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

- The literal meaning of the word ‘mole’ is heap.
- One mole of a substance is the amount that contains 6.022 x 10
^{23}constituent particles (atoms, molecules, or ions) of the substance. - Chemical calculations can be done in a simpler and more convenient method by using the concept of mole.
- 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 (N
_{A}). - The value of Avogadro’s number (N
_{A}) is 6.22x 10^{23}. It is independent of both temperature and pressure. - It is very difficult to grasp the enormousness of Avogadro’s number. N
_{A}= 602213670000000000000000. - The value of Avogadro’s constant is 6.022 x 10
^{23}mol^{-23}. - Avogadro’s number correlates with the microscopic and macroscopic world.
- Avogadro’s number finds useful applications in calculations related to physics, chemistry, and biological science.

**Class 9 Physical Science Chapter 4 Concept Of Mole**

**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 10^{23}) 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:**

- 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 10
^{23 }mol^{-1}. It is a universal constant. - Avogadro’s number is a pure number. It has no unit but the unit of Avogadro’s constant is mol
^{-1}.

**Class 9 Physical Science Chapter 4 Short And Long Answer Type Questions**

**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:**

**Avogadro’s number creates a correlation between the macroscopic and microscopic world—**

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 10^{23} number of fundamental particles. This number (6.022 x 10^{23}) 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 10^{23} molecules of water.

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

**Class 9 Physical Science Chapter 4 Short And Long Answer Type Questions**

**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 10^{23}) 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 CO_{2 }= 88g/44g = 2 mol (as a gram-molecular mass of CO_{2} is 44 g).

Hence, number of molecules in 88 g CO_{2} = 2 x 6.022 x 10^{23} = 12.044 x 10^{23}

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 N_{2} molecule = 28 u

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

**Class 9 Physical Science Chapter 4 Short And Long Answer Type Questions**

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 O _{2} and 1 mol O represent the same quantity of oxygen?**

**Answer:**

The molecular formula of oxygen is O_{2} while O represents an atom of oxygen. Hence, 1 mol O_{2 }represents Avogadro’s number of oxygen molecules. On the other hand, 1 mol O_{2} indicates Avogadro’s number of O-atoms.

The mass of Avogadro’s number of O_{2} molecules is 32 g and that of Avogadro’s number of O-atoms is 16 g. Thus, 1 mol O_{2} 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:**

- 22.4 L of any gas at STP contains Avogadro’s number of molecules.
- 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.
- 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:**

**Importance of Avogadro’s number in biology**

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

**Class 9 Physical Science Chapter 4 Short And Long Answer Type Questions**

**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 10^{20}.

## 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?

- Dalton’s number
- Berzelius’ number
- Avogadro’s number
- Faraday’s number

Answer: 3. Avogadro’s number

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

- Many
- Having large volume
- Having large mass
- Heap

Answer: 4. Heap

**Class 9 Physical Science Chapter 4 Very Short And Long Answer Type Questions**

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

- Dalton
- Ostwald
- Avogadro
- Millikan

Answer: 2. Ostwald

**Class 9 Physical Science Chapter 4 Very Short And Long Answer Type Questions**

Question 4. The value of Avogadro’s number is

- 6.024 X 10
^{20} - 0.6023 X 10
^{22} - 0.6022 X 10
^{24} - 0.623 x 10
^{23}

Answer: 3. 0.6022 X 10^{24}

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

- Mole
- Gram-mole
- Gram-atom
- Gram-ion

Answer: 1. Mole

**Class 9 Physical Science Chapter 4 Very Short And Long Answer Type Questions**

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

- Oxygen
- Helium
- Hydrogen
- Chlorine

Answer: 2. Helium

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

- 6.022 x10
^{22} - 1.806 X 10
^{23} - 3.600 x10
^{23} - 1.800 x 10
^{22}

Answer: 2. 1.806 X 102^{23}

**Class 9 Physical Science Chapter 4 Very Short And Long Answer Type Questions**

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

- 6.022 X10
^{28} - 6.022 x 10
^{23} - 6.022 X10
^{-23} - 6.022 x 10
^{18}

Answer: 4. 6.022 x 10^{18}

Question 9. 1 mol of oxygen atom means

- 6.022 x 10
^{23}number of oxygen atoms - 2 x 6.022 x 10
^{23}number of oxygen atoms - 1/2 x 6.022 x 10
^{23}number of oxygen atoms - 6.022 x 10
^{23}number of oxygen molecules

