∴ pOH of the solution = 14 – pH = 14 – 4 = 10

Question 5. If the H3O+ ion concentration in a solution is 1 x 10-9 mol• L-1, then find the pH of the solution.

Answer:

The H3O+ ion concentration in the solution, [H3O+ ] = 1 x 10-9 mol • L-1

∴ pH of the solution = \(-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\) = \(-\log \left(1 \times 10^{-9}\right)=-(-9)=9\)

Question 6. The H3O+ ion concentration of four solutions M, N, O and P are 10-2, 10-3, 10-4 and 10-5 • mol • L-1 respectively. Find the pH of the solutions and arrange the solutions in order of their increasing acid strength.

Answer:

The H3O+ ion concentration of four solutions

M, N, O and P are 10-2, 10-3, 10-4 and 10-5 mol • L-1 respectively.

∴ \(pH of solution M =-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\)

= \(-\log _{10}\left(10^{-2}\right)=-(-2)=2\)

∴ \(pH of solution N =-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\)

= \(-\log _{10}\left(10^{-3}\right)=-(-3)=3\)

∴ \(\mathrm{pH} of solution O =-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\)

= \(-\log _{10}\left(10^{-4}\right)=-(-4)=4\)

∴ \(pH of solution P =-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\)

= \(-\log _{10}\left(10^{-5}\right)=-(-5)=5\)

We know that the acidity of a solution increases as the pH of the solution decreases. Thus, increasing order of acid strength of the solutions is P < O < N < M.

Question 7. The pH of a solution increases by 1 unit. What will be the change in H3O+ ion concentration of the solution?

Answer:

Let us consider that the pH of the given solution is 4. Therefore, the H3O+ ion concentration of the solution will be [H3O+] = 10-pH = 10-5 mol • L-1.

Now, the pH of the solution is increased by 1 unit. So, the pH of the solution now becomes 5.

Hence, the H3O+ ion concentration of the solution will be [H3O+] = 10-pH = 10-5 mol • L-1.

Thus, concentration of a dilute aqueous solution becomes l/10th of its initial concentration when pH is increased by 1 unit.

Similarly, the concentration of a dilute aqueous solution becomes 10 times of its initial concentration if the pH is decreased by 1 unit.

Question 8. State whether the pH of a solution can be less than zero or greater than 14. Justify your answer.

Answer:

At 25°C, the pH of a dilute aqueous solution generally ranges from 0 to 14 because the concentration of H3O+ or OH  ions in dilute aqueous solution usually does not exceed 1 mol • L-1.

However, if the concentration of H3O+ ions in a solution exceeds 1 g-ion L-1, then the pH of the solution becomes less than zero. Similarly, if the concentration of OH ions in a solution exceeds 1 g-ion L-1, then the pH of the solution becomes greater than 14.

Question 9. pH of the solutions P, Q and R are 13, 6 and 2 respectively.

  1. which one of these solutions will form NH3 from (NH4)2SO4?
  2. which one is strongly acidic?
  3. which solution will contain both molecules and ions?

Answer:

  1. Solution P will form NH3 from (NH4)2SO4 as it is strongly basic in nature.
  2. Solution R is strongly acidic as depicted by the given pH value.
  3. Solution Q of pH 6 will contain both molecules and ions.

Question 10. What is pH+-paper? How is it used to determine the pH of a solution?

Answer:

  1. A pH-paper is a paper coated with universal indicator which is used to get an idea about the pH of a solution.
  2. To determine the pH of a solution, the pH-paper is dipped into the solution. The paper attains a definite colour depending on the pH of the solution. The colour of the pH-paper is then compared with the standard pH colour strip and consequently, the pH of the solution is determined.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Topic B Qualitative Concept Of pH pH Paper Is Determining pH Solution

Question 11. Mentionb with reason whether the pH of the aqueous solutions of vinger and soda will be grater or less than 7.

Answer:

We know, pH of acidic and basic solution are less and greater than 7 respectively.

Now the main component of vinegar is acetic acid (CH3COOH) which dissociates partially in the aqueous solution forming H3O and CH3COOΘ ion.

\(\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{H}^{\oplus}+\mathrm{CH}_3 \mathrm{COO}^{\ominus}\)

 

Hence the aqueous solution of vinegar will have pH less than 7.

On the other hand the aqueous solution of soda (Na2CO3) becomes alkaline due to hydrolysis.

\(\mathrm{Na}_2 \mathrm{CO}_3+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NaOH}+\mathrm{H}_2 \mathrm{CO}_3\)

 

Hence the aqueous solution of soda will have pH greater than 7.

Question 12. Why does dental erosion take palce?

Answer:

The enamel of our tooth is made of calcium phosphate. During consumption of food, mainly sweets, the food particles get stuck between our teeth. These food particles are decomposed by bacteria present in our mouth to produce acids which decrease the pH level of the mouth cavity.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Topic B Qualitative Concept Of pH DEntal Erosion Take Place

If the pH level falls below 5.5, then the acids react with calcium phosphate and cause erosion of the tooth enamel.

Question 13. Explain why brushing of teeth is essential after having meals?

Answer:

Food particles, especially the carbohydrate particles get stuck to the dental gap and form acids by the bacterial decomposition. As a consequence, pH inside the mouth decreases.

Dental enamel made of calcium phosphate, starts decaying at a pH less than or equal to 5.5. Most of the toothpastes have a pH around 9.0 which neutralise the formed acid at the dental gap and prevents enamel decay. That is why brushing of teeth after meal is essential.

Question 14. Sometimes lime is added to the water bodies during piscicultyure. Why?

Answer:

If the acidity of a water body increases, it affects the flora and fauna of the water body. The plants and fishes of the water body die due to excess acid. So, alkaline substances like quick lime or calcium oxide is added to water.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Topic B Qualitative Concept Of pH Lime Is Added To The Water Bodies During Pisciculture

It neutralises the excess acids and maintains the acid-base balance of the water body. However, it should also be noted that excess use of lime may increase the alkali level of the water body which also adversely affects its ecosystem.

Question 15. Why is lime added to agricultural lands?

Answer:

Due to acid rain and excessive use of inorganic fertilisers like ammonium sulphate, the acidity of soil increases. As a result, the fertility of soil decreases and it becomes unfit for the growth of crops.

To revive the fertility of soil, alkaline substances like lime is added to the soil. Lime neutralises the excess acid present in the soil and makes the soil fertile.

Chapter 4 Matter Acids Bases And Salts Topic B Qualitative Concept Of pH Very Short Abswer Type Questions Choose The Correct Answers

Question 1. The hardest part of our body is

  1. Enamel of teeth
  2. Bones
  3. Skin
  4. Nails

Answer: 1. Enamel of teeth

Question 2. The pH of two solutions, A and B are 3 and 6 respectively. It means that

  1. Solution A is twice as acidic as solution B
  2. Solution B is twice as acidic as solution A
  3. Solution A is 1000 times more acidic than solution B
  4. Solution B is 1000 times more acidic than solution A

Answer: 3. Solution A is 1000 times more acidic than solution B

Question 3. The chemical nature of commonly used toothpastes is

  1. Acidic
  2. Neutral
  3. Alkaline
  4. Amphoteric

Answer: 3. Alkaline

Question 4. For which of the following solutions, pH will be maximum?

  1. 1 mol • L-1 HCI solution
  2. 0.1 mol • L-1 HCI solution
  3. 0.01 mol • L-1 HCI solution
  4. 0.001 mol • L-1 HCI solution

Answer: 4. 0.001 mol • L-1 HCI solution

Question 5. pH of lemon juice is

  1. > 7
  2. = 7
  3. < 7
  4. > 14

Answer: 3. < 7

Question 6. pH of blood is

  1. 4
  2. 5
  3. 7
  4. 7.35
  5. 7.45

Answer: 4. 7.45

Question 7. pH of milk is

  1. 5.5-5.6
  2. 3.4-3.7
  3. 6.5-6.7
  4. 8.5-9.5

Answer: 3. 6.5-6.7

Question 8. pH of a solution is expressed as

  1. -log[H3O+]
  2. -log[OH]
  3. -log[H ]
  4. log [OH]

Answer: 1. -log[H3O+]

Question 9. If the H+ ion concentration of a solution is 10-5 g-ion/litre, then the pH of the solution will be

  1. 10
  2. 5
  3. 3
  4. 7

Answer: 2. 5

Question 10. The pH at which the enamel of teeth begins to decay is

  1. 3.5
  2. 4.5
  3. 5.5
  4. 6.5

Answer: 3. 5.5

Question 11. Due to acid rain, the pH of agricultural lands

  1. Increases
  2. Decreases
  3. Remains unchanged
  4. Increases in some places and decreases in other places

Answer: 2. Decreases

Question 12. The range of the pH scale at 25°C is

  1. 0-10
  2. 0-14
  3. 0-12
  4. 1-14

Answer: 2. 0-14

Question 13. If acid is added to pure water, pH of the medium will

  1. Increase
  2. Decrease
  3. Remain same
  4. Be zero

Answer: 2. Decrease

Question 14. If a litmus paper is dipped in an alkaline solution, its colour will be

  1. Blue
  2. Red
  3. Yellow
  4. Green

Answer: 1. Blue

Question 15. Which of the solutions has the highest pH ?

  1. Caustic soda
  2. Milk of magnesia
  3. Vinegar
  4. Formic acid

Answer: 1. Caustic soda

Question 16. If a solution turns reddish-pink on addition of phenolphthalein, its pH may be

  1. 4
  2. 7
  3. 10
  4. 5

Answer: 3. 10

Question 17. Concept of pH scale was given by

  1. Arrhenius
  2. Sorensen
  3. Rutherford
  4. Lewis

Answer: 2. Sorensen

Question 18. pH of an alkaline solution may be

  1. 5.4
  2. 6.2
  3. 7
  4. 9.2

Answer: 4. 9.2

Question 19. Which is the correct order of pH?

  1. Tomato juice < water < toothpaste
  2. Water < toothpaste < tomato juice
  3. Water < tomato juice < toothpaste
  4. Toothpaste < tomato juice < water

Answer: 1. Tomato juice < water < toothpaste

Question 20. pH of pure water at 25°C is

  1. 5.7
  2. 6.2
  3. 7
  4. 7.5

Answer: 3. 7

Question 21. pH of an acidic solution may be

  1. 9.7
  2. 7
  3. 14
  4. 5.1

Answer: 4. 5.1

Question 22. pH will be lowest for

  1. NaHCO3 solution
  2. HCI solution
  3. KOH solution
  4. Na2CO3 solution

Answer: 2. HCI solution

Question 23. pH of pure water is 7. Which of the following is true for water?

  1. \(\mathrm{H}^{\oplus}+\mathrm{OH}^{\ominus} \rightleftharpoons \mathrm{H}_2 \mathrm{O}\)
  2. \(\mathrm{H}_2 \mathrm{O}+\mathrm{H}^{\oplus} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{\oplus}\)
  3. \(\mathrm{H}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{\oplus}+\mathrm{OH}^{\ominus}\)
  4. \(2 \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_4 \mathrm{O}_2\)

Answer: 3. \(\mathrm{H}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{\oplus}+\mathrm{OH}^{\ominus}\)

Question 24. Favourable pH of water for piscicullture should be

  1. 5.5-6.5
  2. > 8
  3. 7.5
  4. 7

Answer: 1. 5.5-6.5

Question 25. Which of the solution has the highest pH?

  1. 1 mol • L-1 HCI
  2. 0.1 mol • L-1 HCI
  3. 0.01 mol • L-1 HCI
  4. 0.001 mol • L-1 HCI

Answer: 4. 0.00lmol • L-1 HCI

Question 26. A solution turns yellow with addition of methyl orange. pH of the solution is

  1. 7
  2. 4
  3. > 7
  4. < 7

Answer: 3. > 7

Question 27. pH of vinegar is

  1. 4
  2. 5
  3. 7
  4. 7.35-7.45

Answer: 1. 4

Question 28. To reduce the acidity of soil which one of the following should be added?

  1. Bleaching powder
  2. Blue vitriol
  3. Lime
  4. Compost fertiliser

Answer: 3. Lime

Chapter 4 Matter Acids Bases And Salts Topic B Qualitative Concept Of pH Answer In Brief

Question 1. What is the concentration of H ion in pure water at 25°C?

Answer: 10 mo1. L

Question 2. In which solvent between water and benzene, an acid does not undergo ionisation?

Answer: An acid does not undergo ionisation in benzene because it is a non-polar solvent.

Question 3. What is the nature of an aqueous solution having pH = 4?

Answer: Acidic in nature.

Question 4. What is the nature of an aqueous solution having pH = 7?

Answer: Neutral in nature.

Question 5. What is the nature of an aqueous solution having pH = 9?

Answer: Alkaline in nature.

Question 6. What is the value of pH of pure water at 25°C?

Answer: The pH of pure water at 25°C is 7.

Question 7. What is the pH of rain water?

Answer: The pH of rain water is generally 5.6.

Question 8. Odd one out: litmus paper, pH paper, methyl orange, phenolphthalein.

Answer: pH paper.

Question 9. What is the major component of tooth enamel?

Answer: Calcium phosphate [Ca3(PO4)2].

Question 10. Why does acid form at the base of the tooth?

Answer: The residual food particles, especially the sugar particles get stuck to the space between the teeth and are dissociated by bacteria to form acid. Hence the statement.

Question 11. Which substance should be used to make alkaline agricultural land suitable for farming?

Answer: Compost fertilizer is generally used

Chapter 4 Matter Acids Bases And Salts Topic B Qualitative Concept Of pH Fill In The Blanks

Question 1. Scientist _______ introduced the pH scale.

Answer: Sorensen

Question 2. The term pH originates from the German word ________

Answer: Potenz de Hydrogen

Question 3. At 25°C, the maximum pH value of a dilute solution is _______

Answer: 14

Question 4. pH is the negative logarithm of _______ ion concentration in a solution.

Answer: H3O+

Question 5. Blood is slightly ______ in nature.

Answer: Alkaline

Chapter 4 Matter Acids Bases And Salts Topic B Qualitative Concept Of pH State Whether True Or False

Question 1. The pH of an aqueous solution is 3. The solution is alkaline in nature.

Answer: False

Question 2. If the H3O+ ion concentration of an aqueous solution is 1 x 10-5 mol • L-1, then the pH of the solution will be 5.

Answer: True

Question 3. Saliva is acidic in nature.

Answer: True

Question 4. pH of pure water is 7 at any temperature.

Answer: True

Chapter 4 Matter Acids Bases And Salts Topic C Acidic Basic Amphoteric Oxides And Acid Rain Synopsis

  1. Metallic oxides and hydroxides that react with acids to form salt and water are called bases. Water  soluble metallic hydroxides are called . alkalis. Ail metallic oxides and hydroxides are not water soluble. Hence, they are bases but are not termed as alkalis.  Thus, we can say that all alkalis are bases but all bases are not alkalis.
  2. The mixture consisting of 3 volumes of concentrated HCI and 1 volume of concentrated HNO3 is commonly known as aqua regia. It is exclusively used to dissolve noble metals like gold, silver, platinum etc.
  3. Oxides which react with bases to produce salt and water are known as acidic oxides. All non-metallic oxides are generally considered as acidic oxides.
  4. Oxides which react with acids to produce salt and water are known as basic oxides. In general, all metallic oxides are basic oxides.
  5. An amphoteric oxide reacts with both acids and bases to produce salt and water.
  6. When the acid level of rain water exceeds a certain value, it is called acid rain. Acid rain increases the acidity of soil and thus, decreases its fertility. It also adversely affects changing the aquatic plants and animals by increasing the acid level of water bodies.
  7. Due to acid rain, surfaces of buildings, sculptures and monuments made of marbles get covered with layers of CaSO4 making them appear lustreless. With time, the layers come off and produce scars and pits on the marble buildings. These scars are known as stone cancer.

Chapter 4 Matter Acids Bases And Salts Topic C Acidic Basic Amphoteric Oxides And Acid Rain Short And Long Answer Type Questions

Question 1. Chromium trioxide is an acidic oxide though it is a metallic oxide. Explain.

Answer:

Chromium trioxide (CrO3) dissolves in water to produce chromic acid (H2CrO4). Hence, it is an acidic oxide.

\(\mathrm{CrO}_3+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2 \mathrm{CrO}_4\)

Question 2. What are acidic oxides and give examples.

Answer:

The oxides which react with bases to form salt and water are known as acidic oxides. Generally oxides of non-metals are acidic oxides. Some examples of acidic oxides are carbon dioxide (CO2), sulphur dioxide (SO2), phosphorus pentoxide (P2O5) etc.

⇒ \(\mathrm{S}+\mathrm{O}_2 \rightarrow \mathrm{SO}_2 ; \mathrm{SO}_2+2 \mathrm{NaOH} \rightarrow \mathrm{Na}_2 \mathrm{SO}_3+\mathrm{H}_2 \mathrm{O}\)

Question 3. Define basic oxides with examples.

Answer:

The oxides which react with acids to form salt and water are known as basic oxides. Generally oxides of metals are basic oxides. Some examples re calcium oxide (CaO), ferrous oxide (FeO), magnesium oxide (MgO) etc.

\(2 \mathrm{Ca}+\mathrm{O}_2 \rightarrow 2 \mathrm{CaO} ; \mathrm{CaO}+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_2+\mathrm{H}_2 \mathrm{O}\)

Question 4. Define amphoteric oxides with examples.

Answer:

The oxides which react with both acids and bases to form salt and water are called amphoteric oxides. Some examples are, zinc oxide (ZnO), aluminium oxide (Al2O3) etc.

⇒ \(\mathrm{ZnO}+2 \mathrm{HCl} \rightarrow \mathrm{ZnCl}_2+\mathrm{H}_2 \mathrm{O}\)

⇒ \(\mathrm{ZnO}+2 \mathrm{NaOH} \rightarrow \mathrm{Na}_2 \mathrm{ZnO}_2+\mathrm{H}_2 \mathrm{O}\)

Question 5. why is aluminium oxide called an amphoteric oxide?

Answer:

Aluminium oxide reacts with both acids and bases separately to form salt and water.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Topic C Aluminium Oxide Called And Amphoteric Oxide

So, aluminium oxide is known as an amphoteric oxide.

Question 6. Classify the following oxides: \(\mathrm{ZnO}, \mathrm{SO}_2 \text {, }\mathrm{P}_2 \mathrm{O}_5, \mathrm{~K}_2 \mathrm{O}, \mathrm{Al}_2 \mathrm{O}_3\).

Answer:

Acidic oxides: \(\mathrm{SO}_2, \mathrm{P}_2 \mathrm{O}_5\)

Basic oxides: \(\mathrm{Na}_2 \mathrm{O}, \mathrm{K}_2 \mathrm{O}\)

Amphoteric oxides: \(\mathrm{ZnO}, \mathrm{Al}_2 \mathrm{O}_3\)

Question 7. What are acid anhydrides?

Answer:

The oxide of a non-metal (acidic oxide) which dissolves in water to form its corresponding acid is called an acid anhydride. For example, phosphorus pentoxide (P2O5) dissolves in water to form phosphoric acid (H3PO4). Thus, phosphorus pentoxide is the anhydride of phosphoric acid.

⇒ \(\mathrm{P}_2 \mathrm{O}_5+3 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{H}_3 \mathrm{PO}_4\)

Question 8. Define mixed acid anhydride with example.

Answer:

If a non-metallic oxide dissolves in water to produce more than one acid, then the oxide is considered as the anhydride of both the acids and is called a mixed acid anhydride.

For example, NO2 dissolves in cold water to produce nitrous acid (HNO2) and nitric acid (HNO3). Hence, NO2 is a mixed acid anhydride.

⇒ \(2 \mathrm{NO}_2+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{HNO}_2+\mathrm{HNO}_3\)

Question 9. How is HNO3, formed by lightning?

Answer:

N2 and O2 of the atmosphere reacts with each other during lightning to form nitric oxide.

⇒ \(\mathrm{N}_2+\mathrm{O}_2 \stackrel{\text { lightning }}{\longrightarrow} 2 \mathrm{NO}\)

This nitric oxide then combines with oxygen to form nitrogen dioxide. Nitrogen dioxide then reacts with rain water to form nitric acid.

⇒ \(2 \mathrm{NO}+\mathrm{O}_2 \rightarrow 2 \mathrm{NO}_2\)

⇒ \(3 \mathrm{NO}_2+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{HNO}_3+\mathrm{NO}\)

Question 10. What is acid rain?

Answer:

Normal rainwater is slightly acidic. However, if rain water contains excess amount of acids like H2SO4 , HNO3 and HCI, then the pH of rain water ranges between 3.5-5.6. This is known as acid rain.

Question 11. What are the causes of acid rain?

Answer:

Different causes of acid rain can be classified into two categories—natural and man-made causes. Acidic oxides emitted due to natural causes and different human activities mix with the atmospheric water vapour to cause acid rain.

1. Natural Causes:

  1. SO2 gas is released into the atmosphere during volcanic eruptions
  2. Nitrogen and oxygen present in air combine to form different oxides of nitrogen NOx during lightning discharges.
  3. Bacterial decomposition of ammonium salts present in soil also produces oxides of nitrogen (NOx) that are released into the atmosphere.

2. Man-Made Causes:

  1. Different oxides of sulphur and nitrogen are released into the atmosphere due to combustion of fossil fuels like coal and petroleum used in automobiles, thermal power plants, oil refineries, metal extraction industries etc.
  2. Huge amounts of HCI gas is released from factories where hydrochloric acid is extensively used.
  3. These gases react with atmospheric oxygen, ozone and water vapour to produce different acids which float in air in the form of aerosols and come down on the earth’s surface along with rain water.

Question 12. What is stone cancer?

Answer:

Stone Cancer:

Scars and pits formed on the surface of buildings, sculptures, memorials and monuments made of marble due to acid rain are collectively called stone cancer. The scars are formed due to the reaction of the acid with marble (calcium carbonate, CaCO3).

Question 13. Discuss the harmful effects of acid rain.

Answer:

Harmful Effects Caused By Acid Rain

The harmful effects of acid rain are as follows

WBBSE Solutions For Class 9 Physical Science Chapter 4 Topic C Acidic Basic Amphoteric Oxides And Acid Rain Harmful Effects Of Acid Rain

1. Effect On Soil And Vegetation:

Acid rain increases acidity of soil. The increased acidity changes the solubilities of different metal salts in the soil. Consequently, fertility of the soil decreases which reduces the agricultural productivity. Photosynthetic activities of the plants get disrupted due to acid rain. Soil microorganisms and other living organisms of the soil are also adversely affected due to acid rain.

2. Effect On Aquatic Plants And Animals:

Acid rain decreases normal pH of different water bodies which in turn decreases reproductive capacity of fish leading to decresed production of spawns. Excessive increase in the acid level of water kills the flora and fauna of the water body and disturbs the marine ecosystem.

3. Effect On Human Beings:

Due to acid rain, some metals dissolve in rain water to form toxic salts which on entering the human body cause harmful effects. Acid rain damages our skin, hair and body cells. Acids like H2SO4 and HNO3 present in acid rain enter the human body and adversely affect our nervous system, respiratory system and digestion process.

4. Effect on sculptures and monuments: Acid rain causes extensive damage to buildings, memorials, monuments and sculptures made of marble. Marble (CaCO3) reacts with acids and forms insoluble calcium sulphate that deposits on the surfaces of the buildings. This causes scars, pits on the surface and erodes it.

⇒ \(\mathrm{CaCO}_3+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CaSO}_4 \downarrow+\mathrm{CO}_2 \uparrow+\mathrm{H}_2 \mathrm{O}\)

Chapter 4 Matter Acids Bases And Salts Topic C Acidic Basic Amphoteric Oxides And Acid Rain Vert Short Answer Type Questions Choose The Correct Answer

Question 1. An example of an acidic oxide is

  1. CaO
  2. Na2O2
  3. CO2
  4. MgO

Answer: 3. CO2

Question 2. An example of a basic oxide is

  1. CO2
  2. No
  3. CaO
  4. SO

Answer: 3. CaO

Question 3. An example of an amphoteric oxide is

  1. ZnO
  2. CaO
  3. MgO
  4. FeO

Answer: 1. ZnO

Question 4. Acid rain has adversely affected

  1. Taj Mahal
  2. Victoria Memorial
  3. Sculptures made of marbles
  4. All of these

Answer: 4. All of these

Question 5. As a consequence of acid rain, pH of soil

  1. Increases
  2. Decreases
  3. Remains unaltered
  4. Sometime increases, sometime decreases

Answer: 2. Decreases

Question 6. Which of the following can form each of acidic, basic and amphoteric oxide?

  1. S
  2. Pb
  3. Ca
  4. Cr

Answer: 4. Cr

Question 7. The gas released from Mathura oil refinery which is the reason behind the weathering of the Taj Mahal is

  1. CO2
  2. CH4
  3. CO
  4. SO2

Answer: 4. SO2

Question 8. Number of amphoteric oxides among Cr2O3, ZnO, PbO, Al2O3 is

  1. 1
  2. 2
  3. 3
  4. 4

Answer: 4. 4

Question 9. The gas responsible for acid rain is

  1. CH4
  2. NO2
  3. N2O
  4. O3

Answer: 2. NO2

Question 10. Which of the following elements form acidic oxide?

  1. Na
  2. Mg
  3. P
  4. AI

Answer: 3. P

Question 11. N2O5 is

  1. Acidic oxide
  2. Basic oxide
  3. Amphoteric oxide
  4. Neutral oxide

Answer: 1. Acidic oxide

Question 12. Which one is not an acid anhydride?

  1. CO2
  2. CO
  3. N2O5
  4. SO3

Answer: 2. CO

Question 13. Which one of water, carbon dioxide, calcium oxide and nitrogen dioxide is a neutral oxide?

  1. Water
  2. Carbon dioxide
  3. Calcium oxide
  4. Nitrogen dioxide

Answer: 1. Water

Question 14. Depict the number of acidic oxides among \(\mathrm{SO}_2, \mathrm{~N}_2 \mathrm{O}_5, \mathrm{NO}_2, \mathrm{SO}_3, \mathrm{O}_3, \mathrm{CO}_2, \mathrm{H}_2 \mathrm{O},\mathrm{CO}, \mathrm{BaO}_2 and \mathrm{P}_2 \mathrm{O}_5\)?

  1. 6
  2. 8
  3. 4
  4. 7

Answer: 1. 6

Question 15. Which one of the following is a bicompound of oxygen, but not an oxide?

  1. OF2
  2. NO2
  3. SO2
  4. CO2

Answer: 1. OF2

Question 16. Which one of the following is an acidic oxide?

  1. MnO
  2. Mn2O3
  3. MnO2
  4. Mn2O7

Answer: 4. Mn2O7

Chapter 4 Matter Acids Bases And Salts Topic C Acidic Basic Amphoteric Oxides And Acid Rain Answer In Brief

Question 1. Name some metallic oxides which exhibit acidic property in water.

Answer: Manganese heptoxide (Mn2O7), chromium trioxide (CrO3) etc.

Question 2. Give some examples of acidic oxide.

Answer: Some examples of acidic oxide are carbon dioxide (CO2), sulphur trioxide (SO3) etc.

Question 3. Name two acidic metallic oxide.

Answer: Manganese heptoxide (Mn2O7) and chromium trioxide (CrO3).

Question 4. Which type of oxide is P2O5?

Answer: P2O5 is an acidic oxide.

Question 5. Which class of oxides does calcium oxide belong to?

Answer: Calcium oxide belongs to the class of basic oxide.

Question 6. Give some examples of amphoteric oxide.

Answer: Some examples of amphoteric oxide are zinc oxide (ZnO), aluminium oxide (Al2O3) etc.

Question 7. Which acid is formed when carbon dioxide (CO2) gas dissolves in water?

Answer: Carbonic acid (H2CO3).

Question 8. Which acid is formed when nitrogen dioxide (NO2) gas dissolves in water?

Answer: Nitric acid (HNO3) and nitrous acid (HNO2)

Question 9. Which acid is formed when sulphur dioxide (SO2) gas dissolves in water?

Answer: Sulphurous acid (H2SO3).

Question 10. Which gas dissolves in water to produce sulphuric acid?

Answer: Sulphur trioxide (SO3).

Question 11. Give an example of neutral oxide.

Answer: Water (H2O).

Question 12. Give example of a base which is neither metallic oxide nor metallic hydroxide.

Answer: Ammonia (NH3).

Question 13. What is the pH range of acid rain?

Answer: pH range of acid rain is 3.5 to 5.6;

Question 14. Write down the names of two gases responsible for acid rain.

Answer: SO2 and NO2 gas.

Chapter 4 Matter Acids Bases And Salts Topic C Acidic Basic Amphoteric Oxides And Acid Rain Fill In the Blanks

Question 1. The corrosion of marble buildings and monuments caused by acid rain is known as ______

Answer: Stone cancer

Question 2. An amphoteric oxide reacts with both ______ and _______ to produce salts and water.

Answer: Acids, bases

Question 3. ________ oxides are generally acidic in nature.

Answer: Non-metallic

Question 4. ________ oxides are generally basic in nature.

Answer: Metallic

Question 5. An acid present in acid rain is _______

Answer: H2SO4 or HNO3

Question 6. SO2 dissolves in water to produce ______ acid.

Answer: Sulphurous

Question 7. The product formed due to oxidation of SO2 dissolves in water to produce ________ acid.

Answer: Sulphuric

Question 8. ________ is an example of mixed acid is an anhydride.

Answer: NO

Question 9. Aluminium oxide is an example of ________ oxide.

Answer: Amphoteric

Question 10. Fe3O4 is an example of _______ oxide.

Answer: Mixed

Question 11. Example of neutral oxide is ________

Answer: H2O

Chapter 4 Matter Acids Bases And Salts Topic C Acidic Basic Amphoteric Oxides And Acid Rain State Whether True Or False

Question 1. Nitrogen dioxide dissolves in water to form nitric acid only.

Answer: False

Question 2. Acid rain mainly consists of the acids like HCI, H2SO4 and CH3COOH .

Answer: False

Question 3. CO is an acidic oxide.

Answer: False

Question 4. Example of an amphoteric oxide is Al2O3.

Answer: True

Question 5. PbO is an amphoteric oxide.

Answer: True

Question 6. CO is dissolved in acid rain.

Answer: False

Question 7. NO is an acidic oxide.

Answer: False

Chapter 4 Matter Acids Bases And Salts Topic D Neutralisation Indicator Antacids Salts And Their Classification Synopsis

  1. The substances which indicate the end point of an acid-base neutralisation reaction by changing their colour are known as acid-base indicators.
  2. Equivalent amount of an acid reacts quantitatively with equivalent amount of a base to neutralise each other and produce salt and water. This reaction is known as neutralisation reaction.
  3. Antacids are chemical compounds which are basic in nature and neutralise excess hydrochloric acid secreted in the stomach.
  4. The salt formed when all the replaceble hydrogen atoms of an acid are completely replaced by a metal or some other basic radical is known as a normal salt.
  5. The salt formed when the replaceable hydrogen atoms of a dibasic or polybasic acid are partially replaced by a metal or some other basic radical is known as an acid salt.
  6. The salt formed when the hydroxyl group (OH) of a base or an alkali is partially replaced by an acid radical is known as a basic salt.

