Solutions Class 12 : Important Questions For Board Exams Preparation

इस नोट्स को हिन्दी में पढ़ें

The chapter Solutions is an important chapter for Class 12 Chemistry Board Exams. This question bank contains 100+ important questions specially prepared for English Medium students.

It includes 1-mark very short-answer questions, 2-mark short-answer questions, 3 Mark Questions, 5 Mark Questions and important MCQs, covering key concepts such as concentration terms, Henry’s Law, Raoult’s Law, ideal and non-ideal solutions, colligative properties, osmotic pressure and Van’t Hoff factor.

Use these questions for quick revision, concept practice and board-exam preparation. Regular practice will help you remember important definitions, formulas and concepts and write short, precise and accurate answers in the examination.

Solutions — 25 Very Short Answer Questions(1 Mark Questions)

1. What is a solution?
Answer: A homogeneous mixture of two or more components.

2. What is the solvent?
Answer: The component present in larger amount.

3. What is the solute?
Answer: The component dissolved in the solvent.

4. Define molarity.
Answer: Moles of solute per litre of solution.

5. Define molality.
Answer: Moles of solute per kilogram of solvent.

6. What is mole fraction?
Answer: Ratio of moles of a component to total moles.

7. Which concentration unit is temperature independent?
Answer: Molality.

8. State Henry’s law.
Answer: Solubility of a gas is proportional to its partial pressure.

9. Write Henry’s law equation.
Answer: p = Kₕx

10. What is Raoult’s law?
Answer: pᵢ = xᵢpᵢ°

11. What is an ideal solution?
Answer: A solution obeying Raoult’s law at all concentrations.

12. Give one example of an ideal solution.
Answer: Benzene–toluene mixture.

13. What is positive deviation?
Answer: Vapour pressure is higher than predicted by Raoult’s law.

14. What is negative deviation?
Answer: Vapour pressure is lower than predicted by Raoult’s law.

15. What are colligative properties?
Answer: Properties depending on the number of solute particles.

16. Name any two colligative properties.
Answer: Osmotic pressure and elevation in boiling point.

17. What is osmosis?
Answer: Movement of solvent through a semipermeable membrane.

18. What is osmotic pressure?
Answer: Pressure required to stop osmosis.

19. Write the osmotic pressure equation.
Answer: π = CRT

20. What is reverse osmosis?
Answer: Solvent flow caused by applying pressure above osmotic pressure.

21. What is elevation in boiling point?
Answer: Increase in boiling point due to dissolved solute.

22. Write the formula for elevation in boiling point.
Answer: ΔTᵦ = Kᵦm

23. What is depression in freezing point?
Answer: Decrease in freezing point due to dissolved solute.

24. Write the formula for depression in freezing point.
Answer: ΔT𝒇 = K𝒇m

25. What is Van’t Hoff factor?
Answer: It accounts for association or dissociation of solute particles.

Solutions — 25 Short Answer Type Questions (1 Mark Questions)

1. What is meant by a saturated solution?
Answer: A solution that contains the maximum amount of solute that can dissolve in a solvent at a given temperature is called a saturated solution.

2. Define the term concentration of a solution.
Answer: Concentration represents the amount of solute present in a given amount of solution or solvent.

3. What is the SI unit of molality?
Answer: The SI unit of molality is mol kg⁻¹.

4. Why is molality preferred over molarity in some temperature-dependent calculations?
Answer: Molality does not change with temperature because it is based on the mass of the solvent.

5. What happens to the vapour pressure of a solvent when a non-volatile solute is added to it?
Answer: The vapour pressure of the solvent decreases.

6. State Raoult’s law for a solution containing a volatile component.
Answer: The partial vapour pressure of a component is equal to the product of its mole fraction and vapour pressure in the pure state:
pᵢ = xᵢpᵢ°

7. What is an ideal solution?
Answer: A solution that obeys Raoult’s law over the entire range of concentration is called an ideal solution.

8. Give one example of a nearly ideal solution.
Answer: A mixture of benzene and toluene behaves approximately as an ideal solution.

9. What is meant by positive deviation from Raoult’s law?
Answer: It occurs when the observed vapour pressure of a solution is higher than the value predicted by Raoult’s law.