Answer: 1. 6.022 x 10^{23} number of oxygen atoms

**Class 9 Physical Science Chapter 4 Very Short And Long Answer Type Questions**

Question 10. Number of Cl^{–} ions obtained from 1 mol of CaCI_{2} is

- 2
- 6.022 X 10
^{23} - 12.046 x 10
^{23} - 3 x 6.022 x 10
^{23}

Answer: 3. 12.046 x 10^{23}

Question 11. Unit of Avogadro’s constant is

- Mole
- Per mole
- (mole)
^{2} - Dobson

Answer: 2. Per mole

**Class 9 Physical Science Chapter 4 Very Short And Long Answer Type Questions**

Question 12. Number of molecules in 1 millimol ammonia is

- 6.022 x 10
^{26} - 6.022 x 10
^{23} - 6.022 X 10
^{20} - 6.0222 x 10
^{-3}

Answer: 3. 6.022 X 10^{20}

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

- Boltzman constant
- Atomic radius of solid metals
- Velocity of light in vacuum
- 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 10^{23} molecules.

Hence, number of H-atoms in 1 mol of hydrogen gas = 2 x 6.022 x 10^{23} = 12.044 x 10^{23}.

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

**Answer:** 6.022 x 10^{23} atoms of iron.

**Question 7. How many moles of oxygen atoms do 6.22x 10 ^{24} number of oxygen atoms indicate?**

**Answer:** 6.02 2 x 10^{24} number of oxygen atoms = 10 x 6.022 x 10^{24} oxygen atoms = 10 mol of oxygen atoms.

**Question 8. How many grams of CO _{2} does 1 gram-mole of CO_{2} indicate?**

**Answer:** 1 gram-mole of CO_{2} indicates 44 g of CO_{2}

**Question 9. How many molecules are present in 1 millimole CO _{2}?**

**Answer:** 6.022 x 10^{20} number of molecules are present in 1 millimol CO_{2}.

**Question 10. 2 balloons contain 1.8066 x 10 ^{23} 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 10^{23} 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 10^{23}

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

Answer: mol^{-1}

Question 5. 6.22x 10^{23} number of H^{+} ions will be produced from __________ gram-mol of H_{2}SO_{4}.

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 10^{23} 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 10^{20}

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 10^{23}

Answer: True

Question 6. 1 mol N_{2} and 1 mol N signify the same amount.

Answer: False

Question 7. 1 mol of CO_{2} 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:**

Given

P and Q flasks contain 0.5 mol oxygen and 0.4 mol ozone gas respectively

Number of oxygen atoms in P flask = 2 x 0.5 x 6.022 x 10^{23} = 6.022 x 10^{23}

Number of oxygen atoms in Q flask = 3 X 0.4 X 6.022 X 10^{23} = 1.2 x 6.022 x 10^{23}

∴ 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 gCO _{2}?**

**Answer:**

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

∴ 44 g CO_{2 }= 1 gram-mole of CO_{2}

Now, 1 molecule of CO_{2} 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 (^{1}H) scale, the relative atomic mass of an element

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

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

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

- The relative atomic mass of an element is expressed as a ratio of the masses of two atoms. Hence, it has no unit.
- The atomic mass of an element expressed in gram is known as the gram-atomic mass of the element.
- The actual mass of an atom is expressed in atomic mass unit.
- Atomic mass unit (1u) = 1/12 x actual mass of one C atom = 1.6605 x 10
^{-24}g - One gram atom of an element contains Avogadro’s number of atom.
- Atomic mass is expressed as a ratio of two masses. So it does not have any unit.
- 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.
- 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.
- The molecular mass of a substance expressed in gram is known as its gram-molecular mass.
- One gram-mole of a substance (element or compound) contains Avogadro’s number of molecules.
- Number of moles of a matter = \(\frac{\text { mass of the matter }}{\text { its molar mass }}=\frac{W}{M}\)
- At a given temperature and pressure, the volume occupied by 1 mol of a substance (element or compound) is known as its molar volume.
- At standard temperature and pressure, the molar volume of all the gases is 22.4 L.
- 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:**

**Relative atomic mass with respect to the hydrogen scale**

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 ( ^{12}C) scale.**

**Answer:**

** Relative atomic mass with respect to the carbon ( ^{12}C) scale**

The number of times an atom of an element is heavier than 1/12 th part of the actual mass of an atom of ^{12}C 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 ^{12}C 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 ^{12}C and not hydrogen considered As the standard element to determine as the standard element to determine the relative atomic mass?**

**Answer:**

- 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
^{12}C is taken as the standard element. - 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
^{12}C scale.