Chapter 4 Matter Acids Bases And Salts Topic D Neutralisation Indicator Antacids Salts And Their Classification Short And Long Answer Type Questions

Question 1. What are add-base indicators? Name two acid-base indicators.

Answer:

The chemical substances which indicate the end point of an acid-base neutralisation reaction by changingtheir colour are called acid-base indicators.

Two widely used acid-base indicators are methyl orange and phenolphthalein.

Question 2. What changes will be observed in the colour of a litmus paper when it comes in contact with dry ammonia gas and aqueous solution of ammonia?

Answer:

No change will be observed in the colour of litmus paper when it comes in contact with dry ammonia gas. However, the litmus paper turns blue in aqueous solution of ammonia.

Question 3. What is acid-base neutralisation reaction?

Answer:

Equivalent amount of an acid reacts quantitatively with equivalent amount of a base to produce salt and water and as a result, both the acid and the base lose their individual properties.

This reaction is known as acid-base neutralisation reaction. For example, 1 gram-equivalent HCI reacts with 1 gram-equivalent NaOH to produce salt and water. Both HCI and NaOH lose their individual properties in course of this reaction.

⇒ \(\begin{gathered}
\mathrm{HCl}(a q) \rightarrow \mathrm{H}^{+}(a q)+\mathrm{Cl}^{-}(a q) \\
\mathrm{NaOH}\left((a q) \rightarrow \mathrm{Na}^{+}(a q)+\mathrm{OH}^{-}(a q)\right) \\
\hline \mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \rightarrow \mathrm{NaCl}(a q)+\mathrm{H}_2 \mathrm{O}(l)
\end{gathered}\)

Question 4. What is neutralisation point? How is neutralisation point determined?

Answer:

In an acid-base reaction, the point at which the acid and the base quantitatively react with each other to produce salt and water and as a result, the solution becomes neutral is known as neutralisation point.

The neutralisation point of an acid-base reaction is identified by using an indicator. An indicator indicates the neutralisation point by changing its colour.

Question 5. How wilt you differentiate between Na2CO3 and NaHCO3 solution with one drop of an indicator?

Answer:

Both Na2CO3 and NaHCO3 hydrolyse in their aqueous solutions to form NaOH thereby making the resulting solution alkaline. However, the pH value of Na2CO3 solution is greater than that of NaHCO3.

Hence, on addition of one drop of phenolphthalein to Na2CO3 solution, the solution turns pink. On the other hand, if one drop of phenolphthalein is added to NaHCO3 solution, the solution remains colourless.

Hence one drop of phenolphthalein is sufficient to differentiate between Na2CO3 solution and NaHCO3 solution.

Question 6. Mention the characteristic features of indicators used in neutralisation reactions. Mention the criteria for choosing an indicator for neutralisation reaction with example.

Answer:

An acid-base indicator should be such that it can exhibit different colours in acidic, basic and neutral solutions. For example, methyl orange is yellow in alkaline solution, pinkish-red in acidic solution and orange in neutral solution.

An acid-base indicator is chosen for a particular neutralisation reaction on the basis of the change in pH-value of the solution at the equivalence point or end point.

The selected indicator must exhibit a sharp change in colour in the same pH range as required around the end point. For example, in the neutralisation of HCI and NaOH, a sharp change in pH of the solution from 3.34 to 9.7 is observed around the end point.

The pH range of methyl orange is 3.4-4.4. Hence, methyl orange can be used for the neutralisation reaction of HCI and NaOH.

Question 7. Tabulate the colours exhibited by the following indicators in acidic, basic and neutral medium—methyl orange, litmus, phenolphthalein and methyl red.

Answer:

The colours exhibited by the given indicators in acidic, basic and neutral medium are tabulated below

WBBSE Solutions For Class 9 Physical Science Chapter 4 Topic D Neutralisation Indicator Antacids Salts And Their Classificatio Indicator Colors

WBBSE Solutions For Class 9 Physical Science Chapter 4 Topic D Neutralisation Indicator Antacids Salts And Their Classification

Question 8. How is an indicator chosen for an acid- base neutralisation reaction?

Answer:

At the end point of an acid-base neutralisation reaction, there is a rapid and sharp change in the pH of the solution. The indicators which can change their colour in that pH-range are selected for those acid-base neutralisation reactions.

For example, in the neutralisation of a strong acid and a strong base, the range of pH change is maximum near the end point. The pH range of almost all the indicators lies in this range.

Hence, any indicator is suitable for titration of a strong acid and a strong base. On the other hand, in the neutralisation of a weak acid and a weak base, the range of pH change is minimum near the end point.

No suitable indicator having pH range in this region has been found. Hence, the titration of a weak acid and a weak base is not possible.

Question 9. Mention the colour of the solutions if phenolphthalein is added to them.

  1. NaCl
  2. Ca(OH)2
  3. HCI
  4. MgCI2.

Answer:

WBBSE Solutions For Class 9 Physical Science Chapter 4 Topic D Neutralisation Indicator Antacids Salts And Their Classification Phenolphthalein Is Added To Them

Question 10. What are antacids? Give some examples.

Answer:

The substances which are used to neutralise the excess hydrochloric acid secreted in the stomach and maintain the pH of the gastric juice at an optimum level, are known as antacids.

Gelusil, Digene, Diovol are some of the commonly used antacids.

Question 11. Write the compositions of common antacids, Gelusil and Diovol.

Answer:

Gelusil is composed of aluminium hydroxide, magnesium trisilicate and siloxane.

Diovol is composed of aluminium hydroxide, magnesium carbonate and magnesium hydroxide.

Question 12. What are systemic land non-systemic antacids? Give examples of each.

Answer:

Systemic antacids: The antacids which get readily dissolved and absorbed in blood and hence, disturb the acid-base balance of the body are called systemic antacids.

They follow a definite mechanism to reduce acidity. Common examples include sodium bicarbonate, sodium citrate etc.

Non-systemic antacids: Non-systemic antacids are insoluble and are very poorly absorbed in the blood. Hence, they do not disturb the acid-base balance of the body.

They do not follow any definite mechanism of functioning as antacids. Some common examples include aluminium hydroxide, different calcium salts etc.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Topic D Sytemic ANd Non Systemic Antacids

Question 13. What is milk of magnesia? Write its uses and side effects.

Answer:

Milk of magnesia: Aqueous suspension of magnesium hydroxide is known as milk of magnesia. It contains 7-8% magnesium hydroxide.

Uses:

  1. Magnesium hydroxide is an effective antacid.  It rapidly reacts with HCI produced in stomach and neutralises the acid. Hence, it provides quick relief from acidity.
  2. It is often used along with AI(OH)3 as antacid. This provides quick relief from acidity and eliminates any possibility of constipation due to AI(OH)3.

Side Effects:

  1. HCI reacts with Mg(OH)2 to produce MgCl2 which may cause diarrhoea.
  2. Deposition of magnesium in the kidneys can cause toxic effect in the body.

Question 14. Discuss the uses and side effects of sodium bicarbonate as antacid.

Answer:

Uses: Sodium bicarbonate is highly soluble in water and hence, it works rapidly as an antacid. However, it acts for a relatively short duration.

Side Effects:

  1. Sodium bicarbonate produces CO2 in the stomach which may cause uneasiness and infection in the bile duct.
  2. It disturbs the acid-base balance of the body.
  3. It increases the Na+ ion concentration in blood which may cause high blood pressure.

Question 15. Discuss the uses and side effects of aluminium hydroxide as antacid.

Answer:

Uses: Aluminium hydroxide is used to decrease the hyperacidity of stomach. It works slowly and hence, takes time to relieve the pain. Large amount of acid is produced in the stomach due to peptic ulcer, dyspepsia etc.

These result in chest pain, uneasiness, acidity etc. Aluminium hydroxide readily reacts with the excess acid in the stomach and provides relief.

Side Effects:

  1. Excess use of AI(OH)3 may cause constipation.
  2. Prolonged and continuous use of AI(OH)3 may decrease the phosphate level of blood. As a result, the excretion of Cu through urine increases which results in renal rickets.

Question 16. State whether the aqueous solution of all normal salts are neutral in nature.

Answer:

A normal salt is produced due to complete neutralisation of an acid and a base. So, the aqueous solution of a normal salt should be neutral in nature.

However, there are a number of normal salts whose aqueous solutions are either acidic or alkaline in nature. The cations and anions produced from these salts in their aqueous solutions react with water molecules thereby increasing the concentration of H3O+ or OH  ions. As a result, the solution becomes acidic or alkaline accordingly.

Apart from a salt of strong acid and strong base, the solutions of all other normal salts are either acidic or alkaline.

For example, NH4CI is a normal salt of strong acid, HCI and weak alkali, NH4OH . The hydrolysis of this salt produces excess H3O+ ions in water and the aqueous solution becomes acidic.

On the other hand, Na2CO3 is a salt of weak acid, H2CO3 and strong base, NaOH. It hydrolyses to produce excess OH ions in the solution thus, making it alkaline in nature.

Question 17. Baking soda was used in earlier days as antacid. But nowadays Mg(OH)2, AI(OH)3 are used instead. State the reason behind it.

Answer:

Baking soda can disturb the acid-base equilibrium of the body is it can be absorbed by blood. Baking soda can also interfere with how the body absorbs some medications.

Now-a-days, Mg(OH)2 and AI(OH)3 are used instead. These antacids do not destroy the acid- base equilibrium inside the body as they do not get absorbed by blood. These antacids form corresponding salt and water to decrease the acidity by neutralising the excess amount of HCI present in stomach.

Question 18. What is meant by basicity of an acid?

Answer:

The number of replaceable hydrogen atoms present in 1 molecule of an acid or the number of H+ (or H3O+) ions furnished by 1 molecule of the acid in aqueous solution is called the basicity of the acid.

For example, 1 molecule of HCI dissociates in aqueous solution to produce one H+ ion while 1 molecule of H2SO4 dissociates to give two H+ ions. So, the basicity of HCI and H2SO4 are 1 and 2 respectively.

Question 19. What is meant by acidity of a base?

Answer:

The number of OH ions produced by 1 molecule of a base is called the acidity of the base. In other words, it may be defined as the. number of molecules of a monobasic acid required to neutralise 1 molecule of the base quantitatively.

For example, 1 molecule of NaOH furnishes one OH ion in aqueous solution and 1 molecule of monobasic acid, HCI is required for complete neutralisation of 1 molecule of NaOH. Hence, acidity of NaOH is 1.

Question 20. Define normal salt with examples.

Answer:

The salt produced when all the replaceable hydrogen atoms of an acid are completely replaced by a metal or basic radical is known as normal salt. Some examples are sodium chloride (NaCI), zinc sulphate (ZnSO4) etc.

\(\mathrm{Zn}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{ZnSO}_4+\mathrm{H}_2\)

 

A molecule of H2SO4 has two replaceable hydrogen atoms. Zn reacts with sulphuric acid and replaces both the hydrogen atoms forming ZnSO4. Thus, ZnSO4 is a normal salt.

Question 21. Define acidic salts with examples.

Answer:

Acidic Salts :-

The salt produced when the replaceable hydrogen atoms of a polyprotic acid is partially replaced by a metal or basic radical is known as acidic salt. For example, sodium bisulphate (NaHS04), sodium bicarbonate (NaHCO3) etc.

⇒ \(\mathrm{H}_2 \mathrm{SO}_4+\mathrm{NaOH} \rightarrow \mathrm{NaHSO}_4+\mathrm{H}_2 \mathrm{O}\)

H2SO4 has two replaceable hydrogen atoms. NaHSO4 (sodium bisulphate) is produced when one hydrogen atom of sulphuric acid is replaced by Na-atom. Hence, NaHSO4 is an acidic salt.

Question 22. Define basic salts with examples.

Answer:

Basic Salts :-

The salt produced when the replaceable hydroxyl radicals of a base or an alkali are partially replaced by an acid radical is known as basic salt. Some examples of basic salts are basic lead nitrate [Pb(OH)NO3], basic lead chloride [Pb(OH)CI] etc.

Example 23. Explain ‘sodium bicarbonate is an acid salt’.

Answer:

Sodium Bicarbonate Is An Acid Salt :-

Sodium bicarbonate is the salt of strong alkali, NaOH and weak acid.H2CO3. Now H2CO3 is a dibasic acid (\(\mathrm{H}_2 \mathrm{CO}_3 \rightleftharpoons 2 \mathrm{H}^{\oplus}+\mathrm{CO}_3^{2-}\)) and of the two replaceable hydrogens of H2CO3, one H-atom is replaced by Na-atom to form NaHC03. That is why NaHCO3 is an acid salt.

Question 24. HCI and HNO3 always form normal salts but H2SO4 can form both normal and acid salts—Explain why.

Answer:

HCI and HNO3 are monobasic acids. By the replacement of the only replacable H-atom, corresponding normal salts can only be formed. On the other hand, H2SO4 is a dibasic acid. So there is a probability of formation of both acid salt and normal salt.

When one of the two replacable ‘H’ is replaced by metal ion, acid salt is formed and when both of the replaceable hydrogen atoms are replaced by metal atoms, normal salt is formed.

Question 25. Mention the name and formula of a normal salt and an acid salt which are used in everyday life.

Answer:

Normal Salt And An Acid Salt :-

A normal salt used in everyday life is sodium chloride (NaCI) or common salt.

An acid salt used in everyday life is baking powder or sodium bicarbonate (NaHCO3).

Question 26. Aqueous solution of sodium bicarbonate is alkaline although it Is an acid salt—Explain.

Answer:

Sodium bicarbonate dissociates into Na and HCO3Θ  of ion in aqueous solution.

⇒ \(\mathrm{NaHCO}_3 \rightleftharpoons \mathrm{Na}^{\oplus}+\mathrm{HCO}_3^{\ominus}\)

Now in aqueous solution the tendency of loosing a proton (H+) from HCO3Θ ion is less than the tendency to accept a proton. So HCO3 accepts a proton to form undissociated H2CO3.

⇒ \(\mathrm{HCO}_3^{\ominus}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_2 \mathrm{CO}_3+\mathrm{OH}^{\ominus}\)

So the concentration of OHΘ ion increases in the aqueous solution of NaHCO3. As a result the solution becomes alkaline.

Chapter 4 Matter Acids Bases And Salts Topic D Neutralisation Indicator Antacids Salts And Their Classification Very Short Answer Type Questions Choose The Correct Answer

Question 1. An example of an acid-base indicator is

  1. Methyl alcohol
  2. Methyl orange
  3. Methyl chloride
  4. Methylamine

Answer: 2. Methyl orange

Question 2. In neutral solution, the colour of litmus is

  1. Violet
  2. Red
  3. Orange
  4. Yellow

Answer: 1. Violet

Question 3. An indicator which shows pink colour in alkaline solution is

  1. Litmus
  2. Methyl orange
  3. Phenolphthalein
  4. Methyl red

Answer: 3. Phenolphthalein

Question 4. In neutral solution, the colour of methyl red indicator is

  1. Red
  2. Blue
  3. Yellow
  4. Orange

Answer: 4. Orange

Question 5. In acidic solution, the colour of litmus is

  1. Blue
  2. Red
  3. Yellow
  4. Violet

Answer: 2. Red

Question 6. An aqueous solution of sodium carbonate (Na2CO3) is alkaline because it is a salt consisting of

  1. Strong acid and strong base
  2. Weak acid and weak base
  3. Strong acid and weak base
  4. Weak acid and strong base

Answer: 4. Weak acid and strong base

Question 7. An acid salt is produced when sodium hydroxide (NaOH) reacts with

  1. HCI
  2. HNO3
  3. H2SO4
  4. CH3COOH

Answer: 3. H2SO4

Question 8. The indicator used in the titration of a strong acid and a weak base is

  1. Methyl orange
  2. Litmus
  3. Phenolphthalein
  4. Any indicator

Answer: 1. Methyl orange

Question 9. The indicator which is used in the titration of a weak acid and a strong base is

  1. Methyl orange
  2. Litmus
  3. Phenolphthalein
  4. Any indicator

Answer: 3. Phenolphthalein

Question 10. The indicator which is used to differentiate between aqueous solutions of Na2CO3 and NH4CI is

  1. Methyl orange
  2. Litmus
  3. Phenolphthalein
  4. Any indicator

Answer: 3. Phenolphthalein

Question 11. The indicator used in the titration of a strong acid and a strong base is

  1. Methyl orange
  2. Methyl red
  3. Phenolphthalein
  4. Any indicator

Answer: 4. Any indicator

Question 12. The indicator suitable for titration of a weak acid and a weak base is

  1. Methyl orange
  2. Phenolphthalein
  3. Any indicator
  4. No indicator is suitable

Answer: 4. No indicator is suitable

Question 13. Omeprazole is used as a/an

  1. Pain killer
  2. Antipyretic
  3. Drug to decrease acid secretion in the stomach
  4. Analgesic

Answer: 3. Drug to decrease acid secretion in the stomach

Question 14. An example of a systemic antacid is

  1. NaHCO3
  2. AI(OH)3
  3. NaOH
  4. Ca(OH)2

Answer: 1. NaHCO3

Question 15. An example of a non-systemic antacid is

  1. NaHCO3
  2. Al(OH)3
  3. NaOH
  4. Ca(OH)2

Answer: 2. Al(OH)3

Question 16. An example of a normal salt is

  1. Na2HPO4
  2. NaH2PO4
  3. NaH2PO3
  4. Na2SO4

Answer: 4. Na2SO4

Question 17. Which of the following is the major component of milk of magnesia?

  1. Fe(OH)3
  2. Mn(OH)2
  3. Mg(OH)2
  4. Ca(OH)2

Answer: 3. Mg(OH)2

Question 18. An example of an acidic salt is

  1. Na2SO4
  2. NaH2PO2
  3. NaCl
  4. Na2HPO4

Answer: 4. Na2HPO4

Question 19. An example of a basic salt is

  1. Pb(OH)CI
  2. PbCl2
  3. Pb(OH)2
  4. PbO

Answer: 1. Pb(OH)CI

Question 20. The aqueous solution of which of the following salts is acidic?

  1. NH4CI
  2. NaCl
  3. Na2SO4
  4. CH3COONa

Answer: 1. NH4CI

Question 21. The compound used as an antacid

  1. Ca(OH)2
  2. Mg(OH)2
  3. ZnO
  4. NaOH

Answer: 2. Mg(OH)2

Question 22. Which olne of the following is a tribasic acid?

  1. H3PO3
  2. H3PO4
  3. CH3COOH
  4. H3PO2

Answer: 2. H3PO4

Question 23. Baking powder is

  1. Acid
  2. Base
  3. Acid salt
  4. Salt

Answer: 3. Acid salt

Question 24. To neutralise the excess acid of stomach, which one is used

  1. Antibiotic
  2. Antacid
  3. Analgesic
  4. Antiseptic

Answer: 2. Antacid

Question 25. To form basic salt, minimum acidity of a base should be

  1. 1
  2. 2
  3. 3
  4. None of the above

Answer: 2. 2

Question 26. Number of neutral salts among NaH2PO2, Na2HPO3, Na2SO4, Ca(OH)CI is

  1. 1
  2. 2
  3. 3
  4. 4

Answer: 2. 2

Question 27. A component of digestive juices is

  1. H2SO4
  2. HNO3
  3. HCI
  4. HCOOH

Answer: 3. HCI

Question 28. The colour observed by adding a drop of phenolphthalein to an aqueous solution of washing soda is

  1. Red
  2. Blue
  3. Colourless
  4. Pink

Answer: 4. Pink

Question 29. Which of the following compounds is not used as antacid?

  1. Sodium bicarbonate
  2. Calcium carbonate
  3. Sodium hydroxide
  4. Aluminium silicate

Answer: 3. Sodium hydroxide

Question 30. Which of the following compound is present in baking powder?

  1. Na2CO3
  2. H2CO3
  3. NaHCO3
  4. CaCO3

Answer: 3. NaHCO3

Chapter 4 Matter Acids Bases And Salts Topic D Neutralisation Indicator Antacids Salts And Their Classification Answer In Brief

Question 1. Name two acid-base indicators.

Answer: Methyl orange and phenolphthalein are two widely used acid-base indicators.

Question 2. How do indicators indicate the end point of an acid-base neutralisation reaction?

Answer: An indicator indicates the end point of an acid-base neutralisation reaction by changing its colour.

Question 3. Name an indicator which exhibits orange colour in neutral solution.

Answer: Methyl orange.

Question 4. Name two indicators which exhibit yellow colour in alkaline solution.

Answer: Methyl orange and methyl red exhibit yellow colour in alkaline solution.

Question 5. Give an example of an acid which can produce two types of salts by reacting with a base.

Answer: Sulphuric acid (H2SO4) reacts with a base to produce two types of salts (acid salt and normal salt). It partially reacts with NaOH to produce an acid salt, sodium bisulphate (NaHSO4) and on complete neutralisation it produces a normal salt, sodium sulphate (Na2SO4).

Question 6. Name an indicator that can be used for the neutralisation reaction of CH3COOH and NaOH.

Answer: Phenolphthalein.

Question 7. Name an indicator that can be used for the neutralisation reaction of H2SO4 and NaOH.

Answer: Any indicator.

Question 8. Name an indicator that can be used for the neutralisation reaction of CH3COOH and NH4OH.

Answer: No indicator is suitable for the neutralisation reaction of weak acid, CH3COOH and weak base, NH4OH.

Question 9. what will be the colour of the Na2CO3 solution on addition of methyl orange?

Answer: Yellow.

Question 10. Neutralisation reaction is mainly the reaction of which ions?

Answer: H and OHΘ.

Question 11. State the number of molecules of caustic soda required to neutralise 1 molecule of sulfuric acid.

Answer: 2 molecules of caustic soda are required.

Question 12. What is the chemical name of widely used antacid, Zantac?

Answer: The chemical name of widely used antacid Zantac is ranitidine.

Question 13. How does the stomach get affected if acidity continues for a prolonged period of time?

Answer: If acidity continues for a prolonged period of time, it causes ulcer in the stomach.

Question 14. Name the acid salts produced when orthophosphoric acid reacts with NaOH.

Answer: Disodium hydrogen phosphate (Na2HPO4) and sodium < dihydrogen phosphate (NaH2PO4).

Question 15. Name an acid which always forms normal salt when it reacts with bases.

Answer: Hydrochloric acid (HCI).

Question 16. Give the name and formula of an acid salt which is formed due to partial neutralisation of a monobasic acid.

Answer: An acid salt which is formed due to partial neutralisation of a monobasic acid such as, hydrogen fluoride (HF) is potassium bifluoride (KHF2).

Question 17. What is the reason behind ‘alkalosis disease’?

Answer: Alkalosis disease Can be caused by taking excess NaHCO3 as antacid.

Question 18. Give an example of a salt without a metal atom.

Answer: Ammonium chloride (NH4CI).

Question 19. State about the nature of the aqueous solution of NaHSO4.

Answer: The aqueous solution of NaHSO4 is acidic.

Question 20. Write down the names of two main components of antacid.

Answer: Aluminium hydroxide [AI(OH)3] and magnesium hydroxide [Mg(OH)2].

Question 21. Write down the formula of the acid salt obtained from the reaction of phosphoric acid and magnesium hydroxide.

Answer: MgHPO4, (magnesium hydrogen phosphate).

Question 22. What is ‘milk of magnesia’?

Answer: Aqueous suspension of magnesium hydroxide.

Chapter 4 Matter Acids Bases And Salts Topic D Neutralisation Indicator Antacids Salts And Their Classification Fill In The blanks

Question 1. The indicator which turns colourless in acidic solution is ___________

Answer: Phenolphthalein

Question 2. The colours of methyl orange and methyl red change in _______ solution.

Answer: Acidic

Question 3. ________ indicator turns blue in alkaline medium.

Answer: Litmus

Question 4. Universal indicator is actually a _______ of some selective indicators.

Answer: Mixture

Question 5. _______ types of salts are produced when NaOH reacts with H2SO4.

Answer: Two

Question 6.  HNO3 reacts with NaOH to produce _______ type of salt.

Answer: One

Question 7. The name of the salt, CuSO• Cu(OH)2 is ________

Answer: Basic copper sulphate

Question 8. ______ and _______ are produced in an acid-base neutralisation reaction.

Answer: Salt, water

Question 9. The indicators used in acid-base titrations are generally very weak organic _______ or organic _______

Answer: Acid, bases

Question 10. _______ is an example of a systemic antacid.

Answer: Sodium bicarbonate

Question 11. _______ is an example of a non-systemic antacid.

Answer: Aluminium hydroxide

Question 12. _______ is a normal salt of an inorganic base and an organic acid.

Answer: CH3COONa

Question 13. [Cu(NH3)4]SO4 is a _______ salt.

Answer: Complex

Question 14. The colour of the KCI solution will be _________ phenolphthalein is added to it.

Answer: Colourless

Question 15. NaH2PO2 is a ________ salt.

Answer: Neutral

Question 16. The colour of methyl orange in acid solution is _________

Answer: Pink

Question 17. NaHSO4 is an example of _________ salt.

Answer: Acid

Question 18. Sodium bicarbonate is an example of _________ salt.

Answer: Acid

Question 19. Methyl orange turns _______ in neutral solution.

Answer: Orange

Question 20. Acidity of H3PO4 is ________

Answer: 3

Chapter 4 Matter Acids Bases And Salts Topic D Neutralisation Indicator Antacids Salts And Their Classification State Whether True Or False

Question 1. Methyl orange indicator turns colourless in basic solution.

Answer: False

Question 2. Phenolphthalein can be used as an indicator in the neutralisation reaction of CH3COOH and NH4OH.

Answer: False

Question 3. Sodium chloride and zinc sulphate belong to the category of normal salt.

Answer: True

Question 4. Excessive use of aluminium hydroxide as antacid for a long period of time may cause renal rickets.

Answer: True

Question 5. Excessive use of chemical fertilisers on agricultural lands for a long period of time makes the soil alkaline.

Answer: False

Question 6. Sodium carbonate is an alkali.

Answer: F

Question 7. Oxalic acid is a dibasic acid.

Answer: True

Question 8. Sodium surface is an acidic salt.

Answer: False

Chapter 4 Matter Acids Bases And Salts Topic D Neutralisation Indicator Antacids Salts And Their Classification Miscellaneous Type Questions

Match The column

1.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Topic D Neutralisation Indicator Antacids Salts And Their Classification Match The Column 1

Answer: 1. C, 2. A, 3. D, 4. B

2.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Topic D Neutralisation Indicator Antacids Salts And Their Classification Match The Column 2

Answer: 1. B, 2. D, 3. A, 4. C

WBBSE Solutions for Class 9 Physical Science Chapter 4 Matter Nuclear Force

Chapter 4 Matter Nuclear Force Synopsis

Isotopes: Atoms of same elements with the same atomic number but different mass number are called isotopes,

Example: \({ }_1 H^1\) (protium), \({ }_1 H^2\) (deuterium), \({ }_1 H^3\) (tritium).

Properties Of Isotopes: The chemical properties of the isotopes of an element are identical, but the physical properties which depend on the atomic masses are different.

Uses Of Isotopes:

  1. The age of the earth, antiques, fossils, old plants can be. determined with the help of radioactive \({ }_6 C^14\) isotope.
  2. Radioactive isotopes like \({ }_15 P^32\), \({ }_27 Co^60\), \({ }_53 I^131\) etc., are used in the treatment of cancer, goitre, tumors.
  3. Various isotopes are used in the agricultural field and also used for determining the mechanism of various chemical reactions.

Isobars: Atoms which have the same mass number but different atomic numbers are called isobars.

Read and Learn More WBBSE Solutions for Class 9 Physical Science and Environment

Example: \({ }_18 Ar^40\) and \({ }_20 Ca^40\) (tritium).

Properties Of Isobars: Isobars have different physical and chemical properties.

Isotones: Atoms which have the same number of neutrons but different number of protons are called isotones.

Example: \({ }_1 H^3\), \({ }_2 He^4\) and \({ }_6 C^14\), \({ }_7 N^15\), \({ }_8 O^16\) etc.

Properties Of Isotones: Physical and chemical properties of isotones are different.

Nuclear Force: It is a strong attractive force acting between the nucleons present in an atomic nucleus and it holds the nucleons together. It is created by the continuous exchange of meson particles between one proton and one neutron.

Rules For Distribution Of Electrons In Different Orbits:

  1. Maximum number of electrons in a stationary orbit =2n2, where n is the principal quantum number.
  2. Maximum number of electrons in the outermost shell of an atom is 8.
  3. Electrons do not enter to a higher orbit keeping a lower orbit completely empty.

Electronic Configurations Of Some Elements

WBBSE Solutions for Class 9 Physical Science Chapter 4 Matter Nuclear Force

Chapter 4 Matter Nuclear Force Short And Long Answer Type Questions

Question 1. Define isotope with example.

Answer:

Isotope:-

Isotopes (Greek: iso = same, topos = place) are defined as the atoms of the same element having the same atomic number but different mass number due to the presence of different number of neutrons in the nucleus.

Examples:

There are 3 isotopes of hydrogen – protium or ordinary hydrogen (\({ }_1^1 \mathrm{H}\)), deuterium (\({ }_1^2 \mathrm{H}\)) and tritium (\({ }_1^3 \mathrm{H}\)).

There are 3 isotopes of carbon- \({ }_6^{12} \mathrm{C}\), \({ }_6^{13} \mathrm{C}\) and \({ }_6^{14} \mathrm{C}\).

Question 2. Define isobars with example.

Answer:

Isobars:-

Isobars are atoms of different elements having the same mass number but different atomic numbers.

Example: \({ }_{20}^{40} \mathrm{Ca} \text { and }{ }_{18}^{40} \mathrm{Ar}\) are isobars. They have the same mass number (40) but their atomic numbers are different.

Question 3. Define isotones with example.

Answer:

Isotones:-

Isotones are the atoms of different elements having the same number of neutrons but different atomic numbers and mass numbers.

Example: \({ }_6^{14} \mathrm{C},{ }_7^{15} N,{ }_8^{16} \mathrm{O}\) are isotones as they have the same number of neutrons (8) but different atomic numbers and mass I numbers.

Question 4. The constituents of two nuclei are as follows

WBBSE Solutions for Class 9 Physical Science Chapter 4 Matter Nuclear Force The Constituents Of Two Nuclei Number Of Protons And Neutrons

Find The Relation Between A And B.

Answer:

The atomic number of both the atoms A and B is 17.

Mass number of A = number of protons + number of neutrons = 17 + 18 = 35

Mass number of B = number of protons + number of neutrons = 17 + 20 = 37

Hence, A and B are the isotopes of a particular element.