10. What type of intermolecular interaction generally causes negative deviation from Raoult’s law?
Answer: Stronger interactions between unlike molecules than between like molecules generally cause negative deviation.

11. What are colligative properties?
Answer: Properties that depend only on the number of solute particles present in a solution, and not on their chemical nature, are called colligative properties.

12. Name the colligative property used for determining the molar mass of macromolecules.
Answer: Osmotic pressure is commonly used for determining the molar mass of macromolecules.

13. What is osmotic pressure?
Answer: Osmotic pressure is the minimum external pressure required to stop osmosis.

14. Which direction does the solvent move during osmosis?
Answer: The solvent moves through a semipermeable membrane from the region of lower solute concentration to higher solute concentration.

15. What is a semipermeable membrane?
Answer: It is a membrane that allows solvent molecules to pass through it but restricts the passage of solute particles.

16. Write the equation relating osmotic pressure to concentration.
Answer:
π = CRT

where π is osmotic pressure, C is molar concentration, R is the gas constant and T is absolute temperature.

17. What is reverse osmosis?
Answer: Reverse osmosis is the process in which solvent is forced to move from a solution of higher concentration towards pure solvent or a lower-concentration side by applying pressure greater than osmotic pressure.

18. Mention one important application of reverse osmosis.
Answer: It is widely used for desalination and purification of water.

19. What is elevation in boiling point?
Answer: The increase in the boiling point of a solution compared with that of the pure solvent is called elevation in boiling point.

20. Write the relation between elevation in boiling point and molality.
Answer:
ΔTᵦ = Kᵦm

where Kᵦ is the molal elevation constant and m is molality.

21. Why does the boiling point of a solution increase when a non-volatile solute is added?
Answer: Addition of a non-volatile solute lowers the vapour pressure of the solvent, so a higher temperature is required for boiling.

22. What is depression in freezing point?
Answer: The lowering of the freezing point of a solution compared with that of the pure solvent is called depression in freezing point.

23. Write the mathematical expression for depression in freezing point.
Answer:
ΔT𝒇 = K𝒇m

where K𝒇 is the molal depression constant and m is molality.

24. What is the Van’t Hoff factor?
Answer: The Van’t Hoff factor (i) represents the ratio of the observed colligative property to the theoretically calculated colligative property.

25. Why may the Van’t Hoff factor be less than or greater than one?
Answer: It may be less than 1 due to association of solute particles and greater than 1 due to dissociation of solute particles.

Solutions — 25 Short Answer Questions (1 Mark Questions)

1. Distinguish between molarity and molality.
Answer: Molarity is moles of solute per litre of solution, whereas molality is moles of solute per kilogram of solvent. Molarity changes with temperature, but molality does not.

2. Define mass percentage and write its formula.
Answer: Mass percentage is the mass of solute present in 100 g of solution.
Mass % = (Mass of solute / Mass of solution) × 100

3. What is an ideal solution? Mention two characteristics.
Answer: An ideal solution obeys Raoult’s law throughout the concentration range. For an ideal solution, ΔHmix = 0 and ΔVmix = 0.

4. Why does adding a non-volatile solute lower the vapour pressure of a solvent?
Answer: The solute occupies some surface area and reduces the number of solvent molecules escaping into the vapour phase. Hence, the vapour pressure decreases.

5. State Raoult’s law for a solution of volatile liquids.
Answer: The partial vapour pressure of each component is proportional to its mole fraction.
pᵢ = xᵢpᵢ°

6. What is meant by positive deviation from Raoult’s law? Give one example.
Answer: A solution shows positive deviation when its vapour pressure is higher than predicted by Raoult’s law. Ethanol–acetone is an example.

7. What is negative deviation from Raoult’s law? Give one example.
Answer: A solution shows negative deviation when its vapour pressure is lower than predicted by Raoult’s law. Chloroform–acetone is an example.

8. Why are gases generally less soluble in liquids at higher temperatures?
Answer: Dissolution of most gases in liquids is exothermic. Increasing temperature favours the escape of dissolved gas, thereby decreasing its solubility.

