Chapter 1 Is Matter Around Us Pure Short Answer Type Question And Answers
Question 1. A solution contains 10 g of sugar in 20 g of water. Calculate the concentration in terms of mass-by-mass percentage of the solution.
Answer.
Given:
A solution contains 10 g of sugar in 20 g of water.
Mass of solute (sugar) = 10 g
Mass of solvent (water) = 20 g
⇒ \(\text { Mass of solution }=\frac{\text { mass of solute }}{\text { mass of solution }}\)
= 10 g + 20 g
= 30 g
Mass percentage of solution = mass of solute/mass of solution × 100
= 10/30 × 100
= 33.3 %
Question 2. Suppose you had 1.00 moles of solute dissolved into 1.00 L of solution. What is the molarity?
Answer.
Given:
Suppose you had 1.00 moles of solute dissolved into 1.00 L of solution.
⇒ \(\text { Molarity }=\frac{\text { moles of solute }}{\text { volume of solution in liters }}\)
Molarity = \(\frac{1.00 \mathrm{~mol}}{1.00 \mathrm{~L}}\)
The answer is 1.00 M.
Question 3. A 4 g sugar cube (C12H22O11) is dissolved in a 350 mL water. What is the molarity of the sugar solution?
Answer.
Given:
A 4 g sugar cube (C12H22O11) is dissolved in a 350 mL water.
Molarity (mol/L) = number of moles of solute (mol)/Volume of solvent (Litres)
Step 1: Determine number of moles of sucrose in 4 g
C12H22O11 = (12)(12) + (1)(22) + (16)(11)
C12H22O11 = 144 + 22+ 176
C12H22O11 = 342 g/mol
Divide this amount into the size of the sample
4 g/(342 g/mol) = 0.0117 mol
Step 2: Determine the volume of solvent in litres
350 mL × (1 L/1000 mL) = 0.350 L
Step 3: Determine the molarity of the solution
⇒ \(\text { Molarity }(\mathrm{mol} / \mathrm{L})=\frac{\text { number of moles of solute }(\mathrm{mol})}{\text { Volume of solvent }(\text { Litres })}\)
M = 0.0117 mol/0.350 L
M = 0.033 mol/L
The molarity of the sugar solution is 0.033 mol/L.
Question 4. A mixture of gases was formed by combining 6.3 moles of O2 and 5.6 moles of N2. What is the mole fraction of nitrogen in the mixture?
Answer.
Given:
A mixture of gases was formed by combining 6.3 moles of O2 and 5.6 moles of N2.
Total number of moles with Ntotal = NN2+OO2 .
Ntotal = 6.3 moles + 5.6 moles
= 11.9 moles.
⇒ \(\mathrm{X}_{\text {(mole fraction) }}=\frac{\text { mole } \mathrm{N}_2}{\text { mole } \mathrm{N}_2+\text { moles } \mathrm{O}_2}\)
= (5.6 moles/11.9 moles)
= 0.47
The mole fraction of nitrogen in the mixture is 0.47.
Question 5. The absolute mass of a carbon atom is 12.0 unified atomic mass units. How many grams will a single oxygen atom weigh?
Answer.
Given:
The absolute mass of a carbon atom is 12.0 unified atomic mass units.
The absolute mass of the carbon atom is 12.0 u, Grams in a single oxygen atom weigh
= \(\frac{\text { absolute mass of carbon }}{\text { Avogadro’s Number }}\)
= (12 g/mol) ÷ (6.022 × 1023 mol-1) × (4/3)
= 2.66 × 10-23 g.
Question 6. Water is a compound. Explain.
Answer.
Given:
Water is a compound.
Yes, water is a compound as:
It cannot be separated into its constituents (Hydrogen and oxygen) by physical methods.
Its properties are entirely different from its constituents.
Energy in the form of heat and light is given out when water is formed by burning Hydrogen and Oxygen.
Composition of water is fixed. Water has fixed boiling point.
Question 7. Air is a mixture. Explain.
Answer.
Given:
Air is a mixture
Yes, Air is a mixture as:
Its composition is not always same.