Question 5. Why do isotopes of the same element possess identical chemical properties?

Answer:

Chemical properties of an element depend on its atomic number and the electronic configuration of the element. Isotopes of the same element may have different number of neutrons, but they possess the same number of protons and also have identical electronic configuration.

Thus, isotopes of the same element have identical chemical properties.

Question 6. Write three differences between isotopes and isobars.

Answer:

Differences Between Isotopes And Isobars Are:-

WBBSE Solutions for Class 9 Physical Science Chapter 4 Matter Nuclear Force Three Differences Between Isotopes And Isobars

Question 7. What are the different characteristic features of isotopes?

Answer:

Characteristic Features Of Isotopes Are:-

  1. The isotopes of an element have the same number of electrons and identical electronic configuration. Hence, they possess identical chemical properties.
  2. The atomic masses of isotopes are different. Hence, the physical properties related to mass such as density, melting point, boiling point, rate of diffusion etc., are different for different isotopes.
  3. Due to different nuclear structures, different isotopes may have different nuclear properties. For example, \({ }_6^{12} C\) isotope of carbon is non-radioactive while \({ }_6^{14} C\) isotope of carbon is radioactive.

Question 8. What are the different characteristic features of isobars?

Answer:

Different Characteristic Features Of Isobars Are:-

  1. Isobars are atoms of different elements.
  2. They have different atomic numbers.
  3. They have the same mass number.
  4. They have different physical and chemical properties.
  5. Isobars have different number of protons, neutrons as well as electrons.

Question 9. Can different nuclides be possible for an element?

Answer:

Different Nuclides Are Possible For A Particular Element:-

A nuclide is an atomic species characterised by the specific constitution of its nucleus i.e., by its atomic number and mass number. For different isotopes of an element although mass numbers are different but atomic numbers remain same. In this way different nuclides are possible for a particular element. For example, three nuclides for the three possible isotopes of C are \({ }_6^{12} \mathrm{C},{ }_6^{13} \mathrm{C} \text {, }\) \({ }_6^{14} C\).

Question 10. Which of the following are isotones? \({ }_6^{14} C,{ }_7^{14} N,{ }_9^{19} F,{ }_8^{16} O\)

Answer:

Isotones Have The Same Number Of Neutrons:-

Number of neutrons in \({ }_6^{14} \mathrm{C}\) = 14 – 6 = 8

Number of neutrons in \({ }_7^{14} \mathrm{N}\) = 14 – 7 = 7

Number of neutrons in \({ }_9^{19} \mathrm{C}\)= 19 – 9 = 10

Number of neutrons in \({ }_8^{16} \mathrm{O}\) = 16 – 8 = 8

Hence, \({ }_8^{16} \mathrm{O} \text { and }{ }_6^{14} \mathrm{C}\) are isotones.

Question 11. Describe the structure of nucleus of hydrogen isotopes.

Answer:

The Structure Of Nucleus Of Hydrogen Isotopes:-

Three isotopes of hydrogen are protium (\({ }_1^{1} \mathrm{H}\)), deuterium (\({ }_1^{2} \mathrm{H}\)) and tritium (\({ }_1^{3} \mathrm{H}\)).

Protium nucleus contains one proton but no neutron. Deuterium nucleus consists of one proton and one neutron and tritium nucleus contains one proton and two neutrons.

WBBSE Solutions for Class 9 Physical Science Chapter 4 Matter Nuclear Force Isotopes Of hydrogen

Question 12. Find the number of protons and neutrons in the nuclide \({ }_6^{13} \mathrm{A}\) and write the electronic configuration of the atom. What is the difference between the nuclides \({ }_6^{13} \mathrm{A}\) and \({ }_6^{12} \mathrm{A}\)? Is their chemical property identical?

Answer:

In the nuclide \({ }_6^{13} \mathrm{A}\), number of protons = atomic number = 6.

Number of neutrons = mass number – atomic number = 13 – 6 = 7 .

Number of electrons in the atom = number of protons = 6.

So, the electronic configuration of the atom will be K=2, L = 4.

There are 6 protons and 7 neutrons in the nuclide \({ }_6^{13} \mathrm{A}\). On the other hand, the nuclide \({ }_6^{12} \mathrm{A}\) contains 6 protons and 6 neutrons.

Hence the nuclide \({ }_6^{13} \mathrm{A}\) has 1 neutron more in the nucleus than that in the nuclide \({ }_6^{12} \mathrm{A}\).

Both the nuclides have the same number of protons (they are isotopes). Hence, they will have identical chemical property.

Question 13. What is the relation between F , O2- and Na+ ions? Atomic number of F, O and Na are 9, 8 and 11 respectively.

Answer:

Atomic number of F = 9.

Hence, F-atom contains 9 electrons.

Therefore, number of electrons in F2 ion = 9 + 1 = 10.

Atomic number of O = 8.

Hence, O-atom contains 8 electrons.

Therefore, number of electrons in O2- ion = 8 + 2 = 10 .

Atomic number of Na = 11.

Hence, Na-atom contains llelectrons.

Therefore, number of electrons in Na+ ion = 11 – 1 = 10 .

All these ions contain the same number of electrons.

These are called isoelectronic species.

Question 14. What is nuclear force? How does this force originate?

Answer:

Nuclear Force:-

The powerful attractive force that acts among the nucleons (protons and neutrons) at a very short range (at a distance of about 1.5 fermi) inside the nucleus is known as nuclear force. It is responsible for holding the nucleons together within the nucleus.

In 1935, scientist Hideki Yukawa proposed ‘meson theory’ to explain the origin of nuclear force. According to this theory, protons and neutrons in the nucleus constantly exchange a small particle called n-meson. Due to the exchange of these meson particles, protons get converted to neutrons and vice-versa.

As a result, the repulsion between protons become insignificant and a strong attractive force is created between the nucleons. This is how the nuclear force originates.

Question 15. Mention the different characteristics of nuclear force.

Answer:

Different Characteristics Of Nuclear Force Are:-

  1. Nuclear force is a strong attractive force that holds the nucleons within a close space inside the nucleus.
  2. The nature of nuclear force is quite different from that of gravitational force or electrostatic force of attraction.
  3. Nuclear force is almost 1040 times stronger than gravitational force while it is 100 times stronger than coulombic force.
  4. Nuclear force is charge independent. The nature of forces acting between two protons or two neutrons or between a proton and a neutron are identical.
  5. This is a very short range force. The force acts within a range of approximately 1 x 10-15 m within the nucleus. As a result, the force does not exist outside the nucleus.
  6. With the help of this force, each nucleon gets attracted only by its nearest neighbouring nucleons.

Question 16. What is Bohr-Bury scheme?

Answer:

Bohr-Bury Scheme:-

  1. The Bohr-Bury scheme explains the arrangement of electrons in different orbits around the nucleus. This arrangement is known as electronic configuration. The Bohr-Bury rules regarding electronic configuration of an atom in different orbits are as follows
  2. The n -th orbit of an atom accommodates a maximum of 2n2 electrons where ‘n’ represents the principal quantum number.
  3. The maximum possible number of electrons in the outermost orbit of an atom is 8 irrespective of the principal quantum level.
  4. An electron does not occupy a higher energy orbit while a lower energy orbit is left vacant.

WBBSE Solutions for Class 9 Physical Science Chapter 4 Matter Nuclear Force Bohr-Bury Scheme

Question 17. Is it possible to write the electronic configuration of an atom from the atomic number of the element? Justify your answer.

Answer:

Yes, it is possible to write the electronic configuration of an atom if the atomic number of the element is known.

An atom is electrically neutral and the total positive charge is equal to the total negative charge. Hence, the number of protons in an atom is equal to the number of electrons.

Thus, if atomic number of the element is known, the number of protons as well as number of electrons will be known and hence we can write the electronic configuration of the atom.

Question 18. The atomic number of sodium is 11. Write the electronic configuration of unipositive ion of sodium (Na+).

Answer:

The atomic number of sodium is 11. Hence, there are 11 electrons in a sodium atom and its electronic configuration is 2,8,1. Na atom forms Na+ ion by releasing an electron from its valence shell. Hence, the electronic configuration of Na+ ion is 2,8.

Question 19. Write the electronic configuration of uninegative ion of chlorine (Cl).

Answer:

The atomic number of chlorine is 17. Hence, there are 17 electrons in a Cl -atom and its electronic configuration is 2,8,7. Cl-atom forms Cl ion by accepting an electron and number of  electrons in its valence shell increases by 1. Thus, the electronic configuration of Cl- ion is 2,8,8.

Question 20. Maximum number of electrons that can be accommodated in M-orbit is 18 but the electronic configuration of potassium (atomic number = 19) cannot be written as 2,8,9. Why?

Answer:

The outermost orbit of an atom cannot actommodate more than 8 electrons. So, M orbit being the outermost orbit can accommodate a maximum of 8 electrons. Thus, the 19th electron of potassium atom enters into N orbit.

Therefore, electronic configuration of potassium atom is 2,8,8,1 and not 2,8,9.

Question 21. The valence shell of an atom is M and it contains 2 electrons. Find the atomic number of the atom and give the name of the element.

Answer:

The valence shell is M and it contains 2 electrons. As the inner orbits are completely filled with electrons, the electronic configuration of the atom is K = 2,L = 8,M = 2 i.e., 2,8,2.

Hence, the number of electrons of the element = number of protons =2 + 8 + 2 = 12.

The atomic number of the element is 12. Hence, the element is magnesium (Mg).

Question 22. How is the chemical reactivity of an element related to its electronic configuration? Explain with example.

Answer:

The chemical reactivity of an element is solely dependent on its electronic configuration.

For example, elements having completely filled outermost orbit are chemically inert.

He (K-2), Ne (K = 2, L = 8), Ar (K = 2, L = 8 . M = 8) have their outermost orbits completely filled with electrons.

Hence, they do not take part in chemical reactions.

Elements having 1, 2, 6 or 7 electrons in their outermost orbit of their atoms are relatively more reactive than other elements. This is why, sodium (K=2, L = 8,M = 1), calcium (K = 2 , L = 8 , M = 8 , N = 2), oxygen (K = 2, L = 6) or chlorine (K = 2, L = 8, M = 7) are chemically very reactive.

Question 23. What happens if the electron of outermost orbit acquires sufficient energy to overcome the attraction of nucleus?

Answer:

If the electron of outermost orbit acquires sufficient energy to overcome the attraction of nucleus, it will move out of the orbit and the atom will be converted to the corresponding cation. The minimum energy required by the outermost electron to overcome the attraction of nucleus is known as the ionisation energy.

Question 24. M3+ ion contains 10 electrons and 14 neutrons. Find the atomic number and mass number of the element M. What is the name of the element?

Answer:

The number of electrons in the M3+ ion is 10 Here the atom forms a tri-positive ion. So, the number of electrons lost by the atom is 3.

Hence, the number of electrons in M-atom = 10 + 3 = 13.

Therefore, number of protons in M-atom = 13.

So, the atomic number of the element M = 13.

The number of neutrons in M-atom = 14.

Therefore, mass number of the atom = number of protons + number of neutrons = 13 + 14 = 27.

As the atomic number of the element is 13, the element is aluminium (Al).

Quetsion 25. Write the differences between an atom and an ion.

Answer:

The Differences Between An Atom And An Ion Are As Follows:-

WBBSE Solutions for Class 9 Physical Science Chapter 4 Matter Nuclear Force Differences Between An Atom And An Ion

Question 26. Number of protons in the nucleus of Ar and Ca are 18 and 20 respectively. Again number of neutrons in those atoms are 22 and 20 respectively. What is the relation between them?

Answer:

Mass number of Ar = 18 + 22 = 40 Mass number of Ca = 20 + 20 = 40

∴  The nuclides are \({ }_{18} \mathrm{Ar}^{40}\) and \({ }_{20} \mathrm{Ca}^{40}\).

The two nuclides have the same mass number but their atomic numbers are different. Hence they are isobars.

Question 27. Write down the differences between isotones and isobars.

Answer:

The Differences Between Isotones And Isobars Are As Follows:-

WBBSE Solutions for Class 9 Physical Science Chapter 4 Matter Nuclear Force Differences Between Isotones And Isobars

Question 28. State rationally whether chemical properties of the following nuclides are same or different: \({ }_{Z-1}^A M \text { and }{ }_{Z-1}^{A+1} N\)

Answer:

For the two nuclides, atomic numbers are same, i.e., (Z- 1) but mass numbers are different which indicates that they are isotopes. Hence they possess identical electronic configuration as well as similar chemical properties .

Question 29. Calculate the number of electrons in \({ }_{92} U^{235}\). How will you write the atom if it contains three more neutrons? What will be the relation between the two mentioned atom?

Answer:

  1. In \({ }_{92} U^{235}\), number of protons = 92. Hence, the number of electrons = 92 .
  2. If there is 3 more neutrons, the nuclide will be = \({ }_{92}^{238} U\)
  3. \({ }_{92} U^{235}\) and \({ }_{92}^{238} U\) are isotopes of element U.

Question 30 If \({ }_b^a X \text { and }{ }_d^c Y\) are isotopes to each other, prove that (a2 + c2)(b2 – d2) = 0.

Answer:

Since, \({ }_b^a X \text { and }{ }_d^c Y\) are isotopes to each other, so b = d or, b2 = d2 or, (b2 – d2) = 0

or, (a2 + c2)(b2 – d2) = 0 x (a2 + c2) [(a2 + c2) ≠ 0]

or, (a2 + c2)(b2 – d2) = 0 (Proved)

Question 31. How will you express the nucMtte formed by removal of 3 neutrons from \({ }_{92} U^{238}\)?What will be the number of neutrons in the new nuclide? Depict the relation between the old and new nuclides.

Answer:

Mass no. of the new nuclide = 238 – 3 = 235

∴ The new nuclide will be expressed as \({ }_{92} U^{235}\)

Number of neutrons in \({ }_{92} U^{235}\)=235-92 =143

⇒ \({ }_{92} U^{235}\) and \({ }_{92} U^{238}\) are two isotopes of uranium.

Question 32. If \({ }_a X^c,{ }_b Y^d\) are isotopes and \(m^{A^n}, p^{B^q}\) are isobars to each other, calculate (a2-b2)(n-q) = ?

Answer:

Since \({ }_a X^c,{ }_b Y^d\) are isotopes to each other, so a = b or, (a – b) = 0

Since \({ }_m A^n \text { and }{ }_p B^q\) are isobars to each other, so n = q or, (n – q) = 0

Now, (a2 – b2)(n – q) = (a + b)(a – b)(n – q) = (a + b) x 0 x 0 = 0

Question 33. Mass numbers of the atoms X, Y and Z are 31, 32 and 34 respectively. Their atomic numbers are 15,16 and 16 respectively. Identify the isotopes and isotones among them.

Answer:

Corresponding nuclides are \({ }_{15} X^{31},{ }_{16} Y^{32}\) and \({ }_{16} Z^{34}\).

Atomic numbers of Y and Z are equal but they have different mass numbers.

∴ Y and Z are isotopes.

Again no. of neutrons in \({ }_{15} X^{31}\) = 31 – 15 = 16

and no. of neutrons in \({ }_{16} Y^{32}\) = 32 – 16 = 16

∴ Number of neutrons are same in X and Y but their mass numbers are different.

∴ X and Y isotones.

Question 34. An atom contains 2 electrons in outermost M orbit. What are the atomic number and name of the element?

Answer:

The outermost orbit of the atom is M and it contains 2 electrons.

∴ Electronic configuration will be 2,8,2.

∴ Atomic no. of the element = total no. of protons = total no of electrons = 2 + 8 + 2 = 12

Since atomic number is 12, the corresponding element is Mg, i.e., magnesium.

Question 35. An element contains two electrons in its valence N shell, identify the element.

Answer:

Electronic configuration of the element is K(2), 1(8), M(8), N(2).

∴ Atomic number of the element = 2 + 8 + 8 + 2 = 20.

∴ The corresponding element is calcium (20Ca).

Question 36. Write down the electronic configuration of \({ }_{8} O^{16}\) and \({ }_{12} Mg^{24}\).

Answer:

Electronic configuration of 8O16 is K(2), L(6).

Electronic configuration of 12Mg24 is K(2), L(8), M(2).

Question 37. Atomic number of an element is 20. Write down its electronic configuration. What is the valency of the element?

Answer:

Electronic configuration of the element = K(2) , L(8), M(8), N(2)

Since the outermost N orbit of the element contains 2 electrons, the valency of the element is 2.

Question 38. Which one is more stable in between atoms and ions and why?

Answer:

Ions are more stable than atoms. When an atom gets converted to the corresponding ion it attains a stable electronic configuration similar to that of the nearest inert gas. For example, electronic configuration of Cl -atom is K(2), 1(8), M(7) and electronic configuration of ClΘ ion is K(2), L(8), M(8).

Hence, the electronic configuration of ClΘ ion is identical to the electronic configuration of Ar-atom. Therefore, ClΘ is more stable than Cl -atom.

Question 39. Write down the relation between the nuclides: \({ }_A P^B \text { and }{ }_A Q^{B+1}\) and \({ }_A+1 R^{B+1}\)

Answer:

For \({ }_A P^B \text { and }{ }_A Q^{B+1}\), atomic numbers are equal while mass numbers are different.

∴ \({ }_A P^B \text { and }{ }_A Q^{B+1}\) are isotopes to each other.

For \({ }_A Q^{B+1}\) and \({ }_A+1 R^{B+1}\), mass numbers are equal while atomic numbers are different.

∴ \({ }_A Q^{B+1}\) and \({ }_A+1 R^{B+1}\) are isobars to each other.

For \({ }_A P^B\) and \({ }_A+1 R^{B+1}\), number of neutrons are equal to (B-A).

∴ \({ }_A P^B\) and \({ }_A+1 R^{B+1}\) are isotones to each other.

Question 40. Which one of K atom and K ion acts violently with water and why?

Answer:

Electronic configuration of K -atom = 2, 8, 8, 1.

This K-atom loses one electron to form K ion, electronic configuration of which is similar to that of noble gas Ar. Due to this, K ion is more stable than K-atom. Hence, K-atom reacts more violently with water to form potassium hydroxide and hydrogen gas.

Questiobn 41. Write down the name of a cation and an anion whose electronic configuration is similar to that of Ne.

Answer:

Electronic configuration of Ne = 2,8.

∴ Cation and anion with similar electronic configurations are \({ }_{11} \mathrm{Na}^{\oplus} \text { and }{ }_9 \mathrm{~F}^{\ominus}\) respectively.

Chapter 4 Matter Nuclear Force Very Short Answer Type Questions Choose The Correct Answer

Question 1. \({ }_1^2 \mathrm{H} \text { and }{ }_2^3 \mathrm{He}\) are

  1. Isotope
  2. Isobars
  3. Isotones
  4. None of these

Answer: 3. Isotones

Question 2. The number of isotopes of oxygen is

  1. 3
  2. 4
  3. 1
  4. 2

Answer: 1. 3

Question 3. An element which has no isotope is

  1. Carbon
  2. Chlorine
  3. Fluorine
  4. Oxygen

Answer: 3. Fluorine

Question 4. The electronic configuration of Ca2+ ion is

  1. 2, 8, 8
  2. 2, 8, 8, 2
  3. 2, 8, 8, 1
  4. 2, 8, 7

Answer: 1. 2, 8, 8

Question 5. The isotope of carbon which is used for radiocarbon dating is

  1. \({ }_6^{12} \mathrm{C}\)
  2. \({ }_6^{13} \mathrm{C}\)
  3. \({ }_6^{14} \mathrm{C}\)
  4. \({ }_6^{15} \mathrm{C}\)

Answer: 3. \({ }_6^{14} \mathrm{C}\)

Question 6. Isotopes of an element possess

  1. Identical physical properties
  2. Different chemical properties
  3. Different number of neutrons
  4. Different atomic numbers

Answer: 3. Different number of neutrons

Question 7. Isotopes contain equal number of

  1. Neutrons
  2. Protons
  3. Positrons
  4. Mesons

Answer: 2. Protons

Question 8. Isobars have identical

  1. Atomic number
  2. Number of protons
  3. Number of electrons
  4. Mass number

Answer: 4. Mass number

Question 9. Isotones have identical

  1. Number of electrons
  2. Number of protons
  3. Number of neutrons
  4. Number of positrons

Answer: 3. Number of neutrons

Question 10. For isotopes, a slight difference is observed in case of their

  1. Actual mass
  2. Atomic number
  3. Number of electrons
  4. Number of protons

Answer: 1. Actual mass

Question 11. Protons and neutrons of an atom are collectively known as

  1. Positron
  2. Isotope
  3. Nucleon
  4. Isobar

Answer: 3. Nucleon

Question 12. Which of the following is exchanged between protons and neutrons for the generation of nuclear force

  1. α -particle
  2. β-particle
  3. Positron
  4. Meson

Answer: 4. Positron

Question 13. The range within which nuclear force acts is

  1. 1.5 fermi
  2. 2.5 fermi
  3. 3.5 fermi
  4. 4.5 fermi

Answer: 1. 1.5 fermi

Question 14. The number of electrons in Cl ion is

  1. 19
  2. 18
  3. 20
  4. 15

Answer: 2. 18

Question 15. Maximum number of electrons that can be accommodated in the valence shell of an atom is

  1. 18
  2. 10
  3. 2
  4. 8

Answer: 4. 8

Question 16. Which of the following noble gases does not contain 8 electrons in its valence shell?

  1. Ne
  2. Ar
  3. Xe
  4. He

Answer: 4. He

Question 17. The atoms and ions with equal number of electrons are known as

  1. Isotopes
  2. Isobars
  3. Isotones
  4. Isoelectronic

Answer: 4. Isoelectronic

Question 18. Number of electrons in S2- ion

  1. 15
  2. 16
  3. 17
  4. 18

Answer: 4. 18

Question 19. The element which does not contain 8 electrons in the outermost orbit

  1. Ar
  2. Ne
  3. Pd
  4. None of these

Answer: 3. Pd

Question 20. Mg2+, O2-, Ne, F have identical

  1. Mass number
  2. No. of protons
  3. No. of electrons
  4. No. of neutrons

Answer: 3. No. of electrons

Question 21. Number of electrons in the outermost orbit of \({ }_{15}^{31} p\) is

  1. 2
  2. 3
  3. 5
  4. 6

Answer: 3.5

Question 22. From which orbit of Al, electrons are removed to form Al3+ ?

  1. K
  2. L
  3. M
  4. N

Answer: 3. M

Question 23. Difference in the number of electrons in L and M orbits of Na atom is

  1. 8
  2. 7
  3. 9
  4. 6

Answer: 2. 7

Question 24. Which pair of the following is isoelectronlc?

Na+, O

Na+, Mg2+

He+, K+

Mg2+,O

Answer: 2. Na+, Mg2+

Question 25. Relation between \({ }_{15}^{31} P\) and \({ }_{16}^{32} S\) is

  1. Isotope
  2. Isobar
  3. Isotone
  4. Isomer

Answer: 3. Isotone

Question 26. Electronic configuration of 11Na is

  1. 1, 2, 8
  2. 2, 1, 8
  3. 8, 2, 1
  4. 2, 8, 1

Answer: 4. 2, 8, 1

Question 27. Which one of the following has equal number of electrons to that of Be2+?

  1. H+
  2. He
  3. He2+
  4. Na+

Answer: 2. He

Question 28. Which one of the following is an isotope of hydrogen?

  1. Helium
  2. Deuterium
  3. Beryllium
  4. Calcium

Answer: 2. Deuterium

Chapter 4 Matter Nuclear Force Answer In Brief

Question 1. Who discovered radioactivity?

Answer: French scientist Henri Becquerel.

Question 2. Identify the isotopes and isobars from the following

  1. 8p + 8n,
  2. 8p + 9n,
  3. 18p + 22n,
  4. 20p + 20n (p = proton, n = neutron).

Answer: 1 and 2 have the same number of protons. So, they are isotopes of the same element, 3 and 4 have the same mass number. Hence, they are isobars.

Question 3. Which isotope of carbon is radioactive?

Answer: \({ }_6^{14} C\) isotope of carbon is radioactive.

Question 4. What similarities are observed In the isotopes of an element?

Answer: All the isotopes of an element have the same atomic number and identical chemical properties.

Question 5. What similarities are observed in isobars?

Answer: Isobars have the same mass number.

Question 6. What similarities are observed in isotones?

Answer: Isotones contain same number of neutrons in their nucleus.

Question 7. Exchange of which particle between protons and neutrons is responsible for the generation of nuclear force?

Answer: Exchange of meson particles .

Question 8. Who explained the formation of nuclear force?

Answer: Scientist Yukawa.

Question 9. What is meant by valence shell of an atom?

Answer: The outermost orbit of an atom is known as the valence shell of that atom.

Question 10. Name the inert gas which has two electrons in its valence shell.

Answer: Helium (He) is the inert gas which has two electrons in its valence shell.

Question 11. State the maximum number of electrons that can be accommodated by the valence shell of an atom.

Answer: The maximum number of electrons that can be accommodated by the valence shell of an atom is 8.

Question 12. What is meant by electronic configuration of an atom?

Answer: In a multi-electron atom, the definite arrangement of electrons in different orbits around the nucleus is known as its electronic configuration.

Question 13. What is the maximum possible number of electrons that can be accommodated by the M-orbit of an atom?

Answer: The M-orbit of an atom can accommodate a maximum of 18 electrons.

Question 14. Who was the first to observe the difference in number of neutrons in different atoms of the same element (or the existence of isotopes)?

Answer: Scientist Soddy.

Question 15. Which isotope of hydrogen is radioactive in nature?

Answer: Tritium (\({ }_1^3 \mathrm{H}\)).

Question 16. Which isotope of hydrogen is present in heavy water?

Answer: Deuterium (\({ }_1^2 \mathrm{H}\)) is present in heavy water (D20).

Question 17. Which isotope of chlorine is abundantly found in nature?

Answer: Chlorine has two isotopes, \({ }_{17}^{35} \mathrm{Cl} \text { and }{ }_{17}^{37} \mathrm{Cl}\), out of which \({ }_{17}^{35} \mathrm{Cl}\) is abundantly found in nature (almost 75%).

Question 18. Which isotope of oxygen is abundantly found in nature?

Answer: Oxygen has three isotopes, \({ }_8^{16} O,{ }_8^{17} O\) and \({ }_8^{18} O\), out of which \({ }_8^{16} O\) is abundantly found in nature.

Question 19. Give an example of an element which has no naturally occurring isotope.

Answer: Elements such as fluorine (F) and sodium (Na) have no naturally occurring isotope.

Question 20. Give an example of an isotope in which the number of protons is greater than the number of neutrons.

Answer: In an isotope of helium, \({ }_2^3 \mathrm{He}\) there are two protons and one neutron, i.e., in case of this than the number of neutrons.

Question 21. Which element has the maximum number of stable isotopes?

Answer: Tin (Sn) has the maximum number (10) of stable isotopes.

Question 22. What is the charge of an Na+ ion?

Answer: The charge of an Na+ ion is +1.602 x 10-9 coulomb (it is equal to the charge of an electron but is opposite in nature).

Question 23. What is the charge of an O2- ion?

Answer: The charge of an O2- ion is -2 x 1.602 x 10-19 coulomb or -3.204 x 10-19 coulomb.

Question 24. From which orbit of Al electrons are removed to form Al3+ ion?

Answer: From M-orbit (outermost orbit for Al) of Aluminium 3 electrons are removed to form Al3+ ion.

Question 25. Give an example of isobar of \({ }_7 N^{14}\).

Answer: \({ }_6 C^{14}\)

Question 26. What is the relation between F, O2- and Na ions?

Answer: F, O2- and Na ions are isoelectronic.

Question 27. How does the energy of an orbit vary with increasing distance from nucleus?

Answer: Energy of an orbit increases with the increase in distance from the nucleus.

Question 28. An atom has three orbits. The third orbit contains 5 electrons. What is its atomic number?

Answer: Hence, the electronic configuration of the atom is K(2), 1(8), M(5).

∴ Its atomic number is (2 + 8 + 5) = 15.

Question 29. If \({ }_b^d X \text { and }{ }_d Y\) are isotopes to each other, write down the relation between b and d.

Answer: b = d

Question 30. How many electrons are there in the M- shell of potassium (Z = 19) ?

Answer: Electronic configuration of potassium is K(2), L(8), M(8), A/(l). So number of electron in M-shell is 8.

Question 31. Write down the electronic configuration of S ion.

Answer: K(2), L(8), M(7).

Question 32. Which subatomic particle is responsible for formation of isotope?

Answer: Neutron.

Question 33. Calculate the number of electrons in Mg2+ ion.

Answer: Atomic number of Mg = 12 .

∴ Number of electrons in Mg2+ ion = 12-2 = 10.

Question 34. Write down the electronic configuration of Na+ ion.

Answer: Electronic configuration of 11Na+ is K(2), L(8).

Question 35. What is the possible maximum number of electrons in the n-th orbit of an atom?

Answer: Possible maximum number of electrons in the n-th orbit of an atom is 2n2.

Question 36. What is formed when electrons are removed from an atom?

Answer: Positively charged ion or cation is formed when electrons are removed from the outermost shell of an atom.

Chapter 4 Matter Nuclear Force Fill in the Blanks

Question 1. Isotopes are identical with respect to their ______ properties.

Answer: Chemical

Question 2. Isobars have the same ________ number.

Answer: Mass

Question 3. Isotones contain equal number of ________

Answer: Neutrons

Question 4. Nuclear force operates within a range of ______ m.

Answer: 1.5 x 10-15

Question 5. Nuclear force originates due to the exchange of __________ particles between protons and neutrons.

Answer: Meson

Question 6. Most of the naturally occurring elements exist as a mixture of two or more of their ______

Answer: Isotopes

Question 7. Isotopes of an element have the ________ electronic configuration.

Answer: Same

Question 8. Electronic configurations of isobars are ________

Answer: Different

Question 9. Except helium, all other noble gases contain __________ electrons in their valence shell.

Answer: 8

Question 10. The filling of electrons in their respective orbits obeys the _________ rule.

Answer: Bohr Burry

Chapter 4 Matter Nuclear Force State Whether True Or False

Question 1. The n-th orbit around the nucleus can accommodate a maximum of 2n2 electrons.

Answer: True

Question 2. Isotopes are atoms of the same element with the same atomic number but different mass numbers.

Answer: True

Question 3. The M-orbit of an electron can accommodate a maximum of 8 electrons.

Answer: False

Question 4. Atomic numbers of elements P and Q are 18 and 20 respectively whereas their mass number is 40. P and Q are isotones.

Answer: False

Question 5. Ionisation energy is the minimum amount of energy required by an electron to overcome the attraction of the nucleus.

Answer: True

Question 6. Nuclear force is a long range force.

Answer: False

Question 7. Outermost orbit of neon does not contain 8 electrons.