9. State Henry’s law and write its equation.
Answer: At constant temperature, the solubility of a gas in a liquid is proportional to its partial pressure.
p = Kₕx

10. Why does soda water fizz when the bottle is opened?
Answer: Opening the bottle reduces the pressure of CO₂ above the solution. The solubility of CO₂ decreases, so dissolved gas escapes as bubbles.

11. Define colligative properties. Name any two.
Answer: Colligative properties depend only on the number of dissolved solute particles. Examples are osmotic pressure and depression in freezing point.

12. Why does the boiling point of a solution increase after adding a non-volatile solute?
Answer: The solute lowers the vapour pressure of the solvent. Therefore, a higher temperature is required for the vapour pressure to become equal to atmospheric pressure.

13. Write the relation between elevation in boiling point and molality. Explain Kᵦ.
Answer:
ΔTᵦ = Kᵦm
Here, Kᵦ is the molal elevation constant or ebullioscopic constant.

14. Write the relation between depression in freezing point and molality.
Answer:
ΔT𝒇 = K𝒇m
Here, K𝒇 is the molal depression constant or cryoscopic constant.

15. Why does a solution freeze at a lower temperature than the pure solvent?
Answer: The presence of solute lowers the vapour pressure of the liquid phase. Therefore, the solution must be cooled to a lower temperature before freezing occurs.

16. What is osmotic pressure? How is it represented?
Answer: Osmotic pressure is the pressure required to stop osmosis. It is represented by π.

17. Write the equation for osmotic pressure of a dilute solution.
Answer:
π = CRT
where C is molar concentration, R is gas constant and T is absolute temperature.

18. What is reverse osmosis? Mention one application.
Answer: Reverse osmosis occurs when pressure greater than osmotic pressure is applied on the solution side. It is used in water purification.

19. Why is osmotic pressure useful for finding the molar mass of proteins?
Answer: It can be measured at room temperature and does not require heating. Hence, decomposition of proteins can be avoided.

20. What is Van’t Hoff factor? Why is it important?
Answer: Van’t Hoff factor accounts for changes in the number of solute particles due to association or dissociation. It is represented by i.

21. What happens to the Van’t Hoff factor during dissociation?
Answer: During dissociation, the number of particles increases. Therefore, the Van’t Hoff factor becomes greater than 1.

22. What happens to the Van’t Hoff factor during association?
Answer: During association, the number of particles decreases. Therefore, the Van’t Hoff factor becomes less than 1.

23. Why is molality independent of temperature?
Answer: Molality is based on the mass of the solvent. Since mass does not change with temperature, molality remains constant.

24. A solution contains 1 mole of solute and 9 moles of solvent. Find the mole fraction of the solute.
Answer:
Total moles = 1 + 9 = 10
Xsolute = 1/10 = 0.1

25. What is the effect of increasing pressure on the solubility of a gas in a liquid?
Answer: Increasing pressure generally increases the solubility of a gas in a liquid, according to Henry’s law.

Solutions — 20 Long Answer Questions (3 Mark Questions)

1. Define molarity, molality and mole fraction.

Answer:

Molarity (M): Number of moles of solute present in one litre of solution.
M = Moles of solute / Volume of solution in litre
Molality (m): Number of moles of solute present in one kilogram of solvent.
m = Moles of solute / Mass of solvent in kg
Mole fraction (x): Ratio of moles of one component to the total number of moles of all components.
xᵢ = nᵢ / Σn

2. State Raoult’s law for a solution containing volatile components. Write its mathematical expression.

Answer:
Raoult’s law states that the partial vapour pressure of each volatile component of a solution is directly proportional to its mole fraction.

For component A:
pₐ = xₐpₐ°

For component B:
pᵦ = xᵦpᵦ°

Therefore, total vapour pressure is:
P = pₐ + pᵦ = xₐpₐ° + xᵦpᵦ°

3. What is an ideal solution? Give two characteristics and one example.

Answer:
An ideal solution is one which obeys Raoult’s law over the entire range of concentration.

Characteristics:

ΔHmix = 0
ΔVmix = 0

Example: Benzene and toluene.

4. Explain positive and negative deviations from Raoult’s law.

Answer:

Positive deviation: The vapour pressure is higher than predicted by Raoult’s law. It occurs when A–B interactions are weaker than A–A and B–B interactions. Example: ethanol + acetone.
Negative deviation: The vapour pressure is lower than predicted by Raoult’s law. It occurs when A–B interactions are stronger than A–A and B–B interactions. Example: chloroform + acetone.