Gases present in Air do not lose their identities. Its major constituents can be easily separated by physical method. No energy change takes place among the constituents of air.
Question 8. Is burning of candle a chemical or physical change?
Answer.
Given
Burning of candle a chemical or physical change:
When the candle is burnt, two changes occur:
- Wax gets melts, which is a physical change
- Thread and wax get burned, which is a chemical change.
So, the burning of a candle involves both physical and chemical change.
Question 9. When you mix two clear liquids, what kinds of observations would tell you that a chemical reaction occurred?
Answer.
A gas is produced.
The solid dissolves.
A precipitate forms.
You can see through the mixture.
Answer: A gas is produced and A precipitate forms.
Chapter 2 Is Matter Around Us Pure Master your Test Question And Answers
Question 1. List 4 examples of mixtures.
Answer. 4 examples of mixtures are:
- Sand and Water
- Salt and Water
- Sugar and Water
- Alcohol in Water
Question 2. How are mixtures classified?
Answer.
Mixtures are classified as follows:
- Based on physical state of components: Solutions, Colloids, Suspensions
- Based on their appearance: Homogeneous and Heterogeneous Mixtures
Question 3. Give 2 examples of each homogeneous and heterogeneous mixtures.
Answer.
Homogeneous Mixtures: Rainwater, Vinegar
Heterogeneous Mixture: Muddy Water, Chalk Powder in Water
Question 4. What is the major difference between a pure substance and a mixture?
Answer.
A pure substance is made up of only one type of particles and lacks contamination from any other type of of particle whereas, mixture is made up of two or more type of particles in variable proportion.
Question 5. Define a substance.
Answer.
Substance:
Substance is simply a pure form of matter, i.e., it contains only one type of atom or molecule.
Chapter 2 Is Matter Around Us Pure Master Your Test Question And Answers
Question 1. When 5.0 g of solute is dissolved in 50 g of water at 298 K to get a saturated solution, then what is the solubility of the solute at the given temperature?
Answer.
Given:
When 5.0 g of solute is dissolved in 50 g of water at 298 K to get a saturated solution.
Weight of the solute in saturated solution = 5.0 g
Weight of solvent in saturated solution = 50 g
Solubility = (weight of the solute in saturated solution/weight of solvent in saturated solution) × 100
Solubility = \(\frac{5.0}{50} \times 100=10 \mathrm{~g}\)
Question 2. A solution contains 10 g of sugar in 20 g of water. Calculate the concentration in terms of mass by mass percentage of the solution and also by mass percentage of the solvent.
Answer.
Given:
A solution contains 10 g of sugar in 20 g of water.
Mass of solute (sugar) = 10 g
Mass of solvent (water) = 20 g
Mass of solution = Mass of solute + Mass of solvent
= 10 g + 20 g
= 30 g
Mass percentage of solution = Mass of solute/Mass of solution × 100
= \(\frac{10}{30} \times 100\)
= 33.3 %
Mass percentage of solvent = 100 – mass percentage of solute
= 100 – 33.3
= 66.7%
Question 3. Suppose you had 1.00 moles of solute dissolved into 1.00 L of solution. What’s the molarity?
Answer.
Given:
1.00 moles of solute dissolved into 1.00 L of solution.
⇒ \(\text { Molarity }=\frac{\text { moles of solute }}{\text { volume of solution in litres }}\)
Molarity = \(\frac{1.00 \mathrm{~mol}}{1.00 \mathrm{~L}}\)
The answer is 1.00 M.
Question 4. A 4 g sugar cube (C12H22O11) is dissolved in a 350 mL water. What is the molarity of the sugar solution?
Answer.
Given:
A 4 g sugar cube (C12H22O11) is dissolved in a 350 mL water.