Answer: False

Question 8. \({ }_{14}^{30} S e,{ }_{15}^{31} p,{ }_{16}^{32} S\) are isobars to each other.

Answer: False

Question 9. Isoelectronic anion of Na+ is F.

Answer: True

Question 10. Actual mass of isobars are not same.

Answer: True

Question 11. Number of neutrons in isotopes are same.

Answer: False

Question 12. Chemical properties of isobars are same.

Answer: False

Chapter 4 Matter Nuclear Force Miscellaneous type Questions

Match The Columns

1.

WBBSE Solutions for Class 9 Physical Science Chapter 4 Matter Nuclear Force Match The Column 1

Answer: 1. D, 2. C, 3. B, 4. A

2.

WBBSE Solutions for Class 9 Physical Science Chapter 4 Matter Nuclear Force Match The Column 2

Answer: 1. C, 2. A, 3. D, 4. B

3. 

WBBSE Solutions for Class 9 Physical Science Chapter 4 Matter Nuclear Force Match The Column 3

Answer: 1. C, 2. A, 3. D, 4. B

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Separation Of The Components Of Mixture

Chapter 4 Matter Separation Of The Components Of Mixture Topic A Necessity Of Separation Of The Components Of A Mixture And Purification Of Petroleum Synopsis

  1. A homogeneous mixture is one in which the mixture has the same proportion of the components throughout and has uniform properties at all parts of the mixture.
  2. To separate useful substances from a mixture or to eliminate unwanted or harmful substances from a mixture, the components of a mixture need to be separated. Different methods applied for separation of mixtures are—crystallisation, distillation, fractional distillation, sublimation, filtration, chromatography etc.
  3. Crude petroleum is a liquid mixture of hydrocarbons containing 1-40 carbon atoms (C1– C40). Considering its immense importance in our everyday life, petroleum is often referred to as ‘liquid gold’. Crude petroleum is refined by the process of fractional distillation.

Read and Learn More WBBSE Solutions for Class 9 Physical Science and Environment

The Four Main Parts Obtained From Fractional Distillation Of Petroleum Are:-

  1. Crude naphtha,
  2. Kerosene or paraffin oil,
  3. Fuel oil or diesel,
  4. Residual oil.

Chapter 4 Matter Separation Of The Components Of Mixture Topic A Necessity Of Separation Of The Components Of A Mixture And Purification Of Petroleum Short And Long Answer Type Questions

Question 1. Define mixture. Classify the mixture into different types.

Answer:

Mixture:-

A mixture is the physical combination of two or more substances combined in any possible proportion by mass, in which the identities of the components are retained.

Mixtures may be classified in two types

  1. Homogeneous mixture and
  2. Heterogeneous mixtures.

Question 2. What is a homogeneous mixture? Give example.

Answer:

Homogeneous Mixture:-

  1. A homogeneous mixture is one in which the components of the mixture has the same proportion throughout and has uniform properties at all parts of the mixture.
  2. Aqueous solution of sugar, aqueous solution of alcohol etc. are homogeneous mixtures.

Question 3. Define heterogeneous mixtures.

Answer:

Heterogeneous Mixtures:-

A heterogeneous mixture is a mixture with a non uniform composition. The composition varies from one region to another with at least two phases that remain separate from each other, with clearly identifiable properties.

Example: mixture of sand and water, mixture of chalk-dust and water. Concrete is a heterogeneous mixture of cement and water.

Question 4. Write the differences between homogeneous mixtures and compounds.

Answer:

Differences Between Homogeneous Mixtures And Compounds:-

The major differences between homogeneous mixtures and compounds are as follows

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Separation Of The Components Of Mixture Topic A Differences Between Homogeneous Mixtures And Compound

Question 5. Why is it necessary to separate the components of a mixture?

Answer:

It Is Necessary To Separate The Components Of A Mixture Because:-

  1. Separation methods are necessary for the removal of unwanted and harmful components from a mixture. For example, removal of impurities like soil particles, dirt etc. from water makes it suitable for drinking.
  2. Useful components can be obtained from a mixture by applying different separation techniques. For example, petrol, diesel, kerosene, etc. are obtained from crude petroleum by fractional distillation.

Question 6. What is petroleum? in which process its components can be separated?

Answer:

Petroleum:-

  1. Petroleum is a naturally occurring sticky liquid with characteristic smell, found beneath the earth’s surface and is a mixture of hydrocarbons containing 1 to 40 number of carbon atoms.
  2. In petroleum, a mixture of hydrocarbons with boiling points in the range of 20°C-400°C are present. That is why to separate different components of it, fractional distillation is used.

Question 7. Why is it necessary to separate the components of petroleum?

Answer:

Yes It is Necessary To Separate The Components Of Petroleum:-

Petroleum is a mixture of different hydrocarbons containing 1 to 40 carbon atoms. It is the major source of different fuels like petrol, diesel, kerosene and a number of useful chemicals like lubricating oil, bitumen, paraffin wax etc.

These substances are widely used in different household activities as well as in industries. Thus, refining of crude petroleum i.e., separation of the components of petroleum is necessary.

Question 8. Show how the gaseous components of air are separated with the help of a flowchart.

Answer:

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Separation Of The Components Of Mixture Topic A Gaseous Components Of Air Separated Help Of Flow Chart

Question 9. Mention name and uses of different substances that can be obtained from petroleum.

Answer:

Different substances obtained at different temperature in the process of fractional distillation of crude petroleum are mentioned below

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Separation Of The Components Of Mixture Topic A Name And Uses Of Different Substances Petroleum

Chapter 4 Matter Separation Of The Components Of Mixture Topic A Necessity Of Separation Of The Components Of A Mixture And Purification Of Petroleum Very Short Answer Type Questions Choose The Correct Answer

Question 1. During fractional distillation of crude petroleum into its constituents, petrol distils at

  1. 70°C
  2. 120°C
  3. 250°C
  4. 400°C

Answer: 1. 70°C

Question 2. The correct order of density of

(1)Ether,

(2)Petrol and

(3) Kerosene is

  1. (1) < (3) < (2)
  2. (1) < (2) > (3)
  3. (3) > (2) > (1)
  4. (1) > (2) > (3)

Answer: 3. (3) > (2) > (1)

Question 3. Which of the following is a homogeneous mixture?

  1. Water and oil
  2. Water and alcohol
  3. Water and sand
  4. Water and benzene

Answer: 2. Water and alcohol

Question 4. The component of crude petroleum which gets distilled at the lowest temperature is

  1. Bitumen
  2. Refinery gas
  3. Naphtha
  4. Kerosene

Answer: 2. Kerosene

Question 5. The component of crude petroleum which gets distilled at the highest temperature is

  1. Bitumen
  2. Diesel
  3. Gasoline
  4. Kerosene

Answer: 1. Bitumen

Question 6. Which of the following is not obtained due to fractional distillation of petroleum?

  1. Kerosene
  2. Naphtha
  3. Bitumen
  4. Coal gas

Answer: 4. Coal gas

Question 7. The component of petroleum which is not used as a fuel is

  1. Petrol
  2. Naphtha
  3. Kerosene
  4. Diesel

Answer: 2. Naphtha

Question 8. The temperature at which LPG is obtained during fractional distillation of petroleum is

  1. Below 30°C
  2. Above 30°C
  3. Above 50°C
  4. 95°-100°C

Answer: 1. Below 30°C

Question 9. Which of the following is obtained from the fractional distillation of petroleum

  1. Polythene
  2. Ammonia
  3. Diesel
  4. Teflon

Answer: 3. Diesel

Question 10. Diesel is obtained by fractional distillation of petroleum at

  1. 70°C
  2. 170°C
  3. 270°C
  4. 370°C

Answer: 3. 270°C

Question 11. Which one of the following is not a homogeneous mixture?

  1. Mixture of water and common salt
  2. Mixture of sugar and water
  3. Mixture of water and alcohol
  4. Blood

Answer: 4. Mixture of water and alcohol

Question 12. Which one of the following is not a heterogeneous mixture?

  1. Blood
  2. Milk
  3. Muddy water
  4. Air

Answer: 4. Air

Chapter 4 Matter Separation Of The Components Of Mixture Topic A Necessity Of Separation Of The Components Of A Mixture And Purification Of Petroleum Answer In Brief

Question 1. Between compounds and mixtures, the components of which one can be separated by simple physical processes?

Answer: The components of a mixture can be separated by simple physical processes.

Question 2. Give an example of a homogeneous mixture.

Answer: An aqueous solution of sugar.

Question 3. A mixture consists of nearly ground wheat flour and sugar. Different parts of the mixture are tasted but the sweetness of the mixture is not uniform throughout the mixture. What may be the reason?

Answer: The sweetness of the mixture is not uniform throughout the mixture because it is a heterogeneous mixture.

Question 4. Which process is used in the refining of crude petroleum?

Answer: Fractional distillation.

Question 5. Is it possible to separate the constituents of petroleum by distillation?

Answer: No, it is not possible to separate the constituents of petroleum by distillation.

Question 6. What is the basis of separation of the components of crude petroleum? Or, By which physical property the components of petroleum can be differentiated?

Answer: The different components of crude petroleum are separated on the basis of their difference in boiling points by fractional distillation method.

Question 7. Name a solid substance obtained from the fractional distillation of petroleum.

Answer: Paraffin.

Question 8. What is petroleum gas?

Answer: The hydrocarbon filtrate part containing 1 to 4 carbon atoms, obtained by fractional distillation of petroleum at 20°C to 30°C is termed as petroleum gas.

Question 9. What is the main component of petroleum gas?

Answer: Butane.

Question 10. What is asphalt?

Answer: The black, sticky residue obtained from fractional distillation of petroleum is termed as asphalt.

Question 11. How is gasoline generally known to us?

Answer: Gasoline is known as petrol.

Question 12. Arrange in ascending order of boiling point—petrol, diesel, lubricating oil and kerosene.

Answer: Petrol < kerosene < diesel < lubricating oil.

Question 13. Name some liquid fuels obtained from petroleum by fractional distillation.

Answer: Petrol or gasoline, diesel and kerosene.

Question 14. What is obtained at lowest temperature in the refinement of petroleum?

Answer: Petroleum gas is obtained at lowest (20°C) temperature.

Chapter 4 Matter Separation Of The Components Of Mixture Topic A Necessity Of Separation Of The Components Of A Mixture And Purification Of Petroleum Fill In The Blanks

Question 1. The component of petroleum which is used in construction of roads is ________

Answer: Asphalt

Question 2. The other name of petrol is ________

Answer: Gasoline

Question 3. The main component of petroleum gas is _________

Answer: Butane

Question 4. ________ gas is used as household fuel for cooking.

Answer: Petroleum

Question 5. Petroleum is purified by

Answer: Fractional distillation

Chapter 4 Matter Separation Of The Components Of Mixture Topic A Necessity Of Separation Of The Components Of A Mixture And Purification Of Petroleum State Whether True Or False

Question 1. Naphtha obtained from fractional distillation of petroleum is used as a fuel.

Answer: False

Question 2. Bitumen is obtained as residue in the fractional distillation of petroleum.

Answer: True

Question 3. Milk of magnesia is an example of homogeneous mixture.

Answer: False

Question 4. Coal tar is obtained by the destructive distillation of bituminous coal.

Answer: True

Question 5. Crude petroleum oil can be used as fuel.

Answer: False

Question 6. Petroleum consists of more than 150 types of hydrocarbons.

Answer: True

Question 7. Kerosene is used as a solvent of DDT.

Answer: True

Chapter 4 Matter Separation Of The Components Of Mixture Topic B Distillation Fractional Distillation Separatory Funnel Synopsis

  1. The boiling point of a liquid increases when pressure above the surface of the liquid increases, i.e., the liquid boils at a higher temperature compared to its normal boiling point.
  2. The process of distillation is applied to separate the components of a mixture consisting of a non-volatile solid dissolved in a liquid. It is used mainly to separate the liquid component. For example, iodine and chloroform can be separated from their mixture by distillation. This process is extensively used in the purification of substances by solvent extraction method.
  3. The components of a homogeneous mixture consisting of two liquids can be separated by simple distillation if (1)The difference in boiling points of the two liquids is greater than 30°C and (2) The liquids do not decompose or react with each other on heating. Thus, separation of benzene (boiling point =80°C) and aniline (boiling point = 184°C) from their mixture may be done by simple distillation.
  4. When a mixture of two liquids with different boiling points is heated at a constant pressure, the relative proportion of the more volatile component is greater in the formed vapour than the relative proportion of that component present in the liquid mixture. If the difference in boiling points of the liquids is more than 30°C, the vapour phase mostly contains the more volatile component (i.e. liquid with the lower boiling point) when the liquid mixture is heated around the boiling point of the more volatile component.
  5. The liquids separated by simple distillation are not pure. On heating, the more volatile component gets preferentially vapourised. However, it also contains a small fraction of the less volatile component which gets condensed and mixed with the more volatile liquid.
  6. The components of a homogeneous mixture of two or more miscible liquids can conveniently be separated by fractional distillation if the difference in their boiling points is less than 10°C. For example, acetone (boiling point =56°C) and methanol (boiling point =65°C) are separated from their mixture by the process of fractional distillation.
  7. In fractional distillation, the vapour produced on heating the liquid mixture will contain the more volatile component in greater amount than the other components. The liquid with the lowest boiling point distils first and the liquid with the highest boiling point distils at the end. This process is very useful for separation of benzene from coal tar; oxygen, nitrogen and other gases from liquid air and also in refining of crude petroleum.
  8. During fractional distillation, the relative proportion of the more volatile component in the vapour produced by heating the liquid mixture, increases as it moves up along the fractionating column. The condensed part of the vapour that comes down as liquid in the column, contains mostly the less volatile component.
  9. A mixture of liquids which boils with constant composition at a given temperature and pressure is known as an azeotropic mixture. It is possesses the same composition of components in both liquid phase and vapour phase. A mixture of 95.6% ethyl alcohol and 4.4% water is an example of azeotropic mixture. On heating, the entire mixture changes into the vapour phase at constant temperature. As a result, components of an azeotropic mixture cannot be separated by fractional distillation.
  10. A separatory funnel is used to separate the components of a mixture consisting of two or more immiscible liquids having different densities. This funnel is useful in the separation of kerosene and water, water and mercury, chloroform and water from their mixtures.

Chapter 4 Matter Separation Of The Components Of Mixture Topic B Distillation Fractional Distillation Separatory Funnel Short And Long Answer Type Questions

Question 1. What is distillation?

Answer:

Distillation:-

The process by which a mixture consisting of a liquid and dissolved solid substances is vaporised by applying heat, followed by condensing the vapour back into the liquid state, thereby separating the components of the mixture is called distillation.

Question 2. What is the working principle of distillation?

Answer:

The Working Principle Of Distillation:-

The process of distillation is quite useful for separating the components of a mixture consisting of two miscible liquids, the difference in their boiling points being greater than 30°C. This process is also effective in separating the components from a mixture consisting of a liquid and non-volatile solid impurities.

When the solution is heated around the boiling point of the more volatile component, the vapour phase formed contains the more volatile component in a relatively higher proportion. The vapour is then condensed to the corresponding liquid and collected in a beaker.

When the temperature is raised to the boiling point of the less volatile component, the vapour phase formed contains the less volatile component in relatively higher proportion because the more volatile component has almost completely distilled off at a lower temperature.

Thus in distillation, the liquids distill off from the mixture at their respective boiling points as vapours and the vapours are then condensed to liquid phase and collected in beakers. The most volatile component distills off first while the least volatile component distills off at the end.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Separation Of The Components Of Mixture Topic B Principle Of Fractional Distillation

Question 3. Which type of mixtures are separated by distillation?

Answer:

Types Of Mixtures Are Separated By Distillation Are Given Below:-

Distillation is used to separate a homogeneous mixture of a solid dissolved in a liquid and to separate the components of a mixture consisting of two miscible liquids if the difference in boiling points of the liquids is in the range of 30-50°C and the liquids do not decompose on heating.

Question 4. What is Liebig condenser?

Answer:

The Liebig condenser or straight condenser is a piece of laboratory equipment, consisting of a straight glass tube surrounded by a water jacket in which water is constantly circulated to carry away the heat of vaporisation released by the condensing vapour.

Question 5. Give an example of a natural distillation process.

Answer:

Sea water evaporates leaving behind the salts in the sea. Water vapours obtained as a result of such evaporation go upward and form clouds. From these clouds when water comes down to earth as rain, it does not contain any salt in it. Therefore the water cycle may be cited as an example of natural distillation process.

Question 6. What is vaccum distillation? When this process is applied?

Answer:

Vacuum distillation is distillation performed under reduced pressure, which allows the purification of compounds not readily distilled at ambient pressures or simply to save time or energy. This process allows the components to be distilled at a lower temperature than their boiling points.

Vacuum distillation is ideal for separating mixture of liquids with very high boiling points. The lowering of pressure enables the components to boil at lower temperature.

Once the vapour pressure of the component is equal to the surrounding pressure, it is converted to vapour, which are then condensed and collected as distillate. The vacuum distillation is also used to obtain high-purity samples of compounds that decompose at high temperatures.

Question 7. What is fractional distillation?

Answer:

The process of separation of a mixture consisting of two miscible liquids having a difference of boiling point less than 25°C, by repeated condensation and vapourisation in a fractionating column is known as fractional distillation.

Question 8. Write down the basic principle of fractional distillation.

Answer:

Fractional distillation is used to separate a mixture of two or more miscible liquids which boil without decomposition and for which the difference in boiling points is less than 25°C.

If a mixture of such kind is heated, the liquid with the least boiling point evaporates first and its vapour s transformed to pure liquid by condensation. By increasing the temperature slowly, other components of the mixture are obtained in their pure form.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Separation Of The Components Of Mixture Topic B Working Principle Of Distillation

Question 9. Under which conditions, fractional distillation is used instead of simple distillation?

Answer:

Simple distillation method is not suitable for separation of components of a mixture containing two or more miscible liquids having a difference in boiling point less than 25°C. The more volatile component vapourises when the mixture is heated near its boiling point.

However, as the difference in boiling point is small, other less volatile components also vapourise in small amounts and deposit in the receiving flask. Thus, complete separation of the components of the mixture does not take place by simple distillation. So, fractional distillation is done in which the mixture by repeated vaporisation and condensation in the fractionating column separates into its components.

Question 10. State important applications of fractional distillation.

Answer:

  1. Crude petroleum is refined by fractional distillation to get different components like petrol, diesel, kerosene, lubricating oil etc.
  2. Different gaseous components (oxygen, nitrogen etc.) are separated from liquid air by the process of fractional distillation.
  3. Benzene is separated from coal tar, using fractional distillation.
  4. From a mixture of acetone and methyl alcohol components are separated using fractional distillation.

Question 11. How will you separate benzene (bp 80°C) and toluene (bp 110°C) form a mixture?

Answer:

Benzene and toluene can be separated from their mixture by fractional distillation.The mixture should be taken in a distillation flask and thermometer, fractionating column, condenser etc. should be attached as per standard procedure. The flask should be hosted in an oil bath. AT 80°C boiling starts and the temperature remains constant.

At this temperature benzene vapour passes through the fractionating column and the condenser, condensed to liquid form and collected. After complete distillation of benzene, temperature of the mixture starts to increase. At 110°C toluene starts to vapourise and after passing through the fractionating column and the condenser, collected as liquid toluene.

Question 12. What are the limitations of fractional distillation?

Answer:

The process of fractional distillation cannot be used to separate the components of an azeotropic mixture (constant boiling mixture). For example, a mixture of 95.6% ethyl alcohol and 4.4% water cannot be separated by fractional distillation because it forms a constant boiling mixture or azeotropic mixture which boils at a constant temperature of 78.4°C.

Question 13. What is azeotropic mixture?

Answer:

Azeotropic mixture is the mixture of miscible liquids that has a constant boiling point because the vapour has the same composition as the liquid mixture. The boiling point of an azeotropic mixture may be higher or lower than that of any of its components.

Question 14. Why a mixture of water and ethyl alcohol can not be separated completely by fractional distillation?

Answer:

While concentrating a mixture of ethanol and water through fractional distillation, a mixture of 95.6% ethanol and 4.40% water (by weight) is obtained at some point. On heating, at 78.4°C the whole liquid starts boiling and distilled in the same composition.

In fact, at that composition ethanol and water forms an azeotrope. That is why complete separation of ethanol or ethyl alcohol and water from their mixture can not be done through fractional distillation.

Question 15. How is the boiling point of a liquid related to the pressure above its surface? Write important applications of this property.

Answer:

1. The boiling point of a liquid increases if the pressure above its surface increases. Thus, the liquid boils at a higher temperature than its normal boiling point.

Application:

This property of liquid is applied in pressure cookers.

2. The boiling point of a liquid decreases if the pressure above its surface decreases. Thus, the liquid boils at a lower temperature than its normal boiling point.

Application:

Condensed milk is prepared by using this property. Milk is condensed by boiling it at a low temperature by decreasing the pressure.

Question 16. Write down the effect of boiling point of liquids.

Answer:

Boiling point of liquids depend on the pressure exerted on the liquid. As the pressure increases, boiling point of liquids also increases and decreases with decrease in pressure.

Question 17. The components of which type of mixtures can be separated by using separatory funnel?

Answer:

A mixture of two immiscible liquids having appreciable difference in their densities can be separated by using a separatory funnel. Due to difference in the densities of the constituent liquids, if the mixture is kept undisturbed for some time, it forms two distinct layers. This property is used to separate the components of the liquid mixture by using a separatory funnel.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Sepation Of The Components Of Mixture Topic B Mixtures Can Be Separated By Using Separatory Funnel

Question 18. Give two examples where separatory funnel is used to separate the components of a mixture.

Answer:

  1. Separatory funnel is used to separate the components of a mixture of two immiscible liquids like oil and water, benzene and water, chloroform and water etc.
  2. Separatory funnel is used to extract an organic compound dissolved in water by solvent extraction process.

Question 19. How water and carbon tetrachloride are separated from their mixture?

Answer:

The mixture of water and carbon tetrachloride is poured and kept for sometime in a separatory funnel. Carbon tetrachloride forms the lower layer and water forms the upper layer in the separatory funnel. Carbon tetrachloride is collected first by opening the stopcock and then water is collected in a separate vessel.

Question 20. Is it possible to separate water and alcohol from their mixture with the help of a separatory funnel?

Answer:

Liquids which are completely miscible with each other and forms a homogeneous mixture cannot be separated from their mixture with the help of a separatory funnel. Water and alcohol completely mix with each other in all proportions to form a homogeneous mixture. So, separatory funnel cannot be used to separate water and alcohol from their mixture.

Question 21. Complete the following table.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Sepation Of The Components Of Mixture Topic B Fractional Distillation Separatory Funnel

Answer:

Completing the following table by filling up the blanks, we get

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Sepation Of The Components Of Mixture Topic B Fractional Distillation Separatory Funnel

Question 22. Differentiate between distillation and fractional distillation.

Answer:

Difference between distillation and fractional distillation

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Sepation Of The Components Of Mixture Topic B Differenc Distillation And Fractional Distillation

Question 23. Write down differences between fractional distillation and separation through separatory funnel.

Answer:

The differences between fractional distillation and separation through separatory funnel are

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Sepation Of The Components Of Mixture Fractional Distillation Separation Through Separatory Funnel

Chapter 4 Matter Separation Of The Components Of Mixture Topic B Distillation Fractional Distillation Separatory Funnel Very Short Answer Type Questions Choose The Correct Answer

Question 1. Which of the following is not a compound?

  1. Benzene
  2. Water
  3. Petroleum
  4. Toluene

Answer: 3. Petroleum

Question 2. The process of distillation can be used to separate the

  1. Constituents of a mixture of water and salt
  2. Constituents of a mixture of water and oil
  3. Constituents of a mixture of water and sand
  4. Constituents of a mixture of sugar and salt

Answer: 1. Constituents of a mixture of water and salt

Question 3. Which of the following mixtures cannot be l separated into its constituents by using a separatory funnel?

  1. Water and mercury
  2. Petrol and water
  3. Cchloroform and water
  4. Ethanol and water

Answer: 4. Ethanol and water

Question 4. A fractionating column is used for the process of

  1. Filtration
  2. Distillation
  3. Vapourisation
  4. Fractional distillation

Answer: 4. Fractional distillation

Question 5. enzene and toluene can be separated from their mixture by

  1. Filtration
  2. Distillation
  3. Fractional distillation
  4. Using separatory funnel

Answer: 3. Fractional distillation

Question 6. Which of the following process of separation of a mixture does not require heating?

  1. Distillation
  2. Sublimation
  3. Using separatory funnel
  4. Fractional distillation

Answer: 3. Using separatory funnel

Question 7. Water can be boiled at room temperature by

  1. Decreasing the pressure
  2. Increasing the pressure
  3. Initially increasing the pressure and then decreasing it
  4. Initially decreasing the pressure and then increasing it

Answer: 1. Decreasing the pressure

Question 8. During fractional distillation of two miscible liquids, the distilled vapour produced on heating the mixture is rich in

  1. Less volatile liquid
  2. More volatile liquid
  3. Both the liquids
  4. None of the liquids

Answer: 2. More volatile liquid

Question 9. Which of the following is required in solvent extraction method?

  1. Filter paper
  2. Funnel
  3. Separatory funnel
  4. Beaker

Answer: 3. Separatory funnel

Question 10. Benzene and nitrobenzene can be separated from their mixture by

  1. Fractional distillation
  2. Filtration
  3. Simple distillation
  4. Sublimation

Answer: 3. Simple distillation

Question 11. The glass apparatus which is used to separate the components of a mixture consisting of two or more immiscible liquids is called

  1. Fractionating column
  2. Condenser
  3. Separatory funnel
  4. Buchner funnel

Answer: 3. Separatory funnel

Question 12. A salt can be extracted from its aqueous solution by

  1. Sublimation
  2. Filtration
  3. Distillation
  4. Fractional distillation

Answer: 3. Distillation

Question 13. For every 27 mm increase in atmospheric pressure, the boiling point of water increases by

  1. 1°C
  2. 2°C
  3. 3°C
  4. 4°C

Answer: 1. 1°C

Question 14. The boiling point of water at Darjeeling will be

  1. More than 100°C
  2. Less than 100°C
  3. Equal to 100°C
  4. Cannot be determined

Answer: 2. Less than 100°C

Question 15. The process of separation of components from the mixture of chloroform (boiling point 61°C) and benzene (boiling point 80°C) is

  1. Distillation
  2. Fractional distillation
  3. Separating funnel
  4. Filtration

Answer: 2. Fractional distillation

Question 16. From which of the following mixtures, components cannot be separated by fractional distillation?

  1. Water + ethyl alcohol (bp: 78.2°C)
  2. Water + kerosene (bp: 170°C)
  3. Acetone (bp:56°C)+ methanol (bp: 65°C)
  4. Water + benzene (bp: 80°C)

Answer: 1. Water + ethyl alcohol (bp: 78.2°C)

Question 17. The process by which the components of a mixture of ethanol and water can be separated is

  1. Distillation
  2. Fractional distillation
  3. Sublimation
  4. None of the above

Answer: 4. None of the above

Question 18. The process by which the components of a mixture of ammonium chloride and sand can be separated is

  1. Distillation
  2. Fractional distillation
  3. Sublimation
  4. Chromatography

Answer: 3. Sublimation

Question 19. The process used to separate the gaseous components of air is

  1. Distillation
  2. Fractional distillation
  3. Sublimation
  4. Separating funnel

Answer: 2. Fractional distillation

Question 20. Water can be boiled at 25°C if the standard atmospheric pressure is

  1. Reduced
  2. Increased
  3. Increased at first and then reduced
  4. Reduced at first and then increased

Answer: 1. Reduced

Question 21. The process of separation of a liquid and a non-volatile solid from their mixture is

  1. Fractional distillation
  2. Sublimation
  3. Distillation
  4. Crystallisation

Answer: 3. Distillation

Question 22. The process used to prepare salt from sea water is

  1. Distillation
  2. Evaporation
  3. Fraction distillation
  4. Filtration

Answer: 2. Evaporation

Question 23. Apparatus used in fractional distillation is

  1. Condenser
  2. Liebig’s condenser
  3. Fractional column
  4. 2 and 3 both

Answer: 4. 2 and 3 both

Question 24. Which one of the following cannot be obtained by the fractional distillation of crude petroleum?

  1. Petrol
  2. Diesel
  3. Ethanol
  4. Kerosene

Answer: 3. Ethanol

Question 25. How will you separate the components of a mixture of iodine and ethanol ?

  1. Filtration
  2. Evaporation
  3. distillation
  4. Fractional distillation

Answer: 3. distillation

Question 26. Distillation column is used in

  1. Distillation
  2. Fractional distillation
  3. Condensation
  4. Separatory funnel

Answer: 3. Condensation

Question 27. By which process water and carbon tetrachloride can be separated from their mixture?

  1. Distillation
  2. Fractional distillation
  3. Evaporation
  4. Separatory funnel

Answer: 4. Separatory funnel

Question 28. The components of a mixture of water and kerosene, can be separated by

  1. Distillation
  2. Fractional distillation
  3. Separatory funnel
  4. Sublimation

Answer: 3. Separatory funnel

Chapter 4 Matter Separation Of The Components Of Mixture Topic B Distillation Fractional Distillation Separatory Funnel Answer In Brief

Question 1. What are the processes of separation of component of a mixture of two liquids?

Answer: Distillation, fractional distillation, separating funnel etc.

Question 2. What change is observed in the boiling point of a liquid when the pressure above its surface is decreased?

Answer: When the pressure above the surface of a liquid is decreased, the liquid starts to boil at a lower temperature, i.e., the boiling point of the liquid decreases.

Question 3. What should be the minimum difference in the boiling points of two liquids if they are to be effectively separated by the process of distillation?

Answer: If the difference in the boiling points of two liquids is about 30-50°C, then they can be effectively separated by the process of distillation.

Question 4. Which constituent liquid will be in relatively higher amount in the vapour phase when a mixture containing two miscible liquids is heated during fractional distillation?

Answer: In fractional distillation of a mixture of two miscible liquids, the vapour phase will contain the more volatile liquid in relatively higher amount when the mixture is heated.

Question 5. How can a mixture of sand and iron dust be separated?

Answer: A mixture of sand and iron dust can be effectively separated by using a magnet. The process is called magnetic separation.

Question 6. Define sublimation.

Answer: Sublimation is the transition of a substance directly from the solid to the gaseous state, by the application of heat, without passing through the liquid state.

Question 7. In which process pure water can be obtained from an aqueous solution of salt?

Answer: By the process of distillation.

Question 8. By which process the components of a mixture of methanol (bp: 65°C) and water can be separated?

Answer: By the process of distillation.

Question 9. In which process the components of the mixture of ether (bp: 35°C) and benzene (bp: 80°C) can be separated?

Answer: Distillation.

Question 10. When is fractional distillation used for separating the components of a mixture?