5. State Henry’s law. How does pressure affect the solubility of a gas in a liquid?

Answer:
Henry’s law states that at constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the solution.

p = Kₕx

where p is the partial pressure and x is the mole fraction of the gas.

Thus, increase in pressure increases the solubility of a gas in a liquid.

6. Why does the solubility of gases in liquids decrease with increase in temperature? Give one example.

Answer:

Dissolution of most gases in liquids is an exothermic process.
According to Le Chatelier’s principle, increasing temperature favours the reverse process.
Therefore, the solubility of gases generally decreases with increase in temperature.

Example: Dissolved oxygen in water decreases when water is heated.

7. What are colligative properties? Name and define the four colligative properties.

Answer:
Colligative properties are the properties of dilute solutions that depend only on the number of solute particles, not on their nature.

The four colligative properties are:

Relative lowering of vapour pressure
Elevation in boiling point
Depression in freezing point
Osmotic pressure

8. Derive the relation between elevation in boiling point and molality.

Answer:
The elevation in boiling point is directly proportional to the molality of the solution.

Therefore,

ΔTᵦ ∝ m

or,

ΔTᵦ = Kᵦm

where ΔTᵦ is elevation in boiling point, m is molality and Kᵦ is the molal elevation constant.

For a solution containing w₂ g of solute of molar mass M₂ in w₁ g of solvent:

ΔTᵦ = Kᵦ × (1000w₂ / M₂w₁)

9. Derive the relation between depression in freezing point and molality.

Answer:
The depression in freezing point is directly proportional to the molality of the solution.

Therefore,

ΔT𝒇 ∝ m

or,

ΔT𝒇 = K𝒇m

where K𝒇 is the molal depression constant.

For w₂ g of solute dissolved in w₁ g of solvent:

ΔT𝒇 = K𝒇 × (1000w₂ / M₂w₁)

10. What is osmotic pressure? Write its relation with molar concentration.

Answer:
Osmotic pressure is the minimum external pressure required to stop the flow of solvent through a semipermeable membrane.

For a dilute solution:

π = CRT

where:
π = osmotic pressure
C = molar concentration
R = gas constant
T = absolute temperature

11. What is reverse osmosis? Explain its application in water purification.

Answer:
Reverse osmosis is the process in which solvent is forced to move from the solution side towards the pure solvent side by applying pressure greater than osmotic pressure.

Application:
It is used for desalination of seawater and purification of drinking water. The applied pressure forces water through a semipermeable membrane while dissolved salts are retained.

12. Why is osmotic pressure preferred for determining the molar mass of proteins and other macromolecules?

Answer:
Osmotic pressure is preferred because:

It can be measured at room temperature.
Very small concentrations can produce measurable osmotic pressure.
Proteins and other macromolecules may decompose at higher temperatures, which can be avoided.

13. Explain the effect of a non-volatile solute on the boiling point and freezing point of a solvent.

Answer:
Addition of a non-volatile solute:

Decreases the vapour pressure of the solvent.
Increases its boiling point: ΔTᵦ = Kᵦm
Decreases its freezing point: ΔT𝒇 = K𝒇m

These are colligative properties.

14. What is Van’t Hoff factor? Explain its value for association and dissociation.

Answer:
Van’t Hoff factor (i) accounts for the association or dissociation of solute particles in solution.

i = Observed colligative property / Calculated colligative property

For dissociation, the number of particles increases, so i > 1.
For association, the number of particles decreases, so i < 1.
If there is no association or dissociation, i = 1.