Molarity (mol/L) = number of moles of solute (mol)/Volume of solvent(Litres)
Step 1: Determine number of moles of sucrose in 4 g
C12H22O11 = (12) (12) + (1) (22) + (16) (11)
C12H22O11 = 144 + 22+ 176
C12H22O11 = 342 g/mol
divide this amount into the size of the sample
4 g/(342 g/mol) = 0.0117 mol
Step 2: Determine the volume of solvent in litres
350 mL × (1 L/1000 mL) = 0.350 L
Step 3: Determine the molarity of the solution
Molarity (mol/L) = number of moles of solute (mol)/Volume of solvent(Litres)
M = 0.0117 mol/0.350 L
M = 0.033 mol/L
The molarity of the sugar solution is 0.033 mol/L.
Question 5. A mixture of gases was formed by combining 6.3 moles of O2 and 5.6 moles of N2. What is the mole fraction of nitrogen in the mixture?
Answer.
Given:
A mixture of gases was formed by combining 6.3 moles of O2 and 5.6 moles of N2.
Total number of moles with ntotal = nN2 + nO2.
ntotal = 6.3 moles + 5.6 moles
= 11.9 moles.
X(molefraction) = (moles N2 /moles N2 + moles O2)
= (5.6 moles/11.9 moles)
= 0.47
The mole fraction of nitrogen in the mixture is 0.47.
Question 6. A solution contains 10 mL of alcohol in 20 mL of water. Calculate the volume by volume percentage of the solute.
Answer.
Given:
A solution contains 10 mL of alcohol in 20 mL of water.
Mass of solute (alcohol) = 10 mL
Mass of solvent (water) = 20 mL
Volume of solution = Volume of solute + volume of solvent
= 10 mL + 20 mL
= 30 mL
Volume percentage of solute = Volume of solute/Volume of solution × 100
= \(\frac{10}{30} \times 100\)
= 33.3 %
Question 7. A solution contains 10 g of alcohol in 20 g of water. Calculate the concentration of the solution.
Answer.
Given:
A solution contains 10 g of alcohol in 20 g of water.
Mass of solute (alcohol) = 10 mL
Mass of solvent (water) = 20 mL
Mass of solution = Mass of solute + Mass of solvent
= 10 mL + 20 mL
= 30 mL
Concentration of Solution = Mass of solute/Mass of solution × 100
= \(\frac{10}{30} \times 100\)
= 33.3 %
Question 8. Express the concentration in ppm or ppb, when a solution with mass of 2.00 kg contains 6 mg of solute.
Answer.
Given
A solution with mass of 2.00 kg contains 6 mg of solute
Mass of solute = 6 mg
= 6 × 10-3 g
Mass of solution = 2.00 kg
= 2 × 103 g
Ppm = (mass of solute/mass of solution) × 106
= (6 × 10-3 g / 2 × 102 g) × 106
= 3 ppm
Ppb = (mass of solute/mass of solution) × 109
= (6 × 10-3 g/2 × 103 g) × 109
= 3 × 106 ppm
Chapter 2 Is Matter Around Us Pure Master your Test Question And Answers
Question 1. Define pure substances.
Answer.
Pure substances:
Pure substances in chemistry are defined as the substances that are made up of only one type of particles.
Question 2. How many types of elements are there?
Answer.
There are three types of elements: Metals, Metalloids and Non-Metals
Question 3. List some properties of metals.
Answer.
Properties of metal:
- They have a lustre.
- They conduct heat and electricity.
- High melting point.
- High density.
- They are ductile.
- They are malleable.
- They are sonorous.
Question 4. What are metalloids?
Answer.
Metalloids:
A metalloid is a type of chemical elements which has properties of metals and non-metals.
Chapter 2 Is Matter Around Us Pure Master your Test Question And Answers
Question 1. Write two features of physical and chemical change.
Answer.
Features of physical and chemical change:
- When physical change occurs in a substance, no new substance is created. The substance will remain in its original state. When chemical change occurs in the substance, you will be able to produce a different kind of substance. This means you will lose the original substance and a new one will form.
- A physical change is superficial and can possibly be reversed; a chemical change is complete and permanent.
Question 2. Give two examples of chemical changes.
Answer.
Examples of chemical changes are:
- Burning of wood
- Rusting of Iron
Question 3. Give two examples of physical changes that you observe in nature.
Answer.
Examples of physical changes are:
- Plucking of flower
- Cutting of trees