Answer: If the difference in boiling points of two miscible liquids is less than 25°C, then those liquids are separated from their mixture by fractional distillation.

Question 11. Which apparatus is effectively used in fractional distillation of a mixture containing two or more miscible liquids, the difference in their boiling points being less than 25°C?

Answer: The apparatus which is effectively used in fractional distillation of a mixture containing two or more miscible liquids (the difference in their boiling points being less than 25°C), is a fractional distillation column or fractionating column.

Question 12. Which process is suitable to separate acetone and methanol from their mixture?

Answer: Acetone (boiling point = 56°C) and methanol (boiling point = 65°C) can be separated from their mixture by the process of fractional distillation.

Question 13. Is it possible to separate the components of a mixture containing two or more miscible liquids by distillation if the difference in their boiling points is 25°C?

Answer: No, it is not possible.

Question 14. Mention the percent amount of ethyl alcohol in rectified spirit which is an azeotropic mixture.

Answer: Rectified spirit which in an azeotropic mixture contains 95.6% ethyl alcohol.

Question 15. What is the boiling point of rectified spirit which is an azeotropic mixture?

Answer: The boiling point of rectified spirit which is an azeotropic mixture is 78.4°C.

Question 16. Two immiscible liquids, water and kerosene are kept in a beaker for some time. Which liquid will form the lower layer?

Answer: As water is heavier than kerosene, it will form the lower layer.

Question 17. Which process is suitable to separate water and kerosene from their mixture?

Answer: Water and kerosene can be separated from their mixture by using a separatory funnel.

Question 18. Is it possible to separate water and sand from their mixture by using a separatory funnel?

Answer: Water and sand cannot be separated from their mixture by using a separatory funnel.

Question 19. State whether fractional distillation is a physical or chemical process.

Answer: Fractional distillation is a physical process of separation of two or more miscible liquids.

Question 20. Which one of methanol and acetone will distill first if their mixture is being heated in a distillation flask?

Answer: Boiling points of methanol and acetone are 78.5°C and 56°C respectively. So acetone will distill first if their mixture is being heated in a distillation flask.

Question 21. Why the components of a mixture of methanol and acetone can not be separated by simple distillation?

Answer: The difference in boiling point of methanol (bp: 65°C) and acetone (bp: 56°C) is too small (9°C) to separate them by simple distillation. To separate the components from a mixture by simple distillation, difference in boiling point should be at least 30°C.

Question 22. Give example of azeotropic mixture.

Answer: The mixture of 95.6% ethanol and 4.4% water, is an example of azeotropic mixture.

Question 23. Name the suitable separation technique by which a mixture of two liquids of different boiling points can be separated.

Answer: Distillation or fractional distillation depending upon the difference in boiling points of the component liquids.

Question 24. By which process water and benzene can be separated from their mixture?

Answer: Using separatory funnel.

Question 25. Write down two essential conditions for separation of liquid mixture through separatory funnel?

Answer:

  1. Density of the liquids should be different.
  2. The liquids should be immisscible.

Question 26. How will you separate the components of a mixture containing water and kerosene or water and petrol?

Answer: By using separatory funnel.

Question 27. Name the separation technique by which the mixture of two immissible liquids can be separated.

Answer: By using separatory funnel.

Question 28. Name a liquid mixture the components of which can not be separated by separatory funnel.

Answer: A mixture of water and ethyl alcohol.

Chapter 4 Matter Separation Of The Components Of Mixture Topic B Distillation Fractional Distillation Separatory Funnel Fill In The Blanks

Question 1. A liquid starts to boil when the vapour pressure of the liquid is equal to the ________ pressure.

Answer: Atmospheric

Question 2. Two miscible liquids can be separated from their mixture by distillation if they do not ________ on heating.

Answer: Decompose

Question 3. During separation of a mixture consisting of two or more miscible liquids by fractional distillation, the liquid with the ________ boiling point will distill out first.

Answer: Lowest

Question 4. The different gaseous constituents of air is separated by ________

Answer: Fractional distillation

Question 5. Water can be boiled at room temperature (25°C) if the pressure on water is ________ than the atmospheric pressure.

Answer: Less

Question 6. Ether and toluene can be separated from their mixture by _________

Answer: Distillation

Question 7. The process which is widely used in purification of substances by solvent extraction method is ________

Answer: Distillation

Question 8. Sugar is separated from its aqueous solution _______ by process.

Answer: Distillation

Question 9. Liquids that do not vapourise easily are called _________ liquids.

Answer: Non-volatile

Question 10. In an azeotropic mixture of rectified spirit, the percentage of water is _________

Answer: 4.4%

Question 11. In a fractionating column, as the vapour moves up in the column, the relative amount of the ________ volatile component in the vapour increases.

Answer: More

Question 12. In a fractionating column, the process of condensation and vapourisation take place _______ throughout the column.

Answer: Repeatedly

Question 13. Water and alcohol _______ be separated completely from their mixture by fractional distillation method.

Answer: Cannot

Question 14. The gaseous components of ______ air are separated by fractional distillation.

Answer: Liquid

Question 15. A mixture of water and mustard oil can be separated by using ______

Answer: Separatory funnel

Question 16. Benzene is extracted from coal tar by the process of ________

Answer: Fractional distillation

Question 17. Two liquids cannot be seperated by fractional distillation if their mixture forms an ________

Answer: Azeotrope

Question 18. To cool down the vapours of the distilled liquid in distillation, _________ condenser is used.

Answer: Liebig

Question 19. Distillation = Evaporation + ________

Answer: Condensation

Question 20. In a mixture of water and chloroform, _______ forms the lower layer.

Answer: Chloroform

Question 21. Of the two liquids, the one with _______ density separates first by the separatory funnel.

Answer: More

Question 22. Separate ________ will be formed if a mixture of water and oil is kept for sometime in a separatory funnel.

Answer: Layer

Question 23. Two _______ liquids can be separated from their mixture using separatory funnel.

Answer: Immiscible

Chapter 4 Matter Separation Of The Components Of Mixture Topic B Distillation Fractional Distillation Separatory Funnel State Whether True Or False

Question 1. If the difference in the boiling points of two liquids is 30°C, then those liquids can be separated from their mixture by the process of simple distillation.

Answer: True

Question 2. Fractional distillation is a chemical process of separation of mixtures.

Answer: False

Question 3. A mixture of water and benzene forms a homogeneous mixture.

Answer: False

Question 4. Two immiscible liquids of different densities can be separated from their mixture by a separatory funnel.

Answer: True

Question 5. The different components of liquid air are separated by the process of fractional distillation.

Answer: True

Question 6. Boiling point of a liquid decreases if pressure above the liquid is increased.

Answer: False

Question 7. A mixture of sand and iron dust is separated by magnetic separation.

Answer: True

Question 8. It is not possible to boil water at room temperature.

Answer: False

Question 9. Azeotropic mixtures cannot be separated by fractional distillation.

Answer: True

Question 10.During fractional distillation of two miscible liquids, the less volatile component remains in the vapour phase in greater proportion.

Answer: True

Question 11. Water forms the upper layer when it is kept with kerosene in a beaker.

Answer: False

Question 12. During the process of distillation, the most volatile component distils out last while the least volatile component distils out first.

Answer: False

Question 13. A mixture of water and alcohol can be separated by using a separatory funnel.

Answer: False

Question 14. Fractional distillation is used to separate salt from sea water.

Answer: False

Question 15. Boiling point of methanol is greater than that of water.

Answer: False

Question 16. Boiling point of liquid decreases by decreasing pressure.

Answer: True

Question 17. In a mixture of carbon disulphide and water, carbon disulphide constitutes the lower layer.

Answer: True

Question 18. The mixture of acetone and water can be separated using separatory funnel.

Answer: False

Question 19. Colloid solutions can be separated by separatory funnel.

Answer: False

Chapter 4 Matter Separation Of The Components Of Mixture Topic B Distillation Fractional Distillation Separatory Funnel Miscellaneous Type Questions

Match The Column

1.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Topic B Distillation Fractioanl Distillation Separatory Funnel Match The Column 1

Answer: 1. B, 2. D, 3. A, 4. C

2.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Topic B Distillation Fractioanl Distillation Separatory Funnel Match The Column 2

Answer: 1. C, 2. A, 3. D, 4. B

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Water

Chapter 4 Matter Water Topic A Role Of Water In Development Of Life And Drinking Water Synopsis

  1. Water possesses high specific heat, high boiling point, capillary action and an ability to dissolve almost all substances. Due to these unique properties of water, it helps living organisms to survive and flourish.
  2. Both ionic and covalent compounds existing in solid, liquid or gaseous state can dissolve in water. Thus, water is known as the universal solvent.
  3. Only 1% of the total water content of the earth constitutes of fresh water which can be used for different purposes. Rivers, streams, lakes, ponds, swamps, glaciers and groundwater are the different sources of fresh water on the earth.
  4. The pH range of drinking water should be between 6.5 to 8.5 and the amount of dissolved oxygen must be 4-6 mg • L-1 as recommended by WHO.
    Excessive presence of chlorides, fluorides and arsenic compounds in drinking water makes it unsafe for consumption.
  5. The number of colonies of coliform bacteria such as E. coli present in 100 mL of a water sample is known as the coliform count of that water sample. The coliform count of drinking water must be zero.
  6. Some common methods of purification of water are boiling, chlorination and use of ultraviolet rays.
  7. Proportionate amount of bleaching powder is added to water in a closed vessel to make it germ-free. Then this water is left open for some time when excess chlorine evaporates from the water. The water thus obtained is safe for drinking. Sometimes, chlorine tablets are ‘ also used for the purification of water.
  8. Organic substances present in water react with chlorine to form trihalomethanes (THMs) and trihaloacetic acids. (TAAs). These compounds are highly carcinogenic in nature {i.e., these cause cancer).
  9. Ultraviolet rays are passed through water in aqua filters to make it germ-free. This method is better than any other water purification methods as it does not require any chemicals. Also, this method is more effective than chlorination in killing viruses present in water.

Read and Learn More WBBSE Solutions for Class 9 Physical Science and Environment

Chapter 4 Matter Water Topic A Role Of Water In Development Of Life And Drinking Water Short And Long Answer Type Questions

Question 1. What property of water makes it excellent for fomentation?

Answer:

Property Of Water Makes It Excellent For Fomentation:-

Water has the highest value of specific heat (1 cal • g-1 • C-1). Hence, it can absorb large amount of heat with small increase in temperature and can similarly release large amount of heat with small decrease in temperature.

So, water cools down less readily than other liquids. This is the reason for which water can be used for fomentation (i.e., to provide warmth).

Question 2. Mention one practical application of high specific heat of water.

Answer:

One Practical Application Of High Specific Heat Of Water:-

Water has a very high specific heat 4200 J • kg-1 • K-1 as compared to other substances (solids or liquids). Hence, the amount of heat required by a fixed quantity of water to raise its temperature by 1°C or IK is relatively higher than that required by the same quantity of some other solids or liquids.

Thus, water is used as a coolant in different industries (it helps to cool the machinery) and in radiators of automobiles.

Question 3. Land breeze and sea breeze are formed due to high specific heat of water-explain.

Answer:

Land Breeze And Sea Breeze Are Formed Due To High Specific Heat Of Water:-

Due to high specific heat of water, during daytime the landmass heats up more as compared to sea water. Consequently, air adjacent to the landmass gets heated up, becomes lighter and finally rises up to create a void space. Cold air adjacent to the sea water rushes towards the landmass to fill up the void. This is how sea breeze is resulted.

After sunset, the landmass cools down more rapidly than sea water. Air adjacent to the warm sea water heats up, becomes lighter and then rises up. Cold air adjacent to the landmass rushes towards the sea to fill up the formed void. This is how land breeze is resulted.

Question 4. What is capillary action?

Answer:

Capillary Action:-

Due to cohesive force acting between the molecules of a liquid and adhesive force operating between the liquid molecules and the surface of the vessel, a liquid elevates along a very narrow tube against the force of gravity. This phenomenon is known as capillary action.

Question 5. Discuss the significance of capillary action.

Answer:

Significance Of Capillary Action:-

  1. Water and minerals absorbed by roots of plant reach the leaves by capillary action.
  2. When sufficient water is not present in the upper layers of the earth’s crust, water from the lower layers come up to the upper layers by capillary action and supply water to the plants.
  3. Water comes out through numerous pores on earthen pots due to capillary action. This water evaporates and consequently, cools the water within the pot.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Water Topic A Role Of Water In Development Of Life And Drinking Water Significance Of Capillary Action

Question 6. How do aquatic plants and animals survive in cold countries?

Answer:

Aquatic Plants And Animals Survive In Cold Countries:-

The latent heat of freezing of water is very high. In cold countries, when water at the upper layers of ponds and lakes freezes, large amount of heat is released.

The lower layers of water absorb this heat and consequently the temperature of this part of the water body increases. Thus, water at the lower levels of the water body remain at higher temperature even though the surface of the water freezes. This helps aquatic plants and animals to survive throughout the winter season in cold countries.

Question 7. Discuss the significance of dissolved oxygen for aquatic plants and animals.

Answer:

Significance Of Dissolved Oxygen For Aquatic Plants And Animals:-

Aquatic plants and animals are dependent on dissolved oxygen for carrying out their respiratory activities. A certain amount of dissolved oxygen is necessary for the aquatic organisms to survive.

Dissolved oxygen is very important for fish. If dissolved oxygen in a water body decreases, then the aquatic plants and animals will die due to lack of oxygen.

Question 8. Mention two physical properties of water which are important for evolution of life.

Answer:

Physical Properties Of Water Which Are Important For Evolution Of Life:-

Two physical properties that are very important for evolution of life are:

  1. Higher specific heat of water (4200 J • Kg-1 • K-1)
  2. Polar nature of water molecule.

Question 9. Why is water called a universal solvent?

Answer:

Water Called A Universal Solvent On Following Reasons:-

  1. Water can dissolve most of the solids, liquids and gases.
  2. It has very high value of dielectric constant (80.4) and so, almost all ionic compounds are soluble in water.
  3. Water is a polar compound. Hence, polar covalent compounds are highly soluble in it.
  4. Water exists in the liquid state over a wide range of temperature from 0°C to 100°C.

Question 10. Why is water called a polar solvent?

Answer:

Water Called A Polar Solvent Because:-

Though water molecule is formed by covalent bonds, the oxygen atom being more electronegative than hydrogen atom, becomes partially negatively charged while the hydrogen atoms become partially positively charged.

As a result, the oxygen atom of water molecule can attract the positive part of a solute molecule while hydrogen atoms of water can attract its negative part by electrostatic force of attraction. So water is called a polar solvent.

Question 11. How do electrovalent or ionic compounds easily dissolve in water?

Answer:

Ionic Compounds Easily Dissolve In Water As Follows:-

Water has a very high value of dielectric constant (80.4). Due to this, water as a solvent can separate the positively charged and negatively charged ions in an ionic compound by opposing the electrostatic force of attraction between them. Thus, molecules of these compounds easily dissociate in water yielding the constituent ions. Hence, electrovalent or ionic compounds easily dissolve in water.

Question 12. What are the drinking water?

Answer:

The Drinking Water Is:-

  1. It must be colourless, odourless, tasteless and clear.
  2. It must be free from floating impurities.
  3. It must be free from germs and suspended particulate matter.
  4. It should not contain excess minerals than permissible limit.
  5. It must be free from harmful compounds such as urea, cyanide salts, nitrate salts etc.

Question 13. How do you identify a colourless liquid as water.

Answer:

We Can Identify A Colourless Liquid As Water As Follows :-

If colourless, anhydrous copper sulphate turns blue by coming in contact of a colourless liquid, then it can be told that the colourless liquid must be water.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Water Topic A Role Of Water In Development Of Life And Drinking Water Colorless Liquid Of Water

Question 14. What will be the harmful effects if excess chloride salts are present in drinking water?

Answer:

The nature and extent of pollution caused by excess chloride salts in water largely depends on the cations of the salts. Prolonged intake of water containing excess sodium chloride may cause high blood pressure.

It must be noted that, this adverse effect is related to the presence of sodium ion in water and has no relation with the chloride ion concentration of water.

Question 15. Mention the permissible limits of the given substances in drinking water as recommended by WHO—dissolved solids, chlorides, nitrates and nitrites, fluorides, arsenic and lead.

Answer:

The permissible limits as recommended by WHO are mentioned below

  1. Dissolved solids: 500 mg • L-1
  2. Chlorides: 250 mg – L-1
  3. Nitrates and nitrites: 45 mg • L-1
  4. Fluorides: 1 mg • L-1
  5. Arsenic: 0.01-0.05 mg – L-1
  6. Lead: 0.01 mg • L-1

Question 16. How does the pH value of water affect our health?

Answer:

The pH of drinking water should be ranging from 6.5 to 9.2 as recommended by WHO. If the pH of water exceeds the upper limit, then it may cause burning sensation in eyes, skin and mucous membrane.

Water having a pH value more than 11 is very harmful to eyes and skin. If pH of water is in the range of 10-12.5, then it may cause inflation of hair fibres and reduce our digestive capacity.

When pH of water is less than 4, it causes redness and burning sensation in eyes. If pH of water is less than 2.5, it causes extensive and irreparable damage to our skin. The immunity of animals is also largely dependent on the pH level of water.

Question 17. What is meant by coliform count of water?

Answer:

Coliform Count Of Water Means:-

The number of colonies of coliform bacteria such as E. coli present in 100 mL of a sample of water is known as coliform count of that sample of water.

The coliform count of drinking water should always be zero (0). Water used for bathing or water used in swimming pools should have a coliform count less than 200. Maximum coliform count of water used for navigation or fishing should not be greater than 1000.

Question 18. How water is purified or disinfected before distribution in township?

Answer:

Before distribution in towns, water is disinfected. This disinfection or purification is done by adding chlorine tablets or bleaching powder. This process is called chlorination. Chlorine, by oxidation, destroys the harmful microbes present in water and thus disinfected drinking water.

Question 19. Mention the limitations of purification of water by boiling.

Answer:

The limitations of purification of water by boiling are as follows

  1. Though most of the germs are killed at a temperature near the boiling point of water, some germs still remain active at that temperature.
  2. Harmful metals and some toxic chemical compounds cannot be removed from water by boiling.
  3. Most of the essential minerals present in water are removed by boiling.

Question 20. Write down two advantages of chlorination process.

Answer:

Two advantages of chlorination process are:

  1. Application of this process is easier and advantageous.
  2. Since chlorine remains in water even after a long time from the time of application, the disinfection property also remains active.

Question 21. Mention two limitations of purification of water by chlorination.

Answer:

Two limitations of purification of water by chlorination are as follows

  1. Chlorine reacts with organic substances dissolved in water to form trihalomethanes (THMs) and trihaloacetic acids (TAAs). Among the trihalomethanes, bromoform adversely affects the brain and decreases its efficiency thereby causing drowsiness. Prolonged exposure to bromoform & dibromochloro- methane may cause cancer in liver & kidneys.
  2. Chlorine being volatile in nature quickly evaporates from water and mixes with the surrounding air thereby causing air pollution.

Question 22. What are the limitations of purification of water by UV-rays?

Answer:

Two major limitations of purification of water by UV-rays are as follows

  1. Water which is to be purified by UV-rays should always be clear and transparent. UV- rays are ineffective in turbid water. If some suspended impurities are present in water, it decreases the penetrating power and thus the efficiency of UV-rays.
  2. The effect of UV-rays is temporary. In presence of light, the pathogens again become active in UV-purified water. In chlorinated water, chlorine persists in water long after chlorination is done. So, water .retains its germ-killing capacity.

Question 23. Mention the advantages of purification of water by UV-rays.

Answer:

  1. The purification of water by UV-rays is fast and effective. It does not affect the taste or smell of water. So, it is very effective in
    purification of drinking water and water used in food processing industries.
  2. Purification of water by this method is safe. As no harmful chemicals are used in this method, there is no chance of pollution.
  3. UV-rays are more effective than chlorine in removing viruses from water.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Water Topic A Role Of Water In Development Of Life And Drinking Water purification Of Water By UV Rays

Chapter 4 Matter Water Topic A Role Of Water In Development Of Life And Drinking Water Very Short Answer Type Questions Choose The Correct Answer

Question 1. In SI unit, the specific heat of water is

  1. 4200 J • kg-1 • K-1
  2. 2100 J • kg-1 • K-1
  3. 6300 J • kg-1 • K-1
  4. 4500 J • kg-1 • K-1

Answer: 1. 4200 J • kg-1 • K-1

Question 2. Water is a

  1. Non-polar solvent
  2. Polar solvent
  3. Organic solvent
  4. None of these

Answer: 2. Polar solvent

Question 3. An organic compound that dissolves in water is

  1. Benzene
  2. Alcohol
  3. Wax
  4. Carbon tetrachloride

Answer: 2. Alcohol

Question 4. An inorganic compound that does not dissolve in water is

  1. Sodium chloride
  2. Potassium nitrate
  3. Calcium carbonate
  4. Zinc sulphate

Answer: 3. Calcium carbonate

Question 5. The tendency of a liquid to flow against gravity in a narrow tube is called

  1. Surface tension
  2. Viscosity
  3. Capillary action
  4. None of these

Answer: 1. Surface tension

Question 6. The water content in a fully grown human being with respect to his total body weight is almost

  1. 65%
  2. 30%
  3. 40%
  4. 10%

Answer: 1. 65%

Question 7. The word ‘coliform’ is related to

  1. Bacteria
  2. Virus
  3. Algae
  4. Fungus

Answer: 1. Bacteria

Question 8. pH of a solution indicates the concentration of

  1. OH ions
  2. Cl ions
  3. H+ ions
  4. Na+ ions

Answer: 3. H+ ions

Question 9. pH of an acidic solution is

  1. Less than 7
  2. More than 7
  3. Equal to 7
  4. None of these

Answer: 1. Less than 7

Question 10. pH of an alkaline solution is

  1. Less than 7
  2. More than 7
  3. Equal to 7
  4. None of these

Answer: 2. More than 7

Question 11. The maximum permissible limit of dissolved oxygen in drinking water (in mg • L-1) as recommended by WHO is

  1. 1-2
  2. 4-6
  3. 8-10
  4. 12-15

Answer: 2. 4-6

Question 12. The maximum permissible limit of dissolved chlorides in drinking water (in mg • L-1) as recommended by WHO is

  1. 100
  2. 200
  3. 300
  4. 250

Answer: 4. 250

Question 13. The maximum permissible limit of dissolved fluorides in drinking water (in mg • L-1) as recommended by WHO is

  1. 1-1.5
  2. 11-20
  3. 21-30
  4. 31-40

Answer: 1. 1-1.5

Question 14. The maximum permissible limit of dissolved arsenic in drinking water (in mg • L-1) as recommended by WHO is

  1. 0.05
  2. 0.25
  3. 0.5
  4. 1

Answer: 1. 0.05

Question 15. Which of the following is used for the purification of water?

  1. Infrared rays
  2. X-rays
  3. Ultraviolet rays
  4. Gamma rays

Answer: 3. Ultraviolet rays

Question 16. Chlorination of water is done to

  1. Destroy the microbes
  2. Precipitate the suspended impurities
  3. Improve the taste of water
  4. Increase the clarity of water

Answer: 1. Destroy the microbes

Question 17. Which property of water helps to control the temperature of atmosphere?

  1. Specific heat
  2. Dielectric constant
  3. Bad conductivity of heat
  4. Polar nature

Answer: 1. Specific heat

Question 18. Specific heat of which of the following liquids is the highest?

  1. Water
  2. Kerosene
  3. Petrol
  4. Mercury

Answer: 1. Water

Question 19. Example of a water soluble organic compound is

  1. Toluene
  2. Acetic acid
  3. Xylene
  4. Chloroform

Answer: 2. Acetic acid

Question 20. Example of universal solvent is

  1. Water
  2. Kerosene
  3. Benzene
  4. Alcohol

Answer: 1. Water

Question 21. Value of the dielectric constant of water is

  1. 79.4
  2. 80.4
  3. 81.4
  4. 82.4

Answer: 2. 80.4

Question 22. The temperature range where the anomalous expansion of water can be seen

  1. 4°C-8°C
  2. 0°C-10°C
  3. 0°C-4°C
  4. 0°C-2°C

Answer: 3. 0°C-4°C

Chapter 4 Matter Water Topic A Role Of Water In Development Of Life And Drinking Water Answer In Brief

Question 1. What is meant by the statement—’Specific heat of water is 1 cal • g-1 • °C-1?

Answer: The above statement means that leal of heat is required to raise the temperature of 1 g of pure water by 1°C.

Question 2. At what temperature, the density of water is maximum?

Answer: The density of water is maximum at 4°C (1 g – cm-3).

Question 3. What is a capillary tube?

Answer: A capillary tube is a long and thin tube made of rigid materials like glass or plastic. The diameter of a capillary tube generally ranges from 0.5 mm to 3 mm.

Question 4. Which property of water makes it a coolant for different machinery used in the industries?

Answer: High specific heat of water makes it an excellent coolant for different machinery used in the industries.

Question 5. Give an example of a covalent compound which is soluble in water.

Answer: Hydrogen chloride (HCI) is a covalent compound which is soluble in water.

Question 6. Give an example of a covalent compound which is insoluble in water.

Answer: Benzene (C6H6) is a covalent compound which is insoluble in water.

Question 7. What must be the coliform count of drinking water?

Answer: The coliform count of drinking water must be zero (0).

Question 8. What is meant by the statement—’Coliform count of a sample of water is zero’?

Answer: The above statement means that the sample of water does not contain any colony of coliform bacteria (Escherichia coli).

Question 9. What is the value of dielectric constant of water?

Answer: The value of dielectric constant of water is 80.4.

Question 10. How can water be purified without using any chemicals?

Answer: Water can be purified by using ultraviolet rays (UV-rays) as no chemicals are required in this method.

Question 11. Which halogen is used for the purification of water?

Answer: The halogen used for the purification of water is chlorine.

Question 12. Name two absorbents of water.

Answer: Concentrated sulphuric acid (H2SO4) and anhydrous calcium chloride (CaCI2) are two absorbents of water.

Question 13. How does chlorine destroy the microbes present in water?

Answer: Chlorine destroys the microbes present in water by the process of oxidation.

Question 14. How pH of water changes with increase in temperature?

Answer: pH of water generally decreases with increase in temperature.

Question 15. Give an example of coliform bacteria?

Answer: E.coli.

Question 16. Why chlorination of water is done?

Answer: To destroy the microbes present in water.

Question 17. What should be the permissible limit of oxygen in drinking water as per order of WHO?

Answer: 4-6 mg – L-1

Question 18. Give the formula of heavy water?

Answer: The formula of heavy water is D2O

Question 19. For which property, water can be used for cooking purposes?

Answer: Boiling point of water is comparatively higher, that is why it is used for cooking.

Question 20. What type of harm occur if pH of drinking water becomes more than 10.5?

Answer: It causes disturbances in digestion.

Question 21. What type of harm occur if the pH of drinking water becomes less than 4?

Answer: It causes damages of nervous system, lungs, and respiratory tracks and also disturbs the digestion system.

Question 22. How the surgical apparatus and instruments are disinfected?

Answer: The surgical instruments are boiled at a higher temperature more than 100°C, at a higher pressure in autoclave and thus disinfected.

Question 23. For which property of water, different types of salt dissolve in it?

Answer: For polar nature of water molecule.

Question 24. Due to which physical property of water, the water bodies like ponds and lakes do not freeze immediately on a cold day in winter- countries?

Answer: Latent heat of solidification of water has a higher value (336 J/g). This property of water helps in the mentioned process.

Chapter 4 Matter Water Topic A Role Of Water In Development Of Life And Drinking Water Fill In the Blanks

Question 1. Specific heat of water is ________ than ice.

Answer: Greater

Question 2. Water is a ________ solvent.

Answer: Polar

Question 3. Water absorbed by the roots of plants reach the leaves through stem due to __________

Answer: Capillary

Question 4. __________ is generally used to precipitate the impurities suspended in water.

Answer: Alum

Question 5. Only __________ % of total water content found on the earth’s surface is usable.

Answer: 1

Question 6. pH is the negative logarithm of ______ ion concentration of a solution.

Answer: Hydrogen

Question 7. pH of pure water is __________

Answer: 7

Question 8. UV-rays destroy the ________ present in water.

Answer: Microbes

Question 9. Taste of water depends on the amount of ________ dissolved in it.

Answer: Salts

Question 10. Purification of water by boiling also leads to the removal of essential _________ dissolved in water.

Answer: Minerals

Question 11. __________ % of total water content on the earth’s surface constitutes of sea water.

Answer: 97

Question 12. The latent heat required for solidification of water is ___________ J • g-1

Answer: 336

Question 13. The permissible range of pH value of drinking water is ______

Answer: 6.5-8.5

Question 14. Purification of water by ______ radiation is not effective for turbid water.

Answer: Ultraviolet

Question 15. Chlorine reacts with organic materials present in water to produce _______ and _______

Answer: Trichloromethane, trihaloacetic acid

Question 16. The approved level of dissolved oxygen in drinking water, as per WHO recommendation, is ________ mg • L-1.

Answer: 4-6

Question 17. The admissible level of fluoride in drinking water as per WHO norms is ________ mg • L-1.

Answer: 1

Chapter 4 Matter Water Topic A Role Of Water In Development Of Life And Drinking Water State Whether True Or False

Question 1. The pH of drinking water should range between 4.5 to 6.5.

Answer: False

Question 2. The coliform count of drinking water must be zero.

Answer: True

Question 3. Chlorination is an effective method for removing permanent hardness of water.

Answer: False

Question 4. The density of water is minimum at 0°C.

Answer: False

Question 5. Consumption of deionised water is good for the human body.

Answer: False

Question 6. Fluoride salts can be eliminated from water by passing through activated alumina.

Answer: True

Question 7. Distilled water is less pure than deionised water.

Answer: False

Question 8. Water is used as a coolant because of its low value of specific heat.

Answer: False

Question 9. Magnesium reacts with water at ordinary temperature.

Answer: False

Question 10. Capillary action of water is responsible for the absorption of water by roots and its subsequent transportation to the leaves.

Answer: True

Question 11. Specific heat of water is the highest among known liquids.

Answer: True

Question 12. The colloidal substance floating in water gets precipitated on addition of alum.

Answer: True

Question 13. The density of water at 4°C is lower than that at 0°C.

Answer: False

Question 14. Drinking water can be disinfected by boiling.