15. Calculate the molality of a solution containing 18 g of glucose (C₆H₁₂O₆) dissolved in 500 g of water.

Answer:
Molar mass of glucose = 180 g mol⁻¹

Moles of glucose:

n = 18/180 = 0.1 mol

Mass of water = 500 g = 0.5 kg

Therefore,

Molality = 0.1/0.5 = 0.2 mol kg⁻¹

Answer: 0.2 mol kg⁻¹

16. Calculate the mole fraction of ethanol and water in a solution containing 46 g ethanol and 54 g water.

Answer:
Moles of ethanol:

46/46 = 1 mol

Moles of water:

54/18 = 3 mol

Total moles = 1 + 3 = 4 mol

Mole fraction of ethanol:

xₑₜₕₐₙₒₗ = 1/4 = 0.25

Mole fraction of water:

xwater = 3/4 = 0.75

17. A solution contains 5 g of a non-volatile solute in 95 g of water. Calculate the mass percentage of the solute.

Answer:
Mass of solution:

5 + 95 = 100 g

Mass percentage:

Mass % = (Mass of solute / Mass of solution) × 100

= (5/100) × 100

= 5%

Therefore, the mass percentage of solute is 5%.

18. What happens to the boiling point of water when a non-volatile solute is added? Explain with reason and equation.

Answer:
The boiling point of water increases when a non-volatile solute is added.

Reason: The solute lowers the vapour pressure of water. Hence, a higher temperature is required for the vapour pressure to become equal to atmospheric pressure.

The increase is given by:

ΔTᵦ = Kᵦm

Therefore,

Tᵦ(solution) > Tᵦ(pure solvent)

19. Explain why 0.1 M NaCl solution shows a greater colligative effect than 0.1 M glucose solution under comparable conditions.

Answer:
NaCl dissociates in water:

NaCl → Na⁺ + Cl⁻

Thus, one formula unit produces approximately two particles.

Glucose does not dissociate and remains as individual molecules.

Therefore, NaCl produces more solute particles and shows a greater colligative effect than glucose at comparable concentration.

20. Differentiate between osmosis and reverse osmosis.

Answer:

Osmosis	Reverse Osmosis
Solvent moves naturally through a semipermeable membrane.	Solvent is forced to move by applying external pressure.
Movement is from lower solute concentration to higher solute concentration.	Movement is from higher solute concentration towards the lower concentration/pure solvent side.
It does not require external pressure.	Pressure greater than osmotic pressure is required.

1. State and explain Raoult’s law for a solution of two volatile liquids. Derive the expression for total vapour pressure.

Answer:
Raoult’s law states that the partial vapour pressure of each volatile component of a solution is directly proportional to its mole fraction.

For component A:

pₐ = xₐpₐ°

For component B:

pᵦ = xᵦpᵦ°

According to Dalton’s law, total vapour pressure is:

P = pₐ + pᵦ

Therefore,

P = xₐpₐ° + xᵦpᵦ°

Since,

xₐ + xᵦ = 1

we can write:

P = pᵦ° + (pₐ° − pᵦ°)xₐ

Thus, the total vapour pressure varies linearly with the mole fraction of either component.

2. What are ideal and non-ideal solutions? Explain positive and negative deviations from Raoult’s law with examples.

Answer:

An ideal solution obeys Raoult’s law over the entire range of concentration.

For an ideal solution:

ΔHmix = 0

and

ΔVmix = 0

Example: benzene + toluene.

A non-ideal solution does not obey Raoult’s law over the entire concentration range.

Positive deviation:
The vapour pressure is higher than predicted by Raoult’s law. This occurs when A–B interactions are weaker than A–A and B–B interactions.

Example: ethanol + acetone.

Negative deviation:
The vapour pressure is lower than predicted by Raoult’s law. This occurs when A–B interactions are stronger than A–A and B–B interactions.

Example: chloroform + acetone.

3. Explain the four colligative properties of solutions.

Answer:
Colligative properties depend only on the number of solute particles present in a solution.

The four colligative properties are:

Relative lowering of vapour pressure: Addition of a non-volatile solute lowers the vapour pressure of the solvent.
Elevation in boiling point: The boiling point of a solution is higher than that of the pure solvent.

ΔTᵦ = Kᵦm

Depression in freezing point: The freezing point of a solution is lower than that of the pure solvent.

ΔT𝒇 = K𝒇m

Osmotic pressure: The minimum pressure required to stop osmosis.

π = CRT

4. Explain Henry’s law. Mention three applications of Henry’s law.

Answer:
Henry’s law states that at constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the solution.

Mathematically:

p = Kₕx

where p is the partial pressure of the gas, x is its mole fraction in solution and Kₕ is Henry’s law constant.