Due to these advantages, nowadays ^{12}C 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 10^{23} atoms of the element. For example, 1 gram-atom of nitrogen =14 g nitrogen, because 14g nitrogen contains 6.22x 10^{23} 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 ^{12}C^{ }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:-**

- 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
^{12}C isotope. On the other hand, the mass of an atom of an element means the actual mass of the atom. - 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.
- 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 10^{23} molecules of the substance. For example, 1 gram-mole of oxygen =32 g oxygen, because, 32 g oxygen contains 6.22x 10^{23} 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 H_{2}O = 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 CO_{2 }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 CO _{2}‘?**

**Answer:**

From ‘1 mol CO_{2}‘ it can be known that

- Mass of the CO
_{2}sample is 44 g. - Number of CO
_{2}molecules in the sample is 6.22 x 10^{23}. - Volume of the CO
_{2 }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 CaCI _{2}.**

**Answer:**

**Formula Mass Of CaCI _{2 }**

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

Hence, formula mass of CaCI_{2} = 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, CO_{2} molecule actually exists. Hence, the term molecular mass is applicable for CO_{2} and both molecular mass and formula mass of CO_{2 }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 CO_{2} 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

- 1.66 X 10
^{-24}kg - 1.66 x 10
^{-25}g - 0.166 x 10
^{-23}g - 1.66 x 10
^{-20}g

Answer: 3. 0.166 x 10^{-23} g

Question 2. The mass of 1 atom of _{17}CI^{35 }is

- 17 u
- 34 u
- 35 u
- 70 u

Answer: 3. 35 u

Question 3. 1 u is equal to

Mass of one H-atom

Mass of one ^{12}C-atom

Mass of one ^{16}O-atom

1/12 th mass of one ^{12}C-atom

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

**Class 9 Physical Science Chapter 4 Concept of Mole**

Question 4. Molar volume of a gas at STP is

- 1 L
- 11.1 L
- 22.4 L
- 5.6 L

Answer: 3. 22.4 L

Question 5. Molar mass of sulphuric acid is

- 98 kg •mol
^{-1} - 98 g •mol
^{-1} - 49 u
- 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?

- NH
_{3} - H
_{2}S - CO
_{2} - CH
_{4}

Answer: 3. CO_{2}

**Class 9 Physical Science Chapter 4 Concept of Mole**

Question 7. The formula mass of MgCI_{2} (Mg = 24) is

- 95 u
- 65 u
- 85 u
- 75 u

Answer: 1. 95 u

Question 8. 117g NaCI is equal to how many gram- formula mass of NaCI?

- 1
- 2
- 3
- 4

Answer: 2. 2

Question 9. The compound whose formula mass is equal to 106 is

- CaCO
_{3} - CaCI
_{2} - NaCI
- Na
_{2}CO_{3}

Answer: 4. Na_{2}CO_{3}

**Class 9 Physical Science Chapter 4 Concept of Mole**

Question 10. The statement ‘molecular mass of sodium chloride is 58.5’ is wrong because

- It is a radioactive substance
- It is a solid
- NaCI molecule has no separate existence
- 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

- x
- 1/12 x x
- 12x
- 12/x

Answer: 1. x

Question 12. Atomic mass of nitrogen in carbon scale is 14, its atomic mass in hydrogen scale will be

- 14 x 1/3
- 14 x 1/2
- 14
- 14 x 2

Answer: 3. 14

**Class 9 Physical Science Chapter 4 Concept of Mole**

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

- m x M
- m/M
- M/m
- \(\frac{(m)^2}{M}\)

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 ^{12}C 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 H _{2}O does 1 gram-mole of H_{2}O indicate?**

**Answer:** 1 gram-mole of H_{2}O 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)

**Class 9 Physical Science Chapter 4 Concept of Mole**

**Question 6. What is the mass of one molecule of water (H _{2}O) 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 CaCI _{2} and H_{2}O, the use of formula mass is appropriate?**

**Answer:** The use of formula mass is appropriate for CaCI_{2} as it is an ionic compound.

**Question 10. Find the percentage of carbon (by mass) in CO _{2}.**

**Answer:** Molar mass of CO_{2} = 12 + 16 x 2 = 44 So, percentage of carbon (by mass) in CO_{2} = 12/44 x 100 =27.27%

**Class 9 Physical Science Chapter 4 Concept of Mole**

**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 C^{12} 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 C^{12 }-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.

**Class 9 Physical Science Chapter 4 Concept of Mole**

**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 NH_{3} is _______ L.

Answer: 22.4

Question 6. At STP, the mass of 44.8 L of CO_{4 }(g) is ________ g.