Answer: True

Chapter 4 Matter Water Topic B Soft Water And Hard Water Synopsis

  1. Water that does not produce lather with soap, or does so only if large amount of soap is used is called hard water. On the other hand, soft water easily produces lather or foam with soap.
  2. Hardness of water which can be removed by simple processes like boiling is called temporary hardness. On the other hand, hardness of water which cannot be removed by boiling is called permanent hardness.
  3. Presence of chloride or sulphate salts of calcium, magnesium and iron cause permanent hardness of water. Bicarbonates of these metals produce temporary hardness in water.
  4. Presence of Zn-salts or Al-salts also causes hardness of water. However, salts of alkali metals like sodium or potassium are not responsible for hardness of water.
  5. Temporary hardness of water can be removed by boiling, while ion-exchange resin method removes both temporary and permanent hardness of water.
  6. A sample of water that does not contain any ions other than H+ and OH ions is called
    deionised water.

Chapter 4 Matter Water Topic B Soft Water And Hard Water Short And Long Answer Type Questions

Question 1. What is meant by hard water and soft water? Give examples.

Answer:

Hard water: The type of water that barely produces lather with soap or does so after consuming a large amount of soap is called hard water. For example, river water, sea water, waterfalls etc., are natural sources of hard water.

Soft water: The type of water that easily forms lather with soap is called soft water. For example, distilled water, deionised water etc.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Water Topic B Soft Water And Hard Water

Question 2. What is the cause of hardness of water? What are the different types of hardness of water?

Answer:

Hardness of water is resulted due to the presence of bicarbonate, sulphate and chloride salts of calcium, magnesium and iron which remain dissolved in water.

On the basis of the nature of dissolved salts, hardness of water is of two types

  1. Temporary hardness and
  2. Permanent hardness.

Question 3. What do you mean by temporary hardness of water?

Answer:

Hardness produced due to the presence of dissolved bicarbonate salts of calcium and magnesium [Ca(HCO3)2 and Mg(HCO3)2] and to some extent iron which can be removed by simply boiling the water is known as temporary hardness.

Question 4. What do you mean by permanent hardness of water?

Answer:

Hardness produced due to the presence of dissolved sulphate and chloride salts of calcium, magnesium and iron [CaCl2, CaSO4, MgCI2, MgSO4, FeSO4] which cannot be removed by simply boiling the water is called permanent hardness or non-carbonate hardness.

Question 5. Name two processes for the removal of hardness of water.

Answer:

Two processes for the removal of hardness of water are: boiling (Process for removal of temporary hardness) and ion-exchange process (Process for removal of both temporary and permanent hardness of water).

Question 6. Which one of these has the higher probability of being hard: spring water or rain water.

Answer:

Probability of being hard water is more in case of spring water. Water from different water bodies get evaporated and forms cloud in the higher part of sky which when condense, form rain where the minerals remain absent as those were not evaporated at all.

That is why rain water remains soft in nature. On the other hand, spring water flows above rocks and minerals. During this flow the metallic salts, responsible for hardness dissolve in it and make the spring water hard.

Question 7. Temporary hardness of water caused by magnesium bicarbonate cannot be removed completely by boiling. Explain with reason.

Answer:

On boiling hard water (containing magnesium bicarbonate), the soluble bicarbonate decomposes to form water insoluble carbonate salt.

⇒ \(\mathrm{Mg}\left(\mathrm{HCO}_3\right)_2 \rightarrow \mathrm{MgCO}_3 \downarrow+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \uparrow\)

As MgCO3 is sparingly soluble in water, temporary hardness of water caused by magnesium bicarbonate cannot be removed completely by boiling.

Question 8. How is the degree of hardness of water expressed?

Answer:

Degree of hardness of water is defined as the number of parts by mass of calcium carbonate (CaCO3) equivalent to various calcium and magnesium salts present in one million (106) parts by mass of water. Degree of hardness of water is expressed in terms of ppm (parts per million).

Question 9. What do you mean by the statement— ‘Hardness of a sample of water is 300 ppm’?

Answer:

The above statement means that 300 parts by mass of calcium carbonate equivalent to various salts causing hardness of water is present in one million parts by mass of that water sample.

Question 10. Mention two disadvantages of using hard water.

Answer:

  1. Hard water can not be used in cooking as the food particles are not boiled properly in hard water.
  2. Hard water is not suitable for drinking as it affects the digestive system.

Question 11. Why does normal water behave as hard water if few drops of mineral acid are added to it?

Answer:

If few drops of mineral acids like HCI or H2SO4 are added to normal water, then it behaves as hard water because the acid react with soaps to precipitate insoluble fatty acids.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Water Topic B Soft Water And Hard Water Normal Water Behave As Hard Water

This is why soaps do not produce lather in this type of water.

Question 12. Describe the boiling processj of temporary hardness of water?

Answer:

When the temporary hard water is boiled, the soluble bicarbonates converted to insoluble carbonates and precipitated out. After filtration one can get the hardness freed water.

⇒ \(\mathrm{Ca}\left(\mathrm{HCO}_3\right)_2 \rightarrow \mathrm{CaCO}_3 \downarrow+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \uparrow\)

⇒ \(\mathrm{Mg}\left(\mathrm{HCO}_3\right)_2 \rightarrow \mathrm{MgCO}_3 \downarrow+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \uparrow\)

Question 13. Why do soaps not produce lather with hard water initially?

Answer:

Soaps are sodium or potassium salts of fatty acids having high molecular mass (e.g. stearic acid, palmitic acid, oleic acid etc.) which dissolve in water to form lather.

However, calcium and magnesium salts of these fatty acids are insoluble in water and do, not produce lather. Calcium and magnesium salts remain dissolved in hard water.

When these metal cations react with soap, initially a white precipitate is formed which is actually the insoluble calcium and magnesium salts of those fatty acids. Lather is produced only when all the Ca2+ and Mg2+ ions present in water are completely precipitated

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Water Topic B Soft Water And Hard Water Soaps Not Produce Lather With Hard Water Initially

Question 14. What are cation exchange resins?

Answer:

A cation exchange resin is an artificially synthesised organic polymer having a complex network structure in which an acidic group, — SO3H (sulphonic acid) is attached to a large hydrocarbon chain. The general formula of a cation exchange resin is R —SO3H.

Question 15. What are anion exchange resins?

Answer:

An anion exchange resin is an artificially synthesised organic polymer having a complex network structure in which an alkaline OH group is present as substituted ammonium hydroxide attached to a large hydrocarbon chain. The general formula of a anion exchange resin is R — \(\mathrm{NH}_3^{+} \mathrm{OH}^{-}\).

Question 16. Resins used in the ion exchange method become inactive after prolonged use. How are they regenerated?

Answer:

The inactive cation exchange resins are reactivated by passing dilute H2SO4 or dilute HCI through the resin layers.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Water Topic B Soft Water And Hard Water Resins Used In Ion Exchange Method

[M = Ca, Mg; R = alkyl radicals]

Similarly, the inactive anion exchange resins are reactivated by passing dilute NaOH solution through the resin layers.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Water Topic B Soft Water And Hard Water Inactive Resins After Used Prolonges Used

Question 17. Why is demineralised water or distilled water not suitable for drinking? How can one make this water suitable for drinking?

Answer:

Calcium, magnesium and iron salts are very essential for the growth and development of living organisms. As these salts are absent in demineralised water or distilled water, it is not suitable for drinking although being very pure in nature.

Demineralised or distilled water can be made suitable for drinking by adding essential minerals in proportionate amounts.

Question 18. What is deionised water? Between distilled water and deionised water, which one is purer and why?

Answer:

Water obtained by consecutively treating it with cation exchange resin and anion exchange resin is free from all ions except H+ ions and OH ions. This is known as deionised water.

Deionised water is free from all ions. Hence, it is soft water. However, this water may contain, (apart from H+ and OH ions) some organic impurities, dissolved gases and even germs.

On the other hand, water obtained by distillation is free from ions as well as organic impurities, dissolved gases and germs. Hence, distilled water is purer than deionised water.

Question 19. Explain why distilled water is not suitable for drinking.

Answer:

Calcium, magnesium and iron salts all are very essential elements for living bodies. Demineralised or distilled water do not contain any of such salts.

That is why demineralised or distilled water is not suitable for drinking. By mixing the essential minerals in definite proportions distilled water can be converted to drinking water.

Question 20. Explain why deionised water is not used for injections.

Answer:

Some organic impurities or some microbes may present in deionised water. That is why deionised water is not used for injections.

Question 21. Differentiate between deionised water and distilled water.

Answer:

The difference between deionised water and distilled water are:

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Water Topic B Soft Water And Hard Water Differences Between Deionised Water And Distilled Water

Chapter 4 Matter Water Topic B Soft Water And Hard Water Vert Short Answer Type Questions Choose The Correct Answer

Question 1. If hardness of water increases, then it

  1. Can easily produce lather
  2. Cannot produce lather
  3. Can produce lather only by consuming large amount of soap
  4. None of these

Answer: 3. Can produce lather only by consuming large amount of soap

Question 2. The hardness of water can be of

  1. 4 types
  2. 3 types
  3. 2 types
  4. 5 types

Answer: 3. 2 types

Question 3. A salt responsible for permanent hardness of water is

  1. CaCl2
  2. Ca(HCO3)2
  3. NaCI
  4. Na2CO3

Answer: 1. CaCl2

Question 4. A salt responsible for temporary hardness of water is

  1. CaCI2
  2. Ca(HCO3)2
  3. NaCI
  4. Na2CO3

Answer: 2. Ca(HCO3)2

Question 5. If acid (H+) is added to water, then water will behave as

  1. Soft water
  2. Hard water
  3. Pure water
  4. Deionised water

Answer: 2. Hard water

Question 6. Boiling can be used to

  1. Remove permanent hardness of water
  2. Deionise water
  3. Remove temporary hardness of water
  4. Decolourise water

Answer: 3. Remove temporary hardness of water

Question 7. The general formula of cation exchange resins is

  1. R —OH
  2. R — SO3H
  3. R —Cl
  4. R —NH2

Answer: 2. R — SO3H

Question 8. The general formula of anion exchange resins is

  1. R —Cl
  2. R — COOH
  3. R — NH3+OH
  4. R —H

Answer: 3. R — NH3+OH

Question 9. Cation exchange resins are regenerated by using a dilute solution of

  1. NaOH
  2. HCI
  3. NaCl
  4. Na2SO4

Answer: 2. HCI

Question 10. Anion exchange resins are regenerated by using a dilute solution of

  1. NaOH
  2. HCI
  3. NaCl
  4. Na2SO4

Answer: 1. NaOH

Question 11. Degree of hardness of water is measured in

  1. g • L-1
  2. mol • L-1
  3. ppm
  4. kg • L

Answer: 3. ppm

Question 12. The type of water used in boilers of factories is

  1. Soft water
  2. Hard water
  3. Deionised water
  4. Germ-free water

Answer: 1. Soft water

Question 13. Which of the following happens when hardness of water increases?

  1. Lather forms easily
  2. Lather does not form at all
  3. Lather forms after rubbing the soap for a long time
  4. None of the above

Answer: 3. Lather forms after rubbing the soap for a long time

Question 14. Hardness of deionised water is

  1. 0
  2. 1
  3. 3
  4. 7

Answer: 1. 0

Question 15. pH of deionised water is

  1. 0
  2. 2
  3. 4
  4. 7

Answer: 4. 7

Question 16. Which of the following cations is present in deionised water?

  1. H+
  2. Na+
  3. Mg2+
  4. Ca2+

Answer: 1. H+

Question 17. Which of the following ions is not responsible for hardness of water?

  1. Ca2+
  2. Mg2+
  3. Fe2+
  4. Na+

Answer: 4. Na+

Question 18. Which of the following processes can remove temporary and permanent hardness of water simultaneously?

  1. Boiling
  2. Chlorination
  3. U.V. ray
  4. Ion-exchange process

Answer: 4. Ion-exchange process

Question 19. Which of the following ions is absent in hard water?

  1. Carbonate
  2. Bicarbonate
  3. Chloride
  4. Sulphate

Answer: 1. Carbonate

Question 20. In deionised water

  1. No ion is present
  2. Only H+ ion is present
  3. Only OH+ ion is present
  4. Both H+ ion and OH ion is present

Answer: 4. Both H+ ion and OH ion is present

Question 21. Which of the following is used for injection?

  1. Soft water
  2. Hard water
  3. Deionised water
  4. Distilled water

Answer: 4. Distilled water

Chapter 4 Matter Water Topic B Soft Water And Hard Water Answer In Brief

Question 1. What is the working principle of pressure cooker?

Answer:

In a pressure cooker, the steam, so produced increases the pressure above the surface of water. As a result, water boils at higher temperature than its normal boiling point due to elevation of boiling point. As a result, food gets boiled properly and rapidly.

Question 2. Hard water is suitable for washing clothes’— Is the statement right or wrong?

Answer: When hard water is used, large amount of soap is required to produce lather which results in wastage of soap. Hence, hard water is not suitable for washing clothes.

Question 3. Name two metals that react with water at room temperature.

Answer: Sodium (Na) and calcium (Ca) react with water at room temperature.

Question 4. Name a salt which causes temporary hardness in water.

Answer: Calcium bicarbonate [Ca(HCO3)2] causes temporary hardness in water.

Question 5. Name a sulphate salt responsible for permanent hardness in water.

Answer: Magnesium sulphate (MgSO4) causes permanent hardness in water.

Question 6. What is meant by reversibility of resins?

Answer: Salts are formed due to exchange of H+ and OH ions of the resins with the cations and anions present in hard water. The activity of resins can be regenerated by treating these salts with dilute acid or alkali solutions. This is known as reversibility of resins.

Question 7. Name a fatty acid from which soap is prepared.

Answer: A fatty acid from which soap is prepared is stearic acid.

Question 8. Which type of hardness can be removed by boiling?

Answer: Temporary hardness of water caused by bicarbonate salts of calcium and magnesium can be removed by boiling the water.

Question 9. State if aquatic organisms can survive in distilled water or not.

Answer: No, aquatic organisms cannot survive in distilled water due to lack of dissolved oxygen in it.

Question 10. Which part of a salt is responsible for causing hardness of water?

Answer: The cation of a salt is responsible for causing hardness of water.

Question 11. Two samples of water contain dissolved Zn- salts and Na-salts respectively. Which one is hard water?

Answer: Zn-salts cause hardness of water. Hence, the sample having dissolved Zn-salt in it will be hard water.

Question 12. Which method is effective in removing all types of hardness of water?

Answer: Ion-exchange method is effective in removing all types of hardness of water.

Question 13. By which process permanent hardness of water can be removed?

Answer: Ion exchange process.

Question 14. Which ions are present in deionised water?

Answer: H+ and OH ions are present in deionised water.

Question 15. If or whenever the efficiency of the cation exchange resin reduces, what should be passed through them to regain the efficiency?

Answer: It Dilute H2SO4 or dilute HCI is passed though the resin layer.

Question 16. Which type of hardness is removed by ion exchange process?

Answer: Both temporary and permanent hardness are removed.

Question 17. Name a calcium salt, responsible for hardness of water.

Answer: Calcium sulphate (CaSO4).

Question 18. Name a magnesium salt, responsible for temporary hardness.

Answer: Magnesium bicarbonate [Mg(HCO3)2].

Question 19. NaCI and K2SO4 are dissolved is a sample of water. Mention whether it is soft or hard water.

Answer: Soft water.

Question 20. Name a halogen whose water-soluble salts cause water pollution.

Answer: Fluorine causes water pollution as it remains in water as soluble fluoride salts.

Chapter 4 Matter Water Topic B Soft Water And Hard Water Fill In The Blanks

Question 1. Cation exchange resins are organic _______ acids having high molecular mass.

Answer: Sulphonic

Question 2. ________ is a useful method for the removal of permanent hardness of water.

Answer: Ion-exchange

Question 3. If the hardness of water increases, then the amount of soap required to form lather in that water also _________

Answer: Increases

Question 4. It is preferable to use _________ over soap in hard water.

Answer: Detergents

Question 5. Detergents can produce lather even in ______ water.

Answer: Hard

Question 6. Water after treating with cation exchange resin is ________ in nature.

Answer: Acidic

Question 7. Permanent hardness is also known as ________ hardness.

Answer: Non-carbonate

Question 8. Temporary hardness is also known as __________ hardness.

Answer: Carbonate

Question 9. Efficiency of anion exchange resin is regained by passing ________ solution.

Answer: NaOH

Question 10. _________ hardness cannot be removed by boiling.

Answer: Permanent

Question 11. _______ water should be used for cooking.

Answer: Soft

Chapter 4 Matter Water Topic B Soft Water And Hard Water State Whether True Or False

Question 1. The hardness which can be removed by simply boiling the water is known as permanent hardness.

Answer: False

Question 2. Both temporary and permanent hardness of water can be removed by using ion-exchange resin.

Answer: True

Question 3. The cationic part of a salt is responsible for imparting hardness to water.

Answer: True

Question 4. Spring water is the example of soft water.

Answer: True

Question 5. Water will be hard if Na or K-salt are present in it.

Answer: False

Question 6. Hardness of distilled water is zero.

Answer: True

Question 7. Deionised water is completely pollution free.

Answer: False

Question 8. Deionised water is used for injections.

Answer: False

Chapter 4 Matter Water Topic C Water Pollution Synopsis

  1. Detergents, pesticides and chemical fertilisers dissolve in water and cause extensive water pollution.
  2. The presence of excess phosphate salts in water causes eutrophication (rapid increase in aquatic plant population) in water bodies. On the other hand, The presence of excess nutrients in water causes rapid increase in population of algae and the entire surface of water gets covered by algae. This phenomenon is known as algal bloom.
  3. Undesired changes in human beings, other living organisms and environment due to arsenic contamination of water is called arsenic pollution. It may cause several health related problems in man and other animals. The maximum permissible limit (as per WHO guidelines) of arsenic in drinking water is 0.05 mg -L-1.
  4. Excess arsenic enters into the human body through drinking water and may lead to black foot disease.
  5. Undesired changes in human beings, other living organisms and the environment due to the contamination of water by fluoride salts is called fluoride pollution. The maximum permissible limit (as per WHO guidelines) of fluorides in drinking water is 1.10 mg • L-1.
  6. Fluoride pollution of water causes dental fluorosis.
  7. Water is made arsenic-free by different methods such as adsorption, coprecipitation, ion-exchange method, reverse osmosis etc.
  8. Fluoride salts are removed from water by passing the water through activated alumina columns.

Chapter 4 Matter Water Topic C Water Pollution Short And Long Answer Type Questions

Question 1. What is meant by water pollution?

Answer:

Water Pollution:-

Undesired changes in the physical, chemical or biological parameters of water that results in contamination of water bodies thus, making it unsafe for living organisms (both terrestrial and marine) is called water pollution.

Question 2. Mention three major causes water pollution.

Answer:

Major Causes Water Pollution:-

Three major causes of water pollution are as follows

  1. Disposal of domestic wastes like detergents, plastics etc., in water bodies.
  2. Industrial wastes like oil, grease, acid etc., released in water bodies.
  3. Mixing of fertilisers, pesticides and insecticides used in agricultural lands with nearby water bodies.

Question 3. Name two pesticides which cause water pollution. Mention their harmful effects.

Answer:

Harmful Effects Caused By The Pesticides:- 

Two commonly used pesticides which cause water pollution are—DDT (p, p’-dichlorodi- phenyl trichloroethane) and gammaxene.

These pesticides affect the function of brain and may cause cancer (carcinogenic in nature).

Question 4. Discuss how pesticides cause water pollution

Answer:

Pesticides Cause Water Pollution As Follows:-

Pesticides are chemical compounds used to preserve crops from the attacks of insects, fungi etc. (collectively called pests). These pesticides are washed off by water into nearby rivers, ponds, lakes etc. causing water pollution.

Pesticides are mostly non-biodegradable compounds and do not decompose biochemically. These substances enter into the human bodies and other animals through food chain and cause biomagnification.

This leads to headache, nervous breakdown, slackening of muscles, convulsions etc. Prolonged exposure to these substances may even cause tumour or cancer.

Question 5. Discuss how fertilisers cause water pollution.

Answer:

Fertilisers Cause Water Pollution As Follows:-

Fertilisers are chemical compounds used to increase the fertility of soil thereby increasing the yield of crops such as, urea, ammonium sulphate, ammonium nitrate etc. When fertilisers are applied to agricultural fields, these are washed away by water into nearby rivers, ponds, lakes etc. causing water pollution.

Excessive nitrate salts in drinking water may cause methaemoglobinaemia in infants. Presence of excessive phosphate salts results in eutrophication in water bodies.

Question 6. How do nitrate fertilisers cause water pollution?

Answer:

Nitrate Fertilisers Cause Water Pollution As Follows:-

Nitrate fertilisers are widely used in agricultural lands. These salts being soluble in water dissolve in it and are washed away to nearby water bodies. Excess quantity of these salts act as pollutants rather than nutrients.

If excess quantity of nitrate salts enter the body through consumption of drinking water, it may cause methaemoglobinaemia in infants. It can also interfere with the ability of red blood corpuscles to transport oxygen to different parts of the body thereby disrupting physiological processes like circulation and respiration.

Question 7. What are detergents? What is the advantage of using detergents in hard water?

Answer:

Detergents And Advantage Of Using Detergents In Hard Water:-

Detergents are artificially synthesised mixtures of two or more substances which have cleansing properties. A detergent contains two major components

  1. A surface active substance (like, alkyl benzene sulphonate) and
  2. A builder or filler (like sodium tripolyphosphate).
  3. Detergents can produce lather or foam even in hard water causing less wastage.

Question 8. Discuss how detergents cause water pollution.

Answer:

Detergents Cause Water Pollution As Follows:-

Detergents are the major water pollutants. They cause water pollution in the following ways

  1. Lather produced by detergents accumulate on the surface of water thereby preventing air and sunlight from entering into the depths of water. Consequently, dissolved oxygen of water decreases.
  2. Detergents contain surface-active compounds which form layers on the organic pollutants present in water and prevent their biochemical decomposition. This increases the pollution level of water.
  3. Phosphate compounds are used as fillers in detergents. These phosphate compounds cause eutrophication.

Question 9. What is eutrophication?

Answer:

Eutrophication:-

The phenomenon of rapid growth in the population of aquatic plants (mainly algae) in water bodies due to enrichment of water with excess phosphate fertilisers, phosphate compounds of detergents etc. (which act as nutrients) is known as eutrophication.

Question 10. Discuss the harmful effects of eutrophication.

Answer:

Harmful Effects Of Eutrophication:-

  1. Due to rapid population growth of aquatic plants, demand of oxygen for their respiration also increases.
  2. Thus, amount of dissolved oxygen in water decreases rapidly thereby threatening the survival of aquatic life.
  3. Rapid decrease in the amount Of dissolved oxygen in water leads to an abnormal increase in population of anaerobic bacteria. These bacteria acts on the different organic and inorganic waste materials present in water to produce gases such as, methane, ammonia, hydrogen sulphide etc. which leads to emission of foul smell in water.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Water Topic C Water Pollution Harmful Effects Of Eutrophication

Question 11. What is algal bloom? Mention its effects.

Answer:

Algal Bloom And Its Effects:-

  1. The rapid increase in the population of algae due to presence of excess nutrients in stagnant water bodies is known as algal bloom.
  2. Due to this increased population growth of algae, the surface of water gets entirely covered with algae. This prevents sunlight from reaching the depths of the water body.
  3. The level of dissolved oxygen also decreases rapidly due to algal bloom. Due to insufficient oxygen, aquatic plants and animals die which disrupts the overall functioning of aquatic ecosystem. Apart from this, bacterial decom-position of dead plants and animals in the water body makes it more polluted.

Question 12. What is arsenic pollution? Is the element arsenic directly responsible for arsenic pollution?

Answer:

Arsenic Pollution:-

The undesired effects on living beings (mainly human beings) caused due to the presence of arsenic compounds in water, in an amount greater than the permissible limit, is known as arsenic pollution.

No, the element arsenic is not directly responsible for arsenic pollution. Presence of certain arsenic salts like arsenate and arsenite in water causes arsenic pollution in water.

Question 13. Mention one natural and one artificial reason for arsenic pollution.

Answer:

Natural And Artificial Reason For Arsenic Pollution:-

Natural Reason: Insoluble layer of arsenic compounds are present in contact with the groundwater layer. Excessive use of this groundwater helps air to come in contact with this insoluble salts. As a result soluble arsenite and arsenate salts are produced which is the main cause of arsenic pollution in water.

Artificial Reason: Different arsenic-containing compounds e.g., sodium arsenite, lead arsenate, calcium arsenate etc. are used as antifungal and pesticidal substance. They are when dissolved in water make the water polluted.

Question 14. Discuss the harmful effects of arsenic pollution on human beings.

Answer:

Harmful Effects Of Arsenic Pollution On Human Beings:-

  1. Consumption of water containing arsenic disturbs the blood circulation in our body.
  2. It makes the skin rough and black patches appear on the skin in the neck, shoulder and back region of the body. Long term exposure to arsenic salts may cause black-foot disease which causes appearance of black spots on palms and feet.
  3. Consumption of water containing arsenic for a long time may lead to cirrhosis of liver and even cancer in lungs and intestine.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Water Topic C Water Pollution Harmul Effects Of Arsenic Pollution On Human Beings

Question 15. Mention any two processes.

Answer:

Two Processes:-

Adsorption Method: Both the salts of arsenic that cause pollution i.e., arsenate and arsenite, are adsorbed by activated alumina (Al2O3).So tube wells are fitted with columns of alumina to remove arsenic from water.

Co-Precipitation Method: If alum and hydrated ferric oxide are added to water and left undisturbed for a long period of time, then the arsenic-containing compounds gradually settle down at the bottom and are separated from water by filtration.

Question 16. In which state chlorine, fluorine and arsenic are present in natural water?

Answer:

Chlorine: Chlorine is found in natural water in elemental state (Cl2), as hypochlorous acid (HOCI), as hypochloride salts and chloride salts dissolved in water.

Fluorine: Fluorine remains dissolved in natural water as fluoride salts.

Arsenic: In natural water, arsenic is found as soluble inorganic salts, arsenite (AsO33-) and arsenate (AsO43-) and as soluble organoarsenic compound, cacodylic acid [(CH3)2AsO2H].

Question 17. What is the maximum permissible limit of fluorides in drinking water? Discuss the harmful effects of excess fluorides in drinking water.

Answer:

As recommended by WHO, the maximum permissible limit of fluoride salts in drinking water is 1.5 mg • L-1.

The presence of 1.0 mg • L-1 of fluoride salts in drinking water is essential as it prevents the decay of tooth enamel. However, if the amount of fluoride salts slightly exceeds this permissible limit, then it may cause dental fluorosis and if the amount largely exceeds the permissible limit, then it may lead to osteoporosis.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Water Topic C Water Pollution Harmul Effects of Excess Fluorides In Drinking Water

Question 18. How is dissolved fluoride salts removed from water?

Answer:

Methods Of Removal of fluorides from water are:-

  1. If water contains large quantities of fluoride salts, then it is passed over activated alumina to eliminate fluoride salts.
  2. Water containing fluoride salts is treated with potash alum and lime and is kept undisturbed for some hours. The fluoride salts settle down and is then removed by filtration.

Chapter 4 Matter Water Topic C Water Pollutoin Very Short Answer Type Questions Choose The Correct Answer

Question 1. The type of salt responsible for algal bloom in water is

  1. Chloride
  2. Arsenic
  3. Phosphate
  4. Fluoride

Answer: 2. Arsenic

Question 2. DDT is used as a

  1. Fertiliser
  2. Detergent
  3. Pesticide
  4. Soap

Answer: 3. Pesticide

Question 3. Which of the following is not a water-borne disease?

  1. Typhoid
  2. Cholera
  3. Malaria
  4. Jaundice

Answer: 3. Malaria

Question 4. The quantity of oxygen (in mg) required for complete oxidation of the dissolved organic substances in 1 litre of water is known as

  1. AOD
  2. BOD
  3. COD
  4. DOD

Answer: 2. BOD

Question 5. The value of COD is generally

  1. Equal to BOD
  2. Less than BOD
  3. Greater than BOD
  4. None of these

Answer: 3. Greater than BOD

Question 6. Degree of hardness of deionised water is

  1. 0
  2. 2
  3. 4
  4. 7

Answer: 1. 0

Question 7. Which of the following is caused in human body due to arsenic pollution in water?

  1. Decay of tooth enamel
  2. Black patches on the skin of mainly feet and palms
  3. Methaemoglobinaemia
  4. Typhoid

Answer: 2. Black patches on the skin of mainly feet and palms

Question 8. What type of salt present in water causes dental erosion?

  1. Chloride
  2. Bromide
  3. Iodide
  4. Fluoride

Answer: 4. Fluoride

Question 9. Which class of compounds present in detergents is responsible for causing water pollution?

  1. Sulphates
  2. Chlorides
  3. Phosphates
  4. Carbonates

Answer: 3. Phosphates

Question 10. Soap is a

  1. Calcium salt of fatty acids having high molecular mass
  2. Sodium salt of fatty acids (organic acid) having high molecular mass
  3. Aluminium salt of fatty acids having low molecular mass
  4. Magnesium salt of fatty acids having high molecular mass

Answer: 2. Sodium salt of fatty acids (organic acid) having high molecular mass

Question 11. In which of the following districts of West Bengal, arsenic contamination of groundwater is maximum?

  1. Purulia
  2. Darjeeling
  3. West Midnapore
  4. Murshidabad

Answer: 4. Murshidabad

Question 12. Which of the following is not responsible for fluoride pollution of groundwater?

  1. Cryolite
  2. Fluorapatite
  3. Fluorspar
  4. Sodium fluoride

Answer: 4. Sodium fluoride

Question 13. Which of the following methods is used to control arsenic pollution?

  1. Adsorption method
  2. Co-precipitation method
  3. Ion-exchange method
  4. All of these

Answer: 4. All of these

Question 14. Cause of minamata disease is

  1. Arsenic pollution
  2. Fluoride pollution
  3. Mercury pollution
  4. Phosphate pollution

Answer: 3. Mercury pollution

Question 15. Which one is the purest form?

  1. De-ionised water
  2. Distilled water
  3. Rain water
  4. Tubewell water

Answer: 2. Distilled water

Question 16. Which of the following is formed when chlorine reacts with water?

  1. HOCI
  2. HCI
  3. Nascent oxygen
  4. All of the above

Answer: 4. All of the above

Question 17. Black-foot disease is caused by

  1. \(\mathrm{NO}_3^{-}\)
  2. \(\mathrm{AsO}_3^{3-}\)
  3. \(\mathrm{PO}_4^{3-}\)
  4. F

Answer: 2. \(\mathrm{AsO}_3^{3-}\)

Question 18. Which one of the following is the process for removal of fluoride pollution?

  1. Passing through Al2O3 column
  2. Addition of lime and alum
  3. Co-precipitation process
  4. All of the above

Answer: 4. All of the above

Question 19. The disease caused due to drinking of water containing nitrate ions is

  1. Black-foot disease
  2. Itai-itai
  3. Methamoglobinemia
  4. Fluorosis

Answer: 3. Methamoglobinemia

Question 20. Which of the following reduces in water as a result of eutrophication?