Answer: 88

**Class 9 Physical Science Chapter 4 Concept of Mole**

Question 7. Formula mass of CaCI_{2} is _______ u.

Answer: 111

Question 8. ______ part of the mass of an atom of a C^{12} 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 (^{1}H) and carbon (^{12}C) scales.

Answer: True

Question 2. The actual mass of one molecule of CO_{2} 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

**Class 9 Physical Science Chapter 4 Concept of Mole**

Question 6. 1u =6.022 x 10^{23} 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**

- 46 g sodium,
- 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

**Class 9 Physical Science Chapter 4 Concept of Mole**

∴ 3g carbon = 3/12 = 0.25 gram-atom carbon

**Question 2. Express in gram**

- 0.25 mol nitrogen atom,
- 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**

- 64 g oxygen,
- 2.2 g carbon dioxide.

**Class 9 Physical Science Chapter 4 Concept of Mole**

**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 (CO_{2}) = 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:**

Given

mass of l gram-atom of oxygen

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 10^{23} 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 = N_{A} 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.

6. Number of molecules in n mol of any substance = n x N_{A}.

## 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 N_{A} = m/M x N_{A}.

**Class 9 Physical Science Chapter 4 Concept of Mole**

**Question 2. What is the actual mass of 1 molecule of NH _{3}?**

**Answer:**

**Actual mass of 1 molecule of NH _{3}**

Molecular mass of NH_{3} = 14 + 1 x 3 = 17

Hence, actual mass of one NH_{3} 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 Cl _{2} and O_{2} at STP, which one has more mass?**

**Answer:**

Given

Among equal volumes of Cl_{2} and O_{2} at STP

We know, the volume of 1 g-mole of any gaseous substance at STP is 22.4 L.

Now 1 g-mole Cl_{2} = 2 X 35.5 g = 71 g Cl_{2}

1 g-mole O_{2} = 2 x 16 g = 32 g O_{2}

∴ At STP, mass of 22.4 L of Cl_{2} gas = 71 g and mass of 22.4 L of O_{2} gas = 32 g

∴ Cl_{2} gas will have more mass than O_{2} at given conditions.

**Class 9 Physical Science Chapter 4 Concept of Mole**

**Question 6. An isotope of chlorine is 35 times heavier than 1/12 th of the mass of a C ^{12} isotope. What is the actual mass of the Cl-leading isotope?**

**Answer:**

Given

An isotope of chlorine is 35 times heavier than 1/12 th of the mass of a C^{12} isotope.

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.

**Class 9 Physical Science Chapter 4 Concept of Mole**

**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:**

Given

Molecular mass and atomic element are x and y respectively.

Since atomic mass of the element is y, hence mass of 1 gram-atom = yg

∴ Total mass of N_{A} number of atoms = yg [N_{A} = 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 N_{A} 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

- 6.022 X 10
^{23} - 12.044 X 10
^{23} - 3.011 x 10
^{23} - 9.0345 X 10
^{23}

Answer: 2. 12.044 X 10^{23}

Question 2. Which of the following contains maximum number of moles?

- 18 g water,
- 34 g ammonia,
- 71 g chlorine,
- 8 g hydrogen.

Answer: 4. 8 g hydrogen.

Question 3. Which of the following contains maximum number of atoms?

- 28g N
_{2}, - 18g H
_{2}O, - 17g NH
_{3}, - 16g CH
_{4}.

Answer: 4. 16gCH_{4}.

**Class 9 Physical Science Chapter 4 Concept of Mole**

Question 4. Number of H-atoms in 2 mol water is

- 1 mol
- 2 mol
- 3 mol
- 4 mol

Answer: 4. 4 mol

Question 5. Number of moles of H^{+} ions obtained from 98 g H_{2}SO_{4} is

- 1 mol
- 2 mol
- 3 mol
- 4 mol

Answer: 2. 2 mol

Question 6. Amount of Ca(OH)_{2} required to produce 1 mol of OH^{–} ions is

- 74 g
- 148 g
- 37 g
- 60 g

Answer: 3. 37 g

Question 7. Which of the following has a volume of 22.4 L at STP?

- 36 g water
- 64 g oxygen
- 71 g chlorine
- 4 g hydrogen

Answer: 3. 71 g chlorine

Question 8. Which of the following contains maximum number of molecules?

- 44 g CO
_{2} - 48 g O
_{3} - 8 g H
_{2} - 64 g SO
_{2}

Answer: 3. 8 g H_{2}

**Class 9 Physical Science Chapter 4 Concept of Mole**

Question 9. At STP, the mass of 44.8 L of a gas is 88 g. The molecular formula of the gas is

- N
_{2} - CO
_{2} - NH
_{3} - O
_{2}

Answer: 2. CO_{2}

Question 10. Which of the following contains maximum number of molecules?

- 10 g O
_{2}(g) - 15 L H
_{2}(g) at STP - 5 L N
_{2}(g) at STP - 0.5 g H
_{2 }(g)

Answer: 2. 15 L H_{2} (g) at STP

Question 11. Number of atoms in 4.25 g of NH_{3} is

- 6.022 x 10
^{23} - 4 x 6.022 x 10
^{23} - 1.7 x 10
^{24} - 4.25 X 6.022 x 10
^{23}

Answer: 1. 6.022 x 10^{23}

Question 12. How many grams of CO_{2} will contain the same number of molecules as that in 44.8 L of NH_{3} at STP?

- 44 g
- 66 g
- 88 g
- 132 g

Answer: 3. 88 g

Question 13. Which of the following contains maximum number of molecules?

- 1 g CO
_{2} - 1 g H
_{2} - 1 g O
_{2} - 1 g CH
_{4}

Answer: 2. 1 g H_{2}

Question 14. Mass of 44.8 L of a gas at STP is 92 g. Formula of the gas is

- N
_{2} - NO
_{2} - NH
_{2} - O
_{2}

Answer: 2. NO_{2}

**Class 9 Physical Science Chapter 4 Concept of Mole**

Question 15. Mass of 2 mol of ammonia is

- 12 g
- 51 g
- 34 g
- 17 g

Answer: 3. 34 g

Question 16. The volume of 34 g ammonia at STP is

- 11.2 L
- 22.4 L
- 44.8 L
- 42.2 L

Answer: 3. 44.8 L

Question 17. If the mass of 6.022 x 10^{20} atoms of an element is 0.012 g, its atomic mass will be

- 11
- 12
- 13
- 24

Answer: 2. 12

Question 18. The volume of 2 g-mol CO_{2} at STP is

- 22.4 L
- 44.8 L
- 11.2 L
- 2.24L

Answer: 2. 44.8 L

Question 19. Number of carbon atoms in 1.71 g of sugar is

- 3.6 x 10
^{22} - 7.2 x 10
^{22} - 6.6 x 10
^{22} - None of the above

Answer: 1. 3.6 x 10^{22}

Question 20. Amount of carbon present in 0.5 mol potassium ferrocyanide (K_{4}[Fe(CN)_{6}]) is ®

- 1.5 mol
- 36 g
- 18 g
- 3.6 g

Answer: 2. 36 g

**Class 9 Physical Science Chapter 4 Concept of Mole**

Question 21. Which of the following contains least number of molecules?

- 1 g H
_{2} - 1 g N
_{2} - 1.1 g O
_{2} - 1.5 g O
_{2}

Answer: 3. 1.1 g O_{2}

Question 22. Number of moles of oxygen atoms present in 6.022 x 10^{24} molecules of CO is

- 10
- 5
- 1
- 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 N _{2}, 32 g O_{2},18 g H_{2}O, and 100 g CaCO_{2}?**

**Answer:** 28g N_{2} = 1 mol N_{2};

32g O_{2} = 1 mol O_{2};

18g H_{2}O = 1 mol H_{2}O;

100 g CaCO_{3} = 1 mol CaCO_{3}

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 10^{23}.

**Question 2. What is the mass of 112 mL hydrogen gas at NTP?**

**Answer:**

**Mass of 112 mL hydrogen gas at NTP**

0.01 g.

**Question 3. A gas jar contains 17 g NH _{3} and another contains 44.8 L NH_{3} at STP. Which gas jar contains more molecules?**

**Answer:** 17g NH_{3} = 17/17 mol = 1 mol NH

Again, 44.8L NH_{3} at STP = \(\frac{44.8}{22.4}\) mol NH_{3} = 2 mol NH_{3}

Hence, the second gas jar contains more number of molecules.