  1. Dissolved hydrogen
  2. Dissolved oxygen
  3. Dissolved nitrate
  4. Dissolved minerals

Answer: 2. Dissolved oxygen

Question 21. Which of the following is present in the toxic chemicals excreted from paper industry?

  1. Cd
  2. Hg
  3. Pb
  4. Nitrate

Answer: 2. Hg

Question 22. Itai-itai disease is caused by

  1. Mercury pollution
  2. Phosphate pollution
  3. Cadmium pollution
  4. Sulphate pollution

Answer: 3. Cadmium pollution

Question 23. Which of the following medium is polluted mostly by arsenic?

  1. Air
  2. Water
  3. Soil
  4. All of the above

Answer: 2. Water

 

Chapter 4 Matter Water Topic C Water Pollutoin Answer In Brief

Question 1. Name three metals pollutants present in industrial wastes.

Answer: Cadmium (Cd), mercury (Hg) and lead (Pb).

Question 2. Name a pesticide that causes water pollution.

Answer: DDT (p, p’-dichlorodiphenyltrichloroethane) is a pesticide that causes water pollution.

Question 3. Which type of fertiliser present in water may cause methaemoglobinaemia in children?

Answer: Presence of nitrate fertilisers in water may cause methaemoglobinaemia in children.

Question 4. Which type of fertiliser is responsible for causing algal bloom in water bodies?

Answer: Mainly phosphate fertilisers are responsible for causing algal bloom in water bodies.

Question 5. State whether eutrophication increases or decreases the amount of dissolved oxygen in water.

Answer: Eutrophication decreases the amount of dissolved oxygen in water.

Question 6. What is the full form of BOD?

Answer: BOD stands for Biochemical Oxygen Demand (or Biological Oxygen Demand).

Question 7. What is the permissible value of BOD for pure water?

Answer: The permissible value of BOD for pure water is 5 ppm.

Question 8. What does a high value of BOD of water indicate?

Answer: A high value of BOD of water indicates the presence of large amount of organic pollutants in water.

Question 9. Waste water released from a factory has a pH value less than 3.7—From this given statement, what can you conclude about the quality of the waste water?

Answer: The waste water released from the factory is highly acidic in nature and thus, will cause water pollution.

Question 10. What are the major causes of thermal pollution of water?

Answer: Hot water released from the thermal power plants, oil refineries and other industries mixes with the nearby water bodies to cause thermal pollution.

Question 11. Presence of which metal in drinking water causes Minamata disease?

Answer: The presence of mercury or mercury containing compounds in drinking water causes Minamata disease.

Question 12. Name a disease caused due to fluoride pollution of water.

Answer: Dental fluorosis is a disease caused due to fluoride pollution of water.

Question 13. Name a disease caused due to arsenic pollution of water.

Answer: Black foot disease is caused due to arsenic pollution of water.

Question 14. Name a pesticide that contains arsenic.

Answer: Lead hydrogen arsenate (PbHAsO4) is an inorganic pesticide that contains arsenic.

Question 15. Name a metal ore found in the earth’s crust that contains arsenic.

Answer: Arsenopyrites (FeAsS) is an ore of iron that contains arsenic and is found in the earth’s crust.

Question 16. Which soluble salts are produced when water insoluble arsenic compounds present in the earth’s crust react with air?

Answer: When water insoluble arsenic compounds present in the earth’s crust react with air, soluble arsenate and arsenate salts are produced.

Question 17. Which substance is used as the semiper- meable membrane during removal of arsenic from water by reverse osmosis?

Answer: During removal of arsenic from water by reverse osmosis method, cellulose triacetate is used as the semipermeable membrane.

Question 18. What amount of arsenic in blood may cause arsenicosis?

Answer: If blood contains more than 60μg • L-1 of arsenic, then it may cause arsenicosis.

Question 19. What is biological magnification?

Answer: Biological magnification or biomagnification is a phenomenon by which the toxic chemical substances (such as DDT, aldrin which are generally not decomposed by biochemical reactions) are continuously deposited in the fat-tissues of living organisms through intake of food (food chain).

Question 20. Which pollutant present in drinking water causes anemia?

Answer: Nitrate (\(\mathrm{NO}_3^{-}\)) causes anaemia.

Question 21. What is the reason behind knock-knee syndrome?

Answer: consumption of fluoride (F) contaminated water for a long period of time causes knock- knee syndrome.

Question 22. Which ores are responsible for fluoride pollution of drinking water?

Answer: Fluorapetite [3Ca3(PO4)2 • CaF2],fluospar (CaF2) and cryolyte (Na3AIF6).

Question 23. Which disease is caused by Arsenic Pollution?

Answer: Arsenicosis.

Question 24. What is the pollution of water caused due to the algal decomposition called?

Answer: Algal bloom.

Question 25. Name two permanent organic pollutant?

Answer: Dichlorodiphenyltrichloroethane (DDT) and methyl murcury.

Question 26. Which metalloid-compound in responsible for water pollution?

Answer: Arsenic.

Question 27. What is fluorosis disease?

Answer: Fluorosis is a crippling disease resulted from deposition of fluorides in the hard and soft tissues of body caused by excess intake of fluoride through drinking water and affects teeth and bones.

Question 28. Due to the presence of which element itai- itai occurs?

Answer: Due to the presence of cadmium in drinking water itai-itai disease occurs.

Chapter 4 Matter Water Topic C Water Pollutoin Fill In The Blanks

Question 1. The metalloid whose compounds are responsible for water pollution is _________

Answer: Arsenic

Question 2. High _________ power of water makes it more vulnerable towards pollution.

Answer: Solvation

Question 3. Water soluble arsenic compounds like arsenite and arsenate salts are adsorbed by active ________

Answer: Alumina(AI2O3)

Question 4. The full form of COD is ________

Answer: Chemical Oxygen Deeman

Question 5. An example of a water-borne disease is _________

Answer: Typhoid

Question 6. Malathion and parathion are ________

Answer: Insecticides

Question 7. _______ is an example of organoarsenic compound.

Answer: Cacodylic acid

Question 8. Continuous lifting of groundwater increases the amount of _______ in water.

Answer: Arsenic

Question 9. ______are used as fillers in detergents.

Answer: Phosphates

Question 10. There are ______ layers of filters in a filtration tank.

Answer: Three

Question 11. The extent of water pollution _______ as the BOD value of water increases.

Answer: Increases

Question 12. The phenomenon of continuous accumulation of toxic chemicals in the fat tissues of living beings is known as ________

Answer: Biomagnification

Question 13. ________ which is a fluoride-containing mineral causes fluoride pollution in water.

Answer: Cryolite

Question 14. DDT, a widely used pesticide is not ________ in nature.

Answer: Biodegradable

Question 15. The amount of dissolved oxygen in water _______ due to eutrophication.

Answer: Decreases

Question 16. Presence of ________ of fluorides in drinking water is favourable for teeth.

Answer: 1 mg • L-1

Question 17. As a result of eutrophication, the extent of dissolved oxygen in water rapidly ________

Answer: Decreases

Question 18. ________ is used for the removal of arsenic by adsorption method.

Answer: Activated alumina

Chapter 4 Matter Water Topic C Water Pollutoin State Whether True Or False

Question 1. Sulphate salts are responsible for algal bloom in water bodies.

Answer: False

Question 2. Blackfoot disease is caused due to arsenic pollution of water.

Answer: True

Question 3. High value of BOD of a water sample indicates decreased level of water pollution.

Answer: False

Question 4. The two major components of detergent are a surface-active agent and a builder or filler.

Answer: True

Question 5. The element arsenic is directly responsible for arsenic pollution of water.

Answer: False

Question 6. The toxic effects of DDT disrupts the food chain and ultimately causes an imbalance in the ecosystem.

Answer: True

Question 7. Eutrophication can lead to a decrease in the depth of a water body.

Answer: True

Question 8. As COD increases, extent of pollution also increases.

Answer: True

Question 9. Value of COD is generally lower than that of BOD for a sample of water.

Answer: False

Question 10. Malaria is a water-born disease.

Answer: False

Question 11. For prolonged intake of arsenic contaminated water arsenicosis occurs.

Answer: True

Chapter 4 Matter Water Topic C Water Pollutoin Miscellaneous Type Questions

Match the columns

1.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Water Topic C Water Pollution Match The Column 1

Answer: 1. B, 2. C, 3. D, 4. A

2.

WBBSE Solutions For Class 9 Physical Science Chapter 4 Matter Water Topic C Water Pollution Match The Column 2

Answer: 1. C, 2. A, 3. D,  4. B

WBBSE Solutions For Class 9 Physical Science Chapter 6 Calorimetry

Chapter 6 Calorimetry Synopsis

Heat is a form of energy and exchange of heat is measurable. The branch of science which deals with the exchange of heat of a body, i.e., principles, methods or matters related to measurement of absorbed heat or emitted heat is known as calorimetry.

Specific Heat is the quantity of heat required to increase the temperature of unit mass of a substance by unit amount.

Units Of Specific Heat in CGS — cal • g-1 • °C-1 and in SI — J • kg-1 • K-1.

\(1 \mathrm{cal} \cdot \mathrm{g}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}=\frac{4.2 \mathrm{~J}}{\frac{1}{1000} \mathrm{~kg} \cdot \mathrm{K}}\) = 4200 J • kg-1 • K-1

Thermal Capacity Or Heat Capacity is the quantity of heat needed to raise the temperature of a body by unity.

Units of thermal capacity in CGS —  cal/°C and in SI —  J/K.

Read and Learn More WBBSE Solutions for Class 9 Physical Science and Environment

Water Equivalent of a body is the mass of water for which the increase in temperature is same as that for the body when the amount of heat supplied is the same for both.

Units Of Water Equivalent in CGS—g and in SI—kg

The Amount Of Heat (H) Absorbed Or Released by a body during a thermal exchange depends on

  1. Mass of the body (m),
  2. Change of temperature (t1 ~ t2) of the body,
  3. Specific heat of the material of the body (s); i.e, H = m • s • (t1 ~ t2)

If two bodies of different temperatures are kept in contact with each other, the hotter body emits, heat and the other one receives that heat. If there is no loss of heat by any means, then at the time of thermal equilibrium, heat emitted by the hotter body = heat received by the colder body This is the fundamental principle of calorimetry.

Units Of Heat in CGS system and SI are calorie (cal) and joule (J), respectively. [Here, 1 cal = 4.1855 J]

Chapter 6 Calorimetry Synopsis Short And Long Answer Type Questions

Question 1. What do you mean by calorimetry?

Answer:

Calorimetry:-

Heat is a kind of energy and exchange of heat of a body is measurable. Calorimetry is that branch of science which deals with the exchange of heat of a body, i.e., principles, methods, or matters related to measurement of absorbed heat or emitted heat.

Question 2. Explain the basic principle of calorimetry.

Answer:

Basic Principle Of Calorimetry:-

If two bodies with different temperatures are kept in contact with each other, the body with the higher temperature releases heat to become colder and the other one accepts this heat to become hotter.

If there is no loss of heat, then at thermal equilibrium, heat emitted by the hotter body = heat received by the colder body This is the basic principle of calorimetry.

Suppose two bodies of masses m1, m2, and respective temperatures t1, t2 (t1 > t2) are kept in contact with each other. The first body emits heat and the second one receives heat. Let us assume that at the time of thermal equilibrium, the common temperature is t.

If the specific heats of the first and the second body are s1 and s2 respectively, then heat emitted by the first body, Q1 = m1s1(t1 -t), and heat received by the second body, Q2 = m2s2(t – t2).

Now as per the basic principle of calorimetry,

Q1 = Q2 or, m1s1(t1 – t) = m2s2(t – t2)

Question 3. What are the conditions for the applicability of the basic principle of calorimetry?

Answer:

Conditions Applicable For Basic Principle Of Calorimetry:-

The conditions for applicability of the basic principle of calorimetry are:

  1. During exchange of heat among different bodies with varied temperature, there should not be any exchange of heat with the surroundings.
  2. There should not be any chemical reaction between the bodies.
  3. If there is any exchange of heat between a liquid and a solid body, then the solid body must not dissolve in the liquid.
  4. There should not be any kind of physical change in the bodies while absorbing or releasing heat.

Question 4. Write down the differences between heat and temperature.

Answer:

The differences between heat and temperature are given below

WBBSE Solutions For Class 9 Physical Science Chapter 6 Calorimetry Differences Between Heat And Temperature

Question 5. Two bodies of different temperatures are bought into thermal contact. What change in temperature of the two bodies will happen?

Answer:

When two bodies at different temperatures are bought in thermal contact, heat is transferred from the body at higher temperature to that at lower temperature. Therefore hotter body starts to cool down whereas the colder body starts to warm up.

The flow of heat continues till both reaches the same temperature. At thermal equilibrium temperature of both the bodies are equal.

Question 6. The temperature of three solid metallic bodies A, B, and C are 10°C, 20°C, and 10°C, respectively. If the three bodies are kept in contact with each other, which one releases heat and which one accepts?

Answer:

In this case, temperature of B is highest. So, B releases heat. The other two bodies A and C have the same temperature which is less than that of B. So, these two accept heat.

Question 7. What do you mean by specific heat of a substance?

Answer:

Specific Heat Of A Substance Means:-

Specific heat is the quantity of heat required to increase the temperature of unit mass of a substance by unit amount.

Question 8. What is thermal equilibrium? When several bodies are in thermal equilibrium which physical quantity is same for all of the bodies?

Answer:

Thermal Equilibrium

When several bodies at different temperatures are bought into thermal contact with each other, they will exchange heat among themselves and ultimately attain the same temperature. This state where all the bodies are at the same temperature is called thermal equilibrium, b When several bodies are at thermal equilibrium the physical quantity ‘temperature’ is same for all of the bodies.

Question 9. Why water is used in hot water bag?

Answer:

Water has a specific heat higher than everything (except ammonia). Hence a fixed mass of water gains more heat than any other liquid of the same mass for the same rise in temperature.

Consequently, it loses more heat during cooling. This heat is used for fomentation. So a hot water bottle remains effective for a long period of time. That is why water is used in hot water bag.

Question 10. What is water equivalent of a substance?

Answer:

The water equivalent of a body is the mass of water for which the increase in temperature is same as that for the body when the amount of heat supplied is the same for both.

Question 11. Write the factors on which the amount of heat absorbed or released by a body depends on.

Answer:

The amount of heat (H) absorbed or released by a body during a thermal exchange depends on

  1. Mass of the body (m),
  2. Change of temperature (t) of the body,
  3. Specific heat of the material of the body (s).

The relation among them is — H = m • s • t.

Question 12. Same amount of heat is supplied to two different bodies of equal mass but of different materials. The increase in temperature of the two bodies be equal or not?

Answer:

We know, given heat H = mst i.e., increase in temperature t = \(\frac{H}{m \cdot s}\).

Since for the two bodies, heat supplied (H) and mass (m) are equal but specific heat (s) of the material of the bodies are different, t ∝ 1/s.

Hence, the body with comparatively less specific heat will have a higher increase in temperature.

Question 13. Same amount of heat is supplied to equal mass of milk and water. Why milk warms quicker than water?

Answer:

Increase in temperature (t) of a substance is inversely proportional to specific heat (s) of the material of the substance when amount of heat (H) absorbed and mass of the body (m) are kept constant.

i.e., t ∝ 1/s, H, and m are constant

Specific heat of milk is less than that of the water. Hence milk warms quicker than water.

Question 14. Temperatures of two liquids of equal masses are t1 and t2 (t1 > t2) respectively. T is the final temperature of their mixture. Find the ratio of specific heat of the two liquids.

Answer:

As per the question, mass (m) of each liquid is same and specific heat of two liquids are s1 and s2 respectively.

According to basic principle of calorimetry, m • (t1 – t) • s1 = m • (t – t2) • s2

∴ \(\frac{s_1}{s_2}=\frac{t-t_2}{t_1-t}\)

or, s1 : s2 = (t – t2) : (t1 – t)

Question 15. What is the specific heat of water during its freezing at 0°C?

Answer:

Specific heat, s = \(\frac{H}{m \cdot t}\). During freezing, the temperature of water remains fixed at 0°C. So the change in temperature t = 0 hence h = ∞.

Chapter 6 Calorimetry Very Short Answer Type Questions Choose The Correct Answer

Question 1. If equal masses of two liquids at temperatures 10°C and 40°C are mixed together, the temperature of the mixture becomes 15°C. The ratio of specific heat of the two liquids is

  1. 5: 1
  2. 1: 5
  3. 2 : 3
  4. 4 : 3

Answer: 1. 5 : 1

Question 2. Heat is the of a material particle.

  1. Changed form of momentum
  2. Changed form of kinetic energy
  3. Changed form of potential energy
  4. Changed form of velocity

Answer: 2. Changed form of kinetic energy

Question 3. Thermometer is used to measure

  1. Temperature of a body
  2. Latent heat
  3. Radiated heat
  4. All of the above

Answer: 1. Temperature of a body

Question 4. Absorbed heat is _____ of a body.

  1. Directly proportional only to the mass
  2. Directly proportional only to the rise of temperature
  3. Directly proportional to the product of mass and increased temperature
  4. Directly proportional to the quotient of mass and increased temperature

Answer: 3. Directly proportional to the product of mass and increased temperature

Question 5. For calorimetric principle to be applicable

  1. Heat has to enter the body or the system from outside
  2. Heat has to go out of the body or the system
  3. A chemical reaction has to take place in the body or the system
  4. One body should not be soluble in another body

Answer: 4. One body should not be soluble in another body

Question 6. Among iron, mercury, water and air, which of the following has the highest specific heat?

  1. Water
  2. Iron
  3. Air
  4. Mercury

Answer: 1. Water

Question 7. A piece of iron and a wooden chair of the same mass are kept in sunlight. After keeping them for the same period of time, the piece of iron feels hotter on touching than the wooden chair due to

  1. Difference in water equivalent
  2. Difference in heat capacity
  3. Difference in heat conduction
  4. Difference in specific heat

Answer: 4. Difference in specific heat

Question 8. Heat accepted or rejected by a body is directly proportional to the increase or decrease of its temperature under the condition that

  1. Mass of the body is variable
  2. Mass of the body is fixed
  3. Water equivalent of the body is variable
  4. Latent heat of change of state of the body is constant

Answer: 2. Mass of the body is fixed

Question 9. According to the formula s = value of the specific heat of a body during melting becomes

  1. Zero
  2. 1
  3. Constant
  4. Infinite

Answer: 3. Constant

Question 10. According to the formula H = mst, which of the following is not true in case of a specific material?

  1. H ∝ m
  2. H ∝ t
  3. H ∝ s
  4. None of these

Answer: 3. H∝ s

Question 11. During fomentation, water bottle is used instead of iron. The main reason behind this is

  1. Specific heat of water is greater than that of iron
  2. Specific heat of iron is greater than that of water
  3. Iron gets heated slowly compared to water
  4. Water releases heat faster than iron

Answer: 1. Specific heat of water is greater than that of iron

Question 12. In the formula H = mst, t indicates

  1. The temperature of the body
  2. The change of temperature of the body
  3. The total time of heating
  4. All of the above

Answer: 2. The change of temperature of the body

Question 13. Mathematical form of the first law of thermodynamics is

  1. H = mL
  2. H = mst
  3. W = JH
  4. W = Fs

Answer: 3. W = JH

Question 14. Specific heat of copper is 0.1 cal • g-1 • °C-1. Water equivalent of a copper calorimeter of mass 0.4 kg is

  1. 40 g
  2. 4000 g
  3. 200 g
  4. 4 g

Answer: 1. 40 g

Question 15. What is the temperature difference between the top and bottom of a 100 m high waterfall, if the total amount of heat produced remains confined within the water?

  1. 0.23°C
  2. 0.46°C
  3. 0.15°C
  4. 0.69°C

Answer: 1. 0.23°C

Question 16. What is the final temperature of a mixture if 19.9 g of water at 30°C and 5g of ice at -20°C are mixed in a calorimeter? (specific heat of ice = 0.5 cal • g-1 • °C-1)

  1. 0°C
  2. -5°C
  3. 5°C
  4. 10°C

Answer: 3. 5°C

Question 17. Same masses of water and ice are mixed together at 0°C temperature. If the entire amount of ice melts, the initial temperature of water becomes

  1. 40°C
  2. 50°C
  3. 60°C
  4. 80°C

Answer: 4. 80°C

Question 18. Heat is supplied to a chunk of ice at the same rate. Ice starts melting after 5 s and in the next 40 s, the entire amount of ice melts. If the specific heat of ice is 0.5 cal • g-1 • °C-1, what is the initial temperature of ice?

  1. -20°C
  2. -22°C
  3. -15°C
  4. -10°C

Answer: 1. -20°C

Question 19. Calorimeters are generally made of

  1. Glass
  2. Copper
  3. Gold
  4. Wood

Answer: 2. Copper

Question 20. Which of the following substances has the highest specific heat?

  1. Water
  2. Alcohol
  3. Kerosene
  4. Milk

Answer: 1. Water

Question 21. cal • g-1 • °C-1 is the unit of

  1. Latent heat
  2. Water equivalent
  3. Thermal capacity
  4. Specific heat capacity

Answer: 4. Specific heat capacity

Question 22. Calorimetry relates to the measurement of

  1. Heat
  2. Temperature
  3. Mechanical energy
  4. Internal energy

Answer: 1. Heat

Question 23. Quantity of heat absorbed by a body depends on

  1. Mass
  2. Specific heat
  3. Increase in temperature
  4. All of them
  5. Answer: 4. All of them

Question 24. Thermal capacity of a body of mass m and specific heat s is

  1. \(\frac{m}{s}\)
  2. \(\frac{s}{m}\)
  3. ms
  4. \(\frac{1}{ms}\)

Answer: 3. ms

Question 25. Mean calorie means

  1. The amount of heat required to increase the temperature of 1 g of water from 0°C to 1°C
  2. The amount of heat required to increase the temperature of 1 g of water from 50°C to 51°C
  3. The amount of heat required to increase the temperature of 1 g of water from 14.5°C to 15.5°C
  4. 1/100 part of the amount of heat required 100 to increase the temperature of 1 g of water from 0°C to 100°C

Answer: 4. 1/100 part of the amount of heat required 100 to increase the temperature of 1 g of water from 0°C to 100°C

Question 26. During boiling of water at 100°C, what will be its specific heat?

  1. Zero
  2. 0.5
  3. 1
  4. Infinite

Answer: 4. Infinite

Question 27. Two bodies of different temperatures are mixed in a calorimeter. Which one of the following will remain conserved?

  1. Sum of the temperatures of the bodies
  2. Total heat of the bodies
  3. Total internal energies of the bodies
  4. Internal energy of each body

Answer: 3. Total internal energies of the bodies

Question 28. Which of the following pairs of physical quantities may have the same unit?

  1. Specific heat capacity and heat
  2. Heat capacity and water equivalent
  3. Specific heat capacity and heat capacity
  4. Heat and work

Answer: 4. Heat and work

Question 29. Which one of the following physical quantity is not relevant in calculation of heat absorbed by a body?

  1. Mass
  2. Density
  3. Specific heat capacity
  4. Increase in temperature

Answer: 2. Density

Question 30. Specific heat of lead in CGS is 0.03 cal • g-1 • °C-1. Specific heat of lead in SI is

  1. 12.6 J • kg-1 • K-1
  2. 126 J • kg-1 • K-1
  3. 1260 J • kg-1 • K-1
  4. 552 J • kg-1 • K-1

Answer: 2. 126 J • kg-1 • K-1

Question 31. Due to release of heat, electrical conductivity of a body

  1. Decreases
  2. Increases
  3. Remains constant
  4. Sometimes increases, sometimes decreases

Answer: 2. Increases

Chapter 6 Calorimetry Answer In Brief

Question 1. What is the unit of heat in CGS system?

Answer: Unit of heat in CGS system is calorie (cal)

Question 2. What is the unit of heat in SI?

Answer: Unit of heat in SI is joule (J).

Question 3. Boiled rice is prepared by boiling grain rice. What type of change takes place in this case due to absorption of heat?

Answer: Due to absorption of heat, chemical reaction takes place in this case.

Question 4. Heat is sometimes given to open the lid of stuck steel tiffin box. What is the role of heat in this case?

Answer: In this case, due to absorption of heat, pressure of gas inside the tiffin box increases, thereby opening the lid of the box.

Question 5. How is the heat received by a body related to the mass of the body?

Answer: For the increase of temperature, heat received by a body is directly proportional to the mass of the body.

Question 6. How is the heat received by a body of definite mass related to the temperature of the body?

Answer: Heat received by a body of definite mass is directly proportional to the increase of temperature of the body.

Question 7. Suppose the velocity of a meteorite of mass 52 kg is reduced from 15 km/s to 5 km/s while passing through the atmosphere of the earth. Calculate the amount of heat produced due to this change in velocity?

Answer: Amount of heat = 109 cal.

Question 8. What is the heat capacity of a body?

Answer: The amount of heat required to increase the temperature of the body by 1 kelvin (K) or 1°C is the heat capacity of the body.

Question 9. What is specific heat of a body?

Answer: Heat required to increase the temperature of a body by a unit (1°C or IK) is the specific heat of that body.

Question 10. If glucose is mixed with water, it gets completely dissolved. Is the basic principle of calorimetry applied here?

Answer: No, when one material is dissolved into another, basic principle of calorimetry is not applied.

Question 11. What does the quantity ms signify in the relationship, H = mst?

Answer: ms signifies heat capacity of the body.

Chapter 6 Calorimetry Fill In The Blanks

Question 1. Value of specific heat at the time of melting Of ice is __________

Answer: Infinte

Question 2. While ________ of heat of a body occurs, its temperature decreases.

Answer: Emission

Question 3. _________ of a body is defined as the amount of heat required to raise the temperature of the body by 1°C.

Answer: Heat capacity

Question 4. Unit of heat capacity in SI is _________

Answer: J.kg-1.K-1

Question 5. Specific heat of water in CGS system is taken as ________

Answer: 1

Question 6. Specific heat is sometimes referred to as _______ per unit mass of body.

Answer: Thermal capacity

Question 7. Specific heat of a substance can not be __________

Answer: Negative

Question 8. Two bodies are said to be in thermal equilibrium if they attain a _________ temperature.

Answer: Common

Chapter 6 Calorimetry State Whether True Or False

Question 1. According to Joule’s law, work done is directly proportional to the heat emitted.

Answer: True

Question 2. Specific heat of water is 4200 J • kg-1 • K-1.

Answer: True

Question 3. Temperature is the external manifestation of heat.

Answer: True

Question 4. Basic principle of calorimetry is not applicable for reaction of lime with water.

Answer: True

Question 5. The quantity of heat absorbed or given out by a body = weight of the body x specific heat capacity x change in temperature.

Answer: False

Question 6. Water equivalent indicates some volume of water.

Answer: False

Question 7. Water equivalent and heat capacity both can be measured by the term mL.

Answer: False

Question 8. Quantity of heat absorbed by a body depends on the medium of the surroundings.

Answer: False

Question 9. Thermal capacity and water equivalent of a body have some value in CGS system.

Answer: True

Chapter 6 Calorimetry Numerical Examples

Useful information

  1. Heat capacity of a body of mass m and specific heat s is = m • s.
  2. Quantity of heat absorbed or given out by a body (H) = mass of the body (m) x specific heat capacity (s) x change in temperature (t1 ~ t2).
  3. If two bodies of masses m1, m2; specific heat capacities s1, s2 and temperatures t1, t2 (t1 > t2) respectively are kept in thermal contact.
  4. At thermal equilibrium, final temperature is t. According to the basic principle of calorimeter, m1s1(t1-t) = m2s2(t-t2)

Question 1. Specific gravities of two liquids are 0.8 and 0.5, respectively. Equal amount of heat is required to raise the temperature of first liquid (3 L) and second liquid (2 L) by 1°C. What is the ratio of the specific heats of these two liquids?

Answer:

Specific gravities of two liquids are 0.8 and 0.5, respectively.

So, the densities of the liquids are 0.8 g/cm3 and 0.5 g/cm3, respectively.

Mass of 3 L or 3000 cm3 of first liquid,

m1 = 3000 x 0.8 = 2400 g

Mass of 2 L or 2000 cm3 of second liquid,

m2 = 2000 x 0.5 = 1000 g

Let us assume, s1 and s2 are the respective specific heats of first and second liquid.

According to the condition, we get

\(m_1 \times s_1 \times 1=m_2 \times s_2 \times 1\)

or, \(\frac{s_1}{s_2}=\frac{m_2}{m_1}\)

or, \(\frac{s_1}{s_2}=\frac{1000}{2400}\)

or, \(\frac{s_1}{s_2}=\frac{5}{12}\)

∴ \(s_1: s_2=5: 12\)

Hence, ratio of specific heats of two liquid is 5 : 12.

Question 2. A saucepan contains 100 g of water at 25°C. If 50 g of water of temperature 60°C is poured in it, the temperature of the mixture becomes 35°C. If the mass of the saucepan is 250 g, what is the specific heat of the constituent of the saucepan?

Answer:

Mass of water in sauce pan, m1 = 100 g

Initial temperature, t1 = 25°C

Mass of an extra amount of water poured in the pan, m2 = 50 g, and initial temperature, t2 = 60°C

Mass of the saucepan, m3 = 250 g

Final temperature of the mixture, t = 35°C

Specific heat of water, s1 = 1 cal • g-1 • °C-1

Now let us assume, specific heat of the constituent of the saucepan = s

So, heat absorbed by the saucepan and water of saucepan,

Q1 =m3 • s(t – t1) + m1 x s1 x (t – t1)

= 250 x s x (35 – 25) + 100 x 1 x (35-25)

=(2500s + 1000) cal

Heat released by water that is poured,

Q2 = m2 x s2 x (t2 -1)

= 50 x 1 x (60 – 25) = 1250 cal

From the basic principle of-calorimetry, we get Q1 = Q2

or, 2500s + 1000 = 1250 or, 2500s = 250

s = \(\frac{250}{2500}\) = 0.1 cal • g-1 • °C-1

Question 3. Temperature of a copper weight of 50 g is 90°C. What is the final temperature if the weight is immersed in 100 g water of temperature 10°C ? Specific heat of copper is 0.09 cal • g-1  • °C-1.

Answer:

Mass of the copper weight, m1 = 50 g

Initial temperature, t1 = 90°C

Specific heat, s1 = 0.09 cal • g-1 • °C-1

Mass of water, m2 = 100 g

Initial temperature, t2 = 10°C

Specific heat, s2 = 1 cal • g-1 • °C-1

Let us assume, final temperature = f.