**Class 9 Physical Science Chapter 4 Concept of Mole**

**Question 4. What is the volume of 7g of nitrogen at NTP?**

**Answer:**

**The volume of 7g of nitrogen at NTP**

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 CO _{2} (g) at STP?**

**Answer:**

**Volume of 88 g CO _{2} (g) at STP**

88gCO_{2 }= \(\frac{88}{44}\)mol CO_{2 }= 2mol CO_{2}

Hence, volume of 88g of CO_{2} 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? 100m ^{3} CO_{2}, 200cm^{3} NH_{3}, 150cm^{3} O_{2}.**

**Answer:** At STP, 200 cm^{3} of NH_{3} 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 10^{23}

As the gas is diatomic, the number of atoms = 2 x 3 X 6.022 x 10^{23} = 36.132 x 10^{23}

**Class 9 Physical Science Chapter 4 Concept of Mole**

**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 10^{23} = 18.066 x 10^{23} = 1.8066 X 10^{24}

**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 10^{23} = 3.011 x 10^{23}

**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 + 2H_{2}O → 2NaOH + H_{2} ↑

∴ 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 10^{23} = 6.022 x 10^{24}

**Question 14. What is the mass of 0.5 mol CO _{2} gas?**

**Answer:** Mass of 0.5 mol of CO_{2} = 44 X 0.5 g = 22 g

**Class 9 Physical Science Chapter 4 Concept of Mole**

## 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 CO_{2} molecules in 88g CO_{2} is _______

Answer: 2 x 6.022 x 10^{23}

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 10^{23}

Question 4. Number of protons in 18 g water is _______

Answer: 10 X 6.022 X 10^{23}

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 H_{2}SO_{4} = _______ mol H_{2}SO_{4}.

Answer: 1

Question 9. Total mass of 3.011 x 10^{23} 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

**Class 9 Physical Science Chapter 4 Concept of Mole**

## 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 CaCO_{3} 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 CH_{4} is 6.022 x 10^{23}.

Answer: True

Question 5. The number of chlorine atoms in 71 g HCI is 0.5 x 6.022 x 10^{23}.

Answer: False

Question 6. 022 x 10^{23} number of molecules are present in 36 g of water.

Answer: False

Question 7. Volume of 32 g SO_{2} at STP is 11.2 L.

Answer: True

Question 8. 4 mol CO_{2} = 88 g CO_{2}.

Answer: False

Question 9. Number of molecules present in 2 millimol of chlorine gas is 2 x 6.022 x 10^{20}.

Answer: True

Question 10. The volume of 64 g oxygen at STP is 10 L.

Answer: False

Question 11. 44 g of CO_{2} contains 6.022 x 10^{23} 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 CO _{2}.**

**Answer:**

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

Now, 1 mol of CO_{2} contains 6.022 x 10^{23} molecules.

Hence, mass of 6.022 x 10^{23} CO_{2} molecules = 44 g

∴ Mass of one CO_{2} molecule = \(\frac{44}{6.022 \times 10^{23}} g\) = 7.3 x 10^{23} 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 10^{23} = 1.5055 X 10^{24}

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 10^{23} = 1.07 x 10^{24}

Therefore, 100 g calcium contains more number of atoms than 100 g iron.

**Question 3. How many grams of H _{2}SO_{4} are required to produce 1 gram-ion of H^{+}?**

**Answer:**

1 molecule of H2S04 gives two H^{+} ions in aqueous solution.

Hence, 1 gram-mole of H_{2}SO_{4 } gives 2 gram-ions of H^{+} ion.

∴ 1 gram-ion of H^{+} ion will be produced from 1/2 gram-mole of H_{2}SO_{4}= 98/2 = 49g H_{2}SO_{4 }[as 1 gram-mole of H_{2}SO_{4} = 98 g]

Thus, 49 g H_{2}SO_{4} is required to produce 1 gram- ion of H^{+} ions.

**Question 4. Which of the following has maximum number of molecules? 0.5 mol H _{2}(g), 10g O_{2}(g) and 15 L Cl_{2} at STP.**

**Answer:**

Number of molecules present in a substance is proportional to its number of moles.

Amount of H_{2} = 0.5 mol

22.4 L Cl_{2} at STP = 1 mol Cl_{2}

∴ 15 L Cl_{2} = \(\frac{15L}{22.4L}\) = 0.67 mol Cl_{2}

Hence, 15 L Cl_{2} at STP will have maximum number of molecules.

**Question 5. Arrange the gases in increasing number of atoms present in given amounts**

- 18gH
_{2}O, - 18gCO
_{2} - 18gCO
_{2}, - 18gCH
_{4}.

**Answer:**

1. 18gH_{2}O = 18/18 = 1 mol H_{2}O = 6.022 x 10^{23} molecules of H_{2}O

As 1 molecule of H_{2}O have 3 atoms, number of atoms = 3 x 6.022 x 10^{23}

2. 18 g O_{2}= 18/32 =0.5625 mol O_{2} = 0.5625 x 6.022 x 10^{23} molecules of O_{2}.

As 1 molecule of O_{2} have 2 atoms, number of atoms = 2 X 0.5625 X 6.022 X 10^{23} = 1.125 x 6.022 x 10^{23}

3. 18 g CO_{2} = 18/44 = 0.409 mol CO_{2} = 0.409 x 6.022 x 10^{23} molecules of CO_{2}

As 1 molecule of CO_{2} have 3 atoms, number of atoms = 3 x 0.409 x 6.022 x 10^{23} = 1.227 x 6.022 x 10^{23}

4. 18 g CH_{4} = 18/16 = 1.125 mol CH_{4} = 1.125 x 6.022 x 10^{23} molecules of CH_{4}

As 1 molecule of CH_{4} has 5 atoms, number of atoms = 5 X 1.125 x 6.022 x 10^{23} = 5.625 X 6.022 X 10^{23}

Hence, correct order is 18g O_{2} < 18g CO_{2} < 18g HO_{2} < 18g CH_{4}

**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 (C_{12}H_{22}O_{11}) = 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 10^{23}

Molecular mass of water (H_{2}O) = 1 x 2 + 16 =18

∴ 18 g water = 1 mol water = 6.022 x 10^{23} molecules of water.

As each molecule of water contains 1 O-atom, number of O-atoms in 18 g water = 6.022 x 10^{23}

∴ Total number of O-atoms in formed solution = 11 x 0.01 x 6.022 X 10^{23} + 1 x 6.022 x 10^{23} = 1.11 X 6.022 x 10^{23} = 6.68 X 10^{23}

**Question 7. Calculate the number of H-atoms present in 9 g water.**

**Answer:**

Molecular mass of H_{2}O = 1 x 2 + 16 = 18

Hence, 9 g water = 9/18 = 0.5 mol water = 0.5 x 6.022 x 10^{23} 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 10^{23} = 6.022 x 10^{23}

**Question 8. State whether number of atoms in**

- 1 g oxygen atom,
- 1 g oxygen molecule and
- 1 g ozone molecule will be same or different? Justify your answer.

**Answer:**

1. 16 g O-atom = N_{A} no. of O-atoms

∴ 1 g O-atom = \(\frac{N_A}{16}\) no. of O-atoms

2. 32 g O_{2} molecule = N_{A} no. of O_{2} molecules

∴ 1 g O_{2 }molecule = \(\frac{N_A}{32}\)no. of O_{2} molecules

= \(\frac{N_A}{32}\) x 2 no. of O-atoms = \(\frac{N_A}{16}\)no. of O-atoms.

3. 48 g O_{3} molecule = N_{A} no. of O_{3} molecules

∴ 1 g O_{3 }molecule = \(\frac{N_A}{48}\) no. of O_{3 }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 10^{23}) 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 10^{23}) of water molecules.

**Question 10. What is the volume of 4.4 g CO _{2} at STP?**

**Answer:**

Gram-molecular mass of CO_{2} = 12 + 16 x 2 = 44 g

Hence, at STP, volume of 44 g of CO_{2} = 22.4 L

∴ Volume of 4.4 g of CO_{2} 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 cm ^{3}. Find the molecular mass of the gas.**

**Answer:**

At STP, mass of 224 cm^{3} of the gas = 0.44 g

∴ At STP, the mass of 22.4L or 22400 cm^{3} 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 (NH_{3}) = 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 SO _{2 }at standard temperature and pressure.**

**Answer:**

Molecular mass of SO_{2} = 32 + 16 x 2 = 64

Hence, at STP, the volume of 64 g SO_{2} = 22.4L

∴ At STP, the volume of 4 g SO_{2 }= \(\frac{22.4}{64}\) 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 NH _{3} and 22.4 L CO_{2} will contain **

- same number of molecules?
- same number of atoms?

**Answer:**

1. At STP, 22.4 L NH_{3} = 1 mol NH_{3} Now, 1 mol NH_{3} = 6.022 x 10^{23} molecules NH_{3}

At STP, 22.4 L carbon dioxide = 1 mol carbon dioxide = 6.022 x 10^{23} molecules of CO_{2}

Hence, both NH_{3} and carbon dioxide contain same number of molecules.

2. 1 molecule of NH_{3} contains 4 atoms.

Hence, number of atoms in 22.4 L NH_{3} at STP = 4 X 6.022 X 10^{23} = 24.088 X 10^{23}

Again, 1 molecule of CO_{2} contains 3 atoms.

Hence, number of atoms in 22.4 L CO_{2} at STP = 3 X 6.022 X 10^{23} = 18.066 X 10^{23}

Therefore, 22.4 L NH_{3} will contain more atoms than that in 22.4 L CO_{2}.