So, heat released by the copper weight,

Q1 = m1s1(t1 – t)

= 50 x 0.09(90 -1) = 4.5(90 – t) cal

and heat absorbed by water,

Q2 = m2s2(t – t1)

= 100 x 1(t – 10) = 100 (t – 10) cal

From the basic principle of calorimetry, Q1 = Q2

or, 4.5(90 – t) = 100(t – 10)

or, 405 – 4.5t= 100t-1000

or, 104.5t = 1405

∴ t = \(\frac{1405}{104.5}\) = 13.44°c

Question 4. A platinum piece of mass 200 g is heated in a furnace and then slowly dipped in 650 g of water at 10° C in a vessel. This vessel has a mass of 500 g and specific heat of 0.1 cal • g-1 • °C-1. If the final temperature of the mixture is 25°C, what is the temperature of the furnace? Specific heat of platinum = 0.3 cal •  g-1 • °C-1.

Answer:

Suppose, temperature of the furnace is t1, then initial temperature of the platinum piece is also t1.

Mass of platinum, m1 = 200 g

Mass of water, m2 = 650 g

Initial temperature, t2 = 10°C

Specific heat, s2 = 1 cal • g-1 • °C-1

Mass of vessel, m3 = 500 g

Specific heat of constituent of vessel, s3 = 0.1 cal • g-1 • °C-1

Final temperature of mixture, t = 25°C

Now, heat released by the platinum piece,

Q1 = m1s1(t1 – t)

= 200 x 0.03 (t1 – 25) = (6t1 – 150) cal

and heat absorbed by vessel and water,

Q2 = m3s3(t – t2) + m2s2(t – t2)

= 500 x 0.1 x (25-10) + 650 x 1 x (25 – 10) = 10500 cal

From the basic principle of calorimetry, we get Q1 = Q2

or, 6t1 – 150 = 10500

∴ t = \(\frac{10650}{8}\)

Question 5. Temperature of three liquids of the same mass A, B, and C are 10°C, 30°C, and 4Q° C, respectively. If A and B are mixed together, the temperature of the mixture becomes 15°C. Again if B and C are mixed together, temperature of the mixture becomes 34°C. What is the ratio of specific heats of these three liquids?

Answer:

Suppose, specific heats of the three liquids A, B, and C are s1, s2, and s3, respectively and each liquid has a mass of m.

If A and B are mixed together, temperature of the mixture becomes 15°C.

From the basic principle of calorimetry, we get heat absorbed by liquid A = heat released by liquid B

∴ m x s1 x (15 – 10) = m x s2 x (30-15) or, 5s1 = 15s2

or, s2 = \(\frac{s_1}{3}\)

Again, if B and C are mixed together, temperature of the mixture becomes 34°C.

Again from basic principle of calorimetry, we get heat absorbed by liquid B = heat released by liquid C

∴ \(m \times s_2 \times(34-30)=m \times s_3 \times(40-34)\)

or, \(4 s_2=6 s_3 \quad \text { or, } s_3=\frac{2 s_2}{3}\)

or, \(s_3=\frac{2}{3} \times \frac{s_1}{3}=\frac{2 s_1}{9}\left[because s_2=\frac{s_1}{3}\right]\)

∴ ratio of specific heats of A , B, and C is

∴ \(s_1: s_2: s_3=s_1: \frac{s_1}{3}: \frac{2 s_1}{9}=9: 3: 2\)

Question 6. A 100 W powerful electric heater raises the temperature of 5 kg of a liquid from 25°C to 31°C in 2 minutes. What is the specific heat of the liquid?

Answer:

In 2 min = 60 x 2s = 120s, heat given by the electric heater,

Q = 100 x 120 J = 12000 J

Mass of the liquid, m = 5 kg

Increase in temperature,

t = (31 – 25)°C = 6°C

As we know, change of 1°C = change of 1 K

∴  increase in temperature, t = 6 K

Let specific heat of the liquid = s.

So, m x s x t = Q

or, 5 x s x 6 = 12000

∴ s = 400 J • kg-1 • K-1

Question 7. 90 g of water at 20°C is kept in a vessel of mass 100 g and specific heat of 0.1 cal • g-1 • °C-1. A piece of metal of mass 50 g is heated and then slowly dropped in this vessel. If decrease in temperature of the piece of metal is 20 times the increase in temperature of the water, then what is the specific heat of the constituent of the metal?

Answer:

Mass of the vessel, m1 = 100 g

Specific heat of the constituent of vessel, s1 = 0.1 cal • g-1  • °C-1

Mass of water, m2 = 90 g

Specific heat of water, s2 = 1 cal • g-1 • °C-1

Mass of metallic piece, m3 = 50 g

If increase in temperature of water is t, then according to the question, decrease of temperature of the piece of metal = 20t.

Now, heat absorbed by the vessel, and water is given by,

Q1 = m1 x s1 x t + m2 x s2 x t

= 100 x 0.1 x t + 90 x 1 x t

= 10t+ 90 = 100t cal

Let s be the specific heat of the constituent of the metal.

So, heat released by the metallic piece is given by

Q2 = m3 x s x 20t = 50 x s x 20t = 100t cal

From basic principle of calorimetry, Q1 =Q2

or, 100t = 1000st

∴ s = 0.1 cal • g-1 – °C-1

Question 8. Three liquids have masses m1, m2, and m3, temperature t1, t2, and t3 (t1 > t2 > t3) and specific heats s1 ,s2 and s3, respectively. If these three liquids are mixed together, what is the final temperature of the mixture?

Answer:

Suppose, final temperature of the mixture = t.

From basic principle of calorimetry,

\(m_1 s_1\left(t-t_1\right)+m_2 s_2\left(t-t_2\right)+m_3 s_3\left(t-t_3\right)=0\)

 

or, \(t\left(m_1 s_1+m_2 s_2+m_3 s_3\right)\)

= \(m_1 s_1 t_1+m_2 s_2 t_2+m_3 s_3 t_3\)

∴ \(t=\frac{m_1 s_1 t_1+m_2 s_2 t_2+m_3 s_3 t_3}{m_1 s_1+m_2 s_2+m_3 s_3}\)

Question 9. How much amount of water at 100°C is to be mixed with some amount of water at 20°C to make a mixture of volume 28 L at a temperature of 40°C?

Answer:

Suppose, volume of water at 100°C = x L and mass = 100x g

So, volume of water at 20°C = (28 – x)L and mass = (28 – x) x 1000 g

Specific heat of water = 1 cal • g-1 • °C-1

∴ heat absorbed by water at 20°C,

Q1 =(28 – x) x 1 x (40-20)

= 1000 x (28 – x) x 20 cal

and heat released by water at 100°C,

Q2 = 1000x x 1 x (100-40) = 1000x x 60 cal

From the basic principle of calorimetry, Q1 = Q2

or, (28 – x) x 1000 x 20 = l000x x 60 or, 28x – x = 3x or, 4x = 28

So, x = 7 and 28 – x = 28 – 7 = 21

Hence, 7 L of water at 100°C is to be mixed with 21 L of water at 20°C to make a mixture of volume 28 L at 40°C.

Question 10. Heat is applied to two different object of same mass at the same rate. 8 minutes and 10 minutes are required for raising the temperature by 20°C of the respective objects. Calculate the ratio of the specific heat of these two objects. If the specific heat of the first object is 0.2 cal • g-1 • °C-1, what is the specific heat of the second object?

Answer:

Let the mass of each object = m g.

Specific heat are s1 cal • g-1 • °C-1 and s2 cal • g-1 • °C-1, respectively.

Heat is applied to these objects at the rate of x cal/min.

So for the first object,

m x s1 x 20 = 8x ……(1)

and for the second object,

m x s2 x 20 = 10x ……(2)

Now by dividing equation (1) by equation (2), we get

\(\frac{m \times s_1 \times 20}{m \times s_2 \times 20}=\frac{8 x}{10 x}\)

 

or, \(\frac{s_1}{s_2}=\frac{4}{5}\)

So, the ratio of the specific heat of these two objects is s1: s2 = 4: 5.

Now, s1 = 0.2 cal • g-1 • °C-1.

Substituting the value in equation (3), we get

\(\frac{0.2}{s_2}=\frac{4}{5}\)

 

∴  s2 = 0.25 cal • g-1 • °C-1.

Question 11. Water from two taps are accumulating in a big vessel at the rate of 1.2 kg/min from the first and 0.8 kg/min from the second. Temperature of water from the first is 10°C and temperature from the second is 90°C. Now if both of these taps are opened simultaneously, then what is the temperature of accumulated water after 5 minutes? (Assume that there is no exchange of heat from the vessel.)

Answer:

After 5 minutes, mass of water emitted from the first tap, m1 = 1.2 x 5 = 6 kg

and mass of water emitted from the second tap, m2 =0.8 x 5 = 4 kg.

Let specific heat of water be s and final temperature of mixed water be t.

From the basic principle of calorimetry,

Q1 = Q2

or, m1 x s x (t – 10) = m2 x s x (90 -t)

or, 6(t – 10) = 4(90 – t) or, 6t – 60 = 360 – 4t

∴ t = 42°C

WBBSE Solutions For Class 9 Physical Science Chapter 6 Saturated Unsaturated Vapour And Anomalous Expansion Of Water

Chapter 6 Saturated Unsaturated Vapour And Anomalous Expansion Of Water Synopsis

If the vapour pressure in a closed space at a particular temperature is equal to the highest possible vapour pressure at that temperature, then that vapour is called
saturated vapour.

  1. If the vapour pressure in a closed space at a particular temperature is less than the highest possible vapour pressure at that temperature, then that vapour is called
    unsaturated vapour.
  2. Saturated vapour remains in equilibrium when in contact with the liquid whereas unsaturated vapour does not remain in equilibrium when in contact with the liquid.
  3. Dew point is that temperature at which a certain volume of air is saturated with the water vapour present in it.
  4. If the temperature of the atmosphere goes below the dew point, then some amount of water vapour of the atmosphere condenses as tiny water particles to accumulate on grass, leaves etc. These tiny particles are known as dew.
  5. If the temperature of a large part of air drops below the dew point, water vapour present in the atmosphere condenses as deposits on dust, coal particles etc. floating in air. A thick layer thus formed is called fog.
  6. Relative humidity is the ratio of the mass of water vapour present in a particular volume of air to the mass of water vapour required to saturate that volume at a particular temperature.
  7. In other words, relative humidity is the ratio of partial pressure of water vapour to the saturated pressure of water vapour at a particular temperature.
  8. ∴ relative humidity = \(\frac{m}{M} \times 100 \%\) = \(\frac{f}{F} \times 100 \%\)

where m is the mass of water vapour present in a particular volume of air at t°C, M is the mass of water vapour required to saturate that volume of air, f is the water vapour pressure in air at t°C and F is the saturated vapour pressure of water at that temperature.

Read and Learn More WBBSE Solutions for Class 9 Physical Science and Environment

Absolute humidity is defined as the amount of water vapour in grams present per cubic meter of air (g/m3).

Generally, if the temperature of any liquid is increased, its volume also increases. But there is an exception to this rule in case of water for a specific range of temperature,i.e., from 0°C to 4°C.

If the temperature of water is increased from 0°C, its volume decreases and density gradually increases. At a temperature of 4°C, density of water is maximum (1 g/cm3) and volume of a particular mass of water is minimum. This abnormal property of water from a temperature range of 0°C to 4°C is called the anomalous expansion of water.

Chapter 6 Saturated Unsaturated Vapour And Anomalous Expansion Of Water Short And Long Answer Type Questions

Question 1. Define saturated and unsaturated vapour.

Answer:

Saturated And Unsaturated Vapour:-

Saturated Vapour: At a particular temperature, if maximum amount of vapour is accommodated in a closed space, then that vapour is called saturated vapour.

Unsaturated Vapour: If the vapour in a closed space at a particular temperature is less than the highest possible vapour that can be accommodated at that temperature, then that vapour is called unsaturated vapour.

Question 2. What is vapour pressure?

Answer:

Vapour Pressure:-

A liquid can vaporise at any temperature. If a liquid is taken to fill a closed vessel partially, then the liquid vaporises and the empty portion of the vessel is filled up by that vapour. This vapour exerts pressure on the inner surface of the vessel. This pressure is called vapour pressure.

Question 3. What is saturated vapour pressure and unsaturated vapour pressure?

Answer:

Saturated Vapour Pressure: If vapour pressure in a closed space at a specific temperature is equal to the maximum vapour pressure at that temperature, then that vapour pressure is called saturated vapour pressure of that closed space.

Unsaturated Vapour Pressure: If vapour pressure in a closed space at a specific temperature is less than the maximum vapour pressure possible at that temperature, then that vapour pressure is called unsaturated vapour pressure of that closed space.

Question 4. Write down the differences between saturated and unsaturated vapour.

Answer:

Differences Between Saturated And Unsaturated Vapours Are:

WBBSE Solutions For Class 9 Physical Science Chapter 6 Saturated Unsaturated Vapour And Anomalous Expansion Of Water Differenes Saturated And Unsaturated Vapour

Question 5. What do you mean by dew point?

Answer:

Dew Point:-

  1. The atmosphere is filled with water vapours almost at all times. If temperature is more, capacity to hold water vapour becomes more. As temperature remains high during the day, capacity to hold water vapour also remains high.So generally during day time, air is not saturated by water vapour present in it.
  2. At night, earth’s surface gets colder by radiation of heat, so atmosphere surrounding the surface gets cold. Then capacity of air to hold water vapour also decreases. Thus when temperature of the atmosphere decreases, air at a specific temperature gets saturated by the water vapour present it.
  3. This temperature is called dew point. In short, the specific temperature at which a certain volume of air is saturated with water vapour present in it is known as dew point.

Question 6. How does dew originate? What are the essential conditions for formation of dew?

Answer:

Dew Originate:-

When temperature of the atmosphere comes down below the dew point, some amount of water vapour of the atmosphere is condensed to form small droplets of water and settle on grass, leaves etc. These droplets are called dew.

Essential Conditions For Formation Of Dew:

  1. Cloud free clear sky;
  2. Calm and still air;
  3. Excess of water vapour in atmosphere;
  4. Closeness to objects which are bad conductors of heat.

Question 7. Explain how fog is formed.

Answer:

Fog Forms As:-

If atmospheric temperature of a vast region falls below the dew point due to some reason, then some amount of water vapour present in the atmosphere condenses as deposits on coal particles, dust etc. and floats in the atmosphere.

This is called fog. When formed over a lake, it is called mist. As air in an industrial area contains considerable amount of dust, coal particles etc., dense fog is formed in those areas.

Question 8. Explain how cloud is formed.

Answer:

Cloud Forms As

Density of water vapour is less than density of air. So, water vapour present in air naturally goes up. As it goes up from the earth’s surface, air pressure decreases. So, volume of water increases due to reduction of air pressure and hence, its temperature decreases.

Now as we know that temperature of upper atmosphere remains low, when water vapours come in contact, the temperature of water vapour decreases further. Now if that temperature becomes less than the dew point, then water vapour condenses on dust particles present there and float around in the form of small water particles which is called cloud.

Question 9. ‘Cloudy sky is not favourable for formation of dews’—explain.

Answer:

Cloudy Sky Is Not Favourable For Formation Of Dews:-

Since radiated heat from the earth gets reflected back by cloud. For this the earth surface does not get cool sufficiently to help the formation of dew. That is why cloudy sky is not favourable for formation of dews.

Question 10. What do you mean by relative humidity?

Answer:

Relative Humidity Means:-

Relative humidity is the ratio of the mass of water vapour present in a particular volume of air to the mass of water vapour required to saturate that volume at a particular temperature. In other words, relative humidity is the ratio of the partial pressure of water vapour to the saturated pres- sure of water vapour at a particular temperature.

Now suppose, m is the mass of water vapour present in a specific volume of air at t°C and M is the mass of water vapour to be present at that temperature to make that volume of air saturated.

∴ relative humidity (RH) = \(\frac{m}{M}\) x 100%

Question 11. What do you mean by absolute humidity? What is its practical unit?

Answer:

Absolute Humidity:-

Absolute humidity (AH) is defined as the amount of water vapour in grams present per cubic meter of air.

Suppose, water vapour of mass m g is present in a specific amount of air of volume V m3 at temperature t°C.

∴ absolute humidity (AH) = \(\frac{m}{M}\)

Practical unit of absolute humidity is g/m3.

Question 12. On a summer day, suppose the temperatures of Delhi and Puri are the same. Which place feels more uncomfortable?

Answer:

As Puri is situated on the seashore, the value of the relative humidity of air in Puri is higher than that in Delhi. Perspiration takes more time to dry up in Puri, as a result, an uncomfortable feeling is there, though the temperature of these two places are the same.

Question 13. If the temperature of a closed room is increased, is there any change of relative humidity?

Answer:

We know that, relative humidity = \(\frac{m}{M}\) x 100%

If room temperature is increased, capacity of air to hold water vapour increases. Hence, water vapour of greater mass is required (say, M1 where M1 > M) to saturate the air of that room. Since amount of water vapour in the room remains unchanged, hence relative humidity decreases.

Question 14. Is there any change of relative humidity when water is sprinkled in a closed room?

Answer:

When water is sprinkled in a closed room, mass of water vapour present in the air of the room increases. Now as the temperature of the room remains unchanged, mass of water vapour necessary to saturate remains unchanged. So, relative humidity of air in the room increases.

Question 15. Is there any change of dew point when water is sprinkled in a dosed room?

Answer:

When water is sprinkled in a closed room, amount of water vapour in the room increases. Now as the temperature of the room remains unchanged, so in order to make the air of the room saturated, comparatively less reduction of temperature has to be brought about. Hence, dew point increases.

Question 16. What do you mean by anomalous expansion of water? Explain it with a graphical representation of volume and temperature.

Answer:

  1. In general, if the temperature of a liquid increases, its volume also increases. But in case of water, there is an exception to this rule in a specific range of temperature, i.e., from 0°C to 4°C temperature.
  2. If temperature of water is increased from 0°C, it volume decreases and density gradually increases. Density of water is maximum (1g • m3) at 4°C, so the volume of a specific mass is lowest. If temperature is increased after 4°C, the volume of water increases and density decreases.
  3. Change of volume of 1 g of water, when temperature in increased from 0 C to 10°C is shown.

WBBSE Solutions For Class 9 Physical Science Chapter 6 Saturated Unsaturated Vapour And Anomalous Expansion Of Water Anomalous expansion Of Water In Graph

This behaviour of water from 0°C to 4°C temperature is known as anomalous expansion of water.

Question 17. Discuss the influence of anomalous expansion of water on aquatic animals.

Answer:

In cold countries, when atmospheric temperature decreases below 0°C, temperature of the upper surface of lakes, water bodies etc. also comes down gradually. During this time, cold and heavy water from the top of the lake goes down and comparatively warm and light water from the bottom of the lake comes upward.

This convection cycle goes on till the temperature of the water at the bottom reach 4°C. At this temperature 4°C, density of water is maximum, i.e., after this temperature, water does not rise any further. Now, temperature of the upper surface of the lakes, water bodies etc. keeps on decreasing.

At this temperature, density of water at the top starts decreasing causing the water not to go down further.

WBBSE Solutions For Class 9 Physical Science Chapter 6 Saturated Unsaturated Vapour And Anomalous Expansion Of Water Anomalous Expansion Of Water On Animals

Hence, temperature of the upper surface comes down to 0°C and finally is converted to ice. As ice is lighter than an equal volume of water, it floats on water. Ice is a bad conductor of heat and so rate of transmission of heat from the lower layer is very low and as a result, the thickness of ice increases gradually.

Some water remains below the ice layer or crust. Temperature of water just below this ice crust is 0°C and temperature increases down the water layers to become 4°C at the bottom. For this reason, though the atmospheric temperature in cold countries comes below 0°C to make the upper surface of lakes frozen, aquatic animals survive easily.

Chapter 6 Saturated Unsaturated Vapour And Anomalous Expansion Of Water Very Short Answer Type Questions Choose The Correct Answer

Question 1. Hygrometer is used to measure

  1. Amount of oxygen in air
  2. Density of air
  3. Increase of density with temperature
  4. Amount of water vapour present in the atmosphere

Answer: 4. Amount of water vapour present in the atmosphere

Question 2. If room temperature is equal to the dew point, relative humidity becomes

  1. 100%
  2. 0%
  3. 70%
  4. 85%

Answer: 1. 100%

Question 3. Which of the following statements regarding saturated vapour is true?

  1. Follows Boyle’s law
  2. Remains in equilibrium when in contact with the liquid
  3. Can be made unsaturated by reducing the temperature
  4. All of the above

Answer: 2. Remains in equilibrium when in contact with the liquid

Question 4. Anomalous expansion is seen for

  1. Mercury
  2. Kerosene
  3. Glycerine
  4. Water

Answer: 4. Water

Question 5. In a cold country if the upper surface of a lake is converted into ice, temperature of water at the bottom of the lake becomes

  1. 0°C
  2. 1°C
  3. 2°C
  4. 4°C

Answer: 4. 4°C

Question 6. On a particular day, temperature of Kolkata and Delhi are the same. But the levels of relative humidity in air are 80% and 60%, respectively. Which of the following statements is correct?

  1. Weather of both the places are equally comfortable
  2. Given information regarding weather is incomplete
  3. The weather of Kolkata is more comfortable
  4. Weather of Delhi is more comfortable

Answer: 4. Weather of Delhi is more comfortable

Question 7. The physical quantity determining the amount of water vapour in a definite volume of air is

  1. Humidity
  2. Dew point
  3. Temperature
  4. Water equivalent

Answer: 1. Humidity

Question 8. If the amount of water vapour increases in air, then

  1. Density of air increases
  2. Density of air decreases
  3. Relative humidity of air decreases
  4. Dew point of air decreases

Answer: 2. Density of air decreases

Question 9. Formation of fog takes place generally in

  1. Summer
  2. Rainy season
  3. Winter
  4. Autumn

Answer: 3. Winter

Question 10. Formation of dew takes place generally in

  1. Summer
  2. Rainy season
  3. winter
  4. Autumn

Answer: 4. Autumn

Question 11. Anomalous expansion of water is observed in the temperature range of

  1. 10°C-14°C
  2. 4°C-14°C
  3. 0°C-4°C
  4. 0°C-10°C

Answer: 3. 0°C-10°C

Question 12. If relative humidity of air decreases from 80% to 60%, then

  1. We feel comfortable
  2. Uneasiness due to perspiration increases
  3. We feel hotter
  4. Temperature comes down from 80°C to 60°C

Answer: 1. We feel comfortable

Question 13. If water is sprinkled in a closed room, then the room’s

  1. Relative humidity decreases
  2. Relative humidity increases
  3. Dew point decreases
  4. Dew point remains unchanged

Answer: 2. Relative humidity increases

Question 14. If temperature of water is decreased from 4°C to 0°C, then

  1. Density increases
  2. Density decreases
  3. Volume decreases
  4. Both density and volume decrease

Answer: 2. Density decreases

Question 15. If the temperature of water is increased from 4°C to 10°C, then

  1. Volume of water decreases
  2. Mass of water decreases
  3. Volume of water increases
  4. Mass of water increases

Answer: 3. Volume of water increases

Question 16. Volume expands due to application of heat. On the basis of this information, water behaves normally

  1. In the range of 0°C-4°C
  2. In the range of 0°C-10°C
  3. In the range of 4°C-10°C
  4. In the range of 3°C-4°C

Answer: 3. In the range of 4°C-10°C

Question 17. Temperature of air and its relative humidity are 30°C and 80%, respectively. If 32 mm Hg is the pressure of saturated water vapour at 30°C, then the pressure of water vapour present in air becomes

  1. 23.2 mm Hg
  2. 24.2 mm Hg
  3. 25.6 mm Hg
  4. 27.2 mm Hg

Answer: 2. 24.2 mm Hg

Question 18. What is the relative humidity if pressure of water vapour is 27 mm Hg and pressure of saturated water vapour at that temperature is 30 mm hg?

  1. 60%
  2. 70%
  3. 80%
  4. 90%

Answer: 4. 90%

Chapter 6 Saturated Unsaturated Vapour And Anomalous Expansion Of Water Answer In Brief

Question 1. What is the value of relative humidity, when temperature of the atmosphere, is equal to the dew point?

Answer: When temperature of the atmosphere is equal to the dew point, value of relative humidity is 100%.

Question 2. Which liquid shows anomalous expansion?

Answer: Water shows anomalous expansion.

Question 3. At what temperature, density of water is maximum?

Answer: At 4°C or 277 K, density of water is maximum.

Question 4. If the upper surface of a lake in a cold country freezes to ice, what is the temperature of water at the bottom surface?

Answer: If the upper surface of a lake in a cold country freezes to ice, the temperature of water at the bottom surface must be 4°C.

Question 5. If the upper surface of a lake in a cold country freezes to ice, what is the temperature of water just below the ice crust?

Answer: If the upper surface of a lake in a cold country freezes to ice, the temperature of water just below the ice crust must be 0°C.

Question 6. Does saturated vapour follow Charles’ law and Boyle’s law?

Answer: No, saturated vapour does not follow Charles’ law and Boyle’s law.

Question 7. What is the range of temperature in which anomalous expansion of water is seen?

Answer: Anomalous expansion of water is seen in the temperature range of 0°C to4°C.

Question 8. The volume of a definite mass of water is minimum at what temperature?

Answer: The volume of a definite mass of water is minimal at 4°C.

Question 9. Due to which property of water, aquatic animals survive in spite of atmosphere temperature going down below the freezing point in cold countries?

Answer: This happens due to anomalous expansion of water.

Question 10. A bowl made up of glass is filled with water up to the brim. What is the result if the temperature of the water is raised and reduced?

Answer: Water spills out for both the rise and fall of temperature.

Question 11. How is unsaturated vapour converted to saturated vapour?

Answer: By increasing the pressure or by decreasing the temperature, unsaturated vapour can be converted to saturated vapour.

Question 12. How is saturated vapour converted to unsaturated vapour?

Answer: By decreasing the pressure or by increasing the temperature, saturated vapour can be converted to unsaturated vapour.

Question 13. Which of the two remains in equilibrium when in contact with the liquid, saturated vapour or unsaturated vapour?

Answer: Saturated vapour remains in equilibrium when in contact with the liquid.

Question 14. What is the required temperature of the atmosphere for the formation of fog?

Answer: If the temperature of the atmosphere comes down below the dew point, the formation of fog takes place.

Chapter 6 Saturated Unsaturated Vapour And Anomalous Expansion Of Water Fill In The Blanks

Question 1. The capacity of a closed space to contain maximum vapour ________ with increasing temperature.

Answer: Increase

Question 2. _______ vapour does not remain in equilibrium when in contact with a liquid.

Answer: Unsaturated

Question 3. _______ vapour may be converted into ________ vapour by increasing temperature or by reducing pressure.

Answer: saturated, unsaturated

Question 4. In general, an increase in temperature of any liquid _______  its volume but in the case of water, there is an __________ to this rule.

Answer: Increase, exception

Question 5. Though Delhi and Puri may have the same temperature on a particular day of summer, we feel warmer in __________

Answer: Puri

Question 6. When the temperature starts _____, the volume of water normally reduces to 4°C. When temperature starts anomalous expansion of water is observed up to 4°C.

Answer: Decreasing

Question 7. When temperature starts _______ anomalous expansion of water is observed up to 4°C.

Answer: Increasing

Question 8. _______ of water at 4°C is higher than its density at 3°C.

Answer: Density

Question 9. The volume of 1 g of water at 4°C is ________ than the volume of 1 g of water at 2°C.

Answer: Less

Question 10. ________ vapour exists on the top surface of a liquid in a closed vessel.

Answer: Saturated

Question 11. ________ indicates the amount of water vapour in air.

Answer: Humidity

Question 12. Air is _______ at the dew point.

Answer: Saturated

Question 13. The weather in Delhi is more dry compared to that of Kolkata. This means the amount of relative humidity in Delhi is comparatively ________

Answer: Lower

Question 14. _______ is defined as the amount of water vapour in grams present in every cubic metre of air.

Answer: Absolute humidity

Chapter 6 Saturated Unsaturated Vapour And Anomalous Expansion Of Water State Whether True Or False

Question 1. The saturated vapour is the maximum amount of vapour that can be accommodated in a closed space at a particular temperature.

Answer: True

Question 2. If the temperature of the atmosphere goes above the dew point, dew is formed.

Answer: False

Question 3. If the temperature of water is increased from 0°C to 4°C, then its volume increases.

Answer: True

Question 4. Clear sky is a favourable condition for the formation of dew.

Answer: True

Question 5. Dew forms only at night.

Answer: True

Question 6. Relative humidity has no unit.

Answer: True

Chapter 6 Saturated Unsaturated Vapour And Anomalous Expansion Of Water Numerical Examples

Useful Information

  1. Relative humidity = \(\frac{M_P}{M_S}\)x 100%
  2. MP = mass of water vapour present in a certain volume of air
  3. MS = mass of water vapour required to saturate the same volume of air.
  4. Relative humidity = \(\frac{f}{F}\) x 100%
  5. Where, f = actual pressure of water vapour present in air.
  6. F = Saturated vapour pressure of water at that temperature of air.

Question 1. At a particular temperature actual pressure of water vapour present in the air is 34 mm of Hg and the saturated vapour pressure of water at that temperature is 40 mm of Hg. Find relative humidity.

Answer:

At a particular temperature, the actual pressure of water vapour present in air (f) = 34 mmHg and saturated vapour pressure at that temperature (F) = 40 mmHg.

∴ Relative humidity = \(\frac{f}{F}\) x 100%

= \(\frac{30}{34}\) x 100% = 85%

Question 2. On a certain day, the room temperature is 20°C and the dew point is 15°C. Saturated vapour pressure at 20°C and 15°C are 20mmHg and 13mmHg respectively. Find the relative humidity of the aforesaid day.

Answer:

Saturated vapour pressure at dew point (f) = 13 mmHg and saturated vapour pressure at room temperature (F) = 20 mmHg.

∴ Relative humidity (RH) = \(\frac{13}{20}\) x 100% = 65%

Question 3. At a particular temperature actual pressure of water vapour present in air is 30 mmHg and relative humidity is 60%. Find saturated water vapour pressure at that temperature.

Answer:

At a particular temperature actual pressure of water vapour present in air (f) = 30mmHg.

Relative humidity (RH) = 60%

Let, saturated water vapour pressure be F at that temperature.

∴ Relative humidity (RH) = \(\frac{f}{F}\) x 100%

or, 60% = \(\frac{30}{F}\) x 100%

or, F = \(\frac{30 \times 100}{60}\) = 50

∴ Saturated water vapour pressure is 50 mmHg.

Chapter 6 Saturated Unsaturated Vapour And Anomalous Expansion Of Water Miscellaneous type Questions

Match the columns

1.

WBBSE Solutions For Class 9 Physical Science Chapter 6 Saturated Unsaturated Vapour And Anomalous Expansion Of Water Match The Column 1

Answer: 1. C, 2. B, 3. D, 4. A

2.

WBBSE Solutions For Class 9 Physical Science Chapter 6 Saturated Unsaturated Vapour And Anomalous Expansion Of Water Match The Column 2

Answer: 1. D, 2. A, 3. B, 4. C