**Question 17. Find the number of constituent ions in 11.1 g calcium chloride.**

**Answer:**

Gram-formula mass of calcium chloride (CaCI_{2}) = 40 + 35.5 X 2 = 111 g

∴ Number of formula units in 111 g CaCl_{2} = 6.022 x 10^{23}

∴ Number of formula units in 11.1 g CaCI_{2}

= \(\frac{6.022 \times 10^{23}}{111}\) x 11.1 = 6.022 x 10^{22}

Each formula unit of calcium chloride contains Ca^{2+} ions and 2CI^{–} ions.

Hence, number of Ca^{2+} ions in 11.1 g CaCI_{2} = 1 x 6.022 X 10^{22} = 6.022 x 10^{22}

Number of Cl^{–} ions in 11.1 g CaCI_{2 }= 2 X 6.022 X 10^{22} = 12.044 x 10^{22}

**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:**

Given

A mixture of hydrogen and oxygen contains 20% hydrogen by weight.

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 10^{23 }= 6.022 x 10^{23}

Number of molecules of oxygen in 0.8 g of oxygen gas

= \(\frac{0.8}{32}\) x 6.022 x 10^{23} = 1.5055 x 10^{22}

∴ Total number of molecules per gram of the mixture

= (6.022 + 1.5055) x 10^{22} = 7.5275 x 10^{22}

**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 Cl ^{Θ }**

**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 Na^{⊕ }ions = number of Cl^{Θ }ions

∴ Number of Na® ions in 2 gram-formula mass of NaCI = number of Cl^{Θ} ions in 2 gram-formula mass of NaCI

= 2 x 6.022 x 10^{23} = 12.044 x 10^{23} = 1.2044 X 10^{24}

**Question 22. You have been given a glass of water mixed with 10 g glucose (C _{6}H_{12}O_{6}). 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 10^{23 }= 3.345 x 10^{22}

∴ 3.345 x 10^{22} number of glucose molecules will be consumed.

**Question 23. Calculate the number of electrons, protons and neutrons in 1 mol O ^{2-} ion.**

**Answer:**

Number of electrons in an O^{2-} ion = (8 + 2) = 10

Number of protons in an O^{2-} ion = 8

∴ Number of neutrons in an O^{2-} ion = (16 – 8) = 8 (assuming _{8}O^{16}16 isotope)

Number of electrons in 1 mol O^{2-} ion = 1 x 10 x 6.022 x 10^{23 } = 6.022 x 10^{24}

∴ Number of protons in 1 mol O^{2-} ion = 1 X 8 x 6.022 X 10^{23} = 4.817 X 10^{24}

∴ Number of neutrons in 1 mol O^{2-} ion = 1 x 8 x 6.022 x 10^{23} = 4.817 x 10^{24}

**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 10^{23} atoms of phosphorus

= 2.4088 x 10^{24 }atoms of phosphorus 124 g phosphorus

= \(\frac{124}{124}\) g-molecule phosphorus

= 1 x 6.022 x 10^{23} molecules of phosphorus = 6.022 x 10^{23} 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 10^{25}

Again, number of H-atoms in 1 molecule of water = 2

∴ Number of H-atoms in 1 L water = 2 X 3.345 X 10^{25} = 6.69 X 10^{25}

**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 NH_{3} = 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 10^{23 }= 5.4198 x 10^{24 }

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 NO _{2 }molecule?**

**Answer:**

Molecular mass of NO = 14 + 16 = 30

∴ Mass of 1 mol NO molecules = 30 g

Again, molecular mass of NO_{2} = 14 + 16 X 2 = 46

∴ Mass of 0.5 mol NO_{2} molecules = 46 X 0.5 = 23 g

∴ 1 mol NO molecules will be heavier than that of 0.5 mol NO_{2} 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 10 ^{21} molecules of CO_{2}are removed from 200 mg of CO_{2}•**

**Answer:**

200 mg of CO_{2} = 0.2 g of CO_{2}

Now, 44g of CO_{23} contains 6.022 x 10^{23 }molecules

∴ 0.2 g of CO_{2} contains

⇒ \(\frac{6.022 \times 10^{23} \times 0.2}{44}=2.7372 \times 10^{21}\) molecules

∴ On removing 10^{21} molecules, number of CO_{2} molecules left

= 2.7372 X 10^{21 }– 10^{21} = 1.7372 X 10^{21}

## 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.

**Answer: **1. B, 2. A, 3. D, 4. C

2.

**Answer:** 1. D, 2. A, 3. B, 4. C