Applications:

Carbonated drinks: CO₂ is dissolved in beverages under high pressure.
Scuba diving: At high pressure, more nitrogen dissolves in the blood. Sudden release of pressure may cause bubbles to form.
High altitudes: Lower atmospheric pressure reduces the solubility of oxygen in blood, which can cause breathing difficulties.

5. Explain osmosis and osmotic pressure. Derive the equation π = CRT.

Answer:
Osmosis is the movement of solvent molecules through a semipermeable membrane from a dilute solution towards a concentrated solution.

The minimum external pressure required to stop this flow is called osmotic pressure (π).

For a dilute solution, osmotic pressure follows the gas equation:

πV = nRT

Therefore,

π = nRT/V

Since,

n/V = C

we get:

π = CRT

where:

π = osmotic pressure
C = molar concentration
R = gas constant
T = absolute temperature

Osmotic pressure is useful for determining the molar masses of proteins and other macromolecules.

6. Explain reverse osmosis and describe its use in the purification of water.

Answer:
In normal osmosis, solvent flows through a semipermeable membrane from a dilute solution towards a concentrated solution.

In reverse osmosis, an external pressure greater than the osmotic pressure is applied on the concentrated solution. This forces the solvent to move in the reverse direction.

Thus, water moves from the concentrated solution towards the pure-water side through the semipermeable membrane.

Application:
Reverse osmosis is widely used for desalination of seawater and purification of drinking water. Dissolved salts and many impurities are retained by the membrane while water passes through.

7. Explain elevation in boiling point and derive its relation with molality.

Answer:
When a non-volatile solute is added to a solvent, the vapour pressure of the solvent decreases. Therefore, the solution must be heated to a higher temperature to reach atmospheric pressure.

The increase in boiling point is called elevation in boiling point.

It is represented by:

ΔTᵦ = Tᵦ − Tᵦ°

For dilute solutions:

ΔTᵦ ∝ m

Therefore,

ΔTᵦ = Kᵦm

For a solution containing w₂ g of solute of molar mass M₂ dissolved in w₁ g of solvent:

m = (1000w₂)/(M₂w₁)

Hence,

ΔTᵦ = Kᵦ(1000w₂)/(M₂w₁)

This relation can be used to determine the molar mass of the solute.

8. Explain depression in freezing point and derive its relation with molality.

Answer:
The lowering of the freezing point of a solution compared with the pure solvent is called depression in freezing point.

It is represented by:

ΔT𝒇 = T𝒇° − T𝒇

For a dilute solution:

ΔT𝒇 ∝ m

Therefore,

ΔT𝒇 = K𝒇m

For w₂ g of solute of molar mass M₂ dissolved in w₁ g of solvent:

m = (1000w₂)/(M₂w₁)

Therefore,

ΔT𝒇 = K𝒇(1000w₂)/(M₂w₁)

This equation can be used to determine the molar mass of an unknown solute.

9. What is Van’t Hoff factor? Explain abnormal molar masses with the help of association and dissociation.

Answer:
The Van’t Hoff factor (i) accounts for the change in the number of solute particles due to association or dissociation.

It is defined as:

i = Observed colligative property / Calculated colligative property

It can also be expressed as:

i = Normal molar mass / Observed molar mass

Dissociation:
When a solute dissociates, the number of particles increases. Therefore:

i > 1

Example:

NaCl → Na⁺ + Cl⁻

Association:
When solute molecules associate, the number of particles decreases. Therefore:

i < 1

Example:

2CH₃COOH ⇌ (CH₃COOH)₂

Thus, association and dissociation cause the experimentally determined molar mass to differ from the normal molar mass.

10. A solution contains 5.85 g of NaCl dissolved in 500 g of water. Calculate its molality. (Molar mass of NaCl = 58.5 g mol⁻¹)

Answer:

Mass of NaCl = 5.85 g

Molar mass of NaCl = 58.5 g mol⁻¹

Moles of NaCl:

n = 5.85/58.5 = 0.1 mol

Mass of water:

500 g = 0.5 kg

Therefore,

Molality = Moles of solute / Mass of solvent in kg

m = 0.1/0.5

m = 0.2 mol kg⁻¹

Answer: 0.2 mol kg⁻¹

11. 18 g of glucose is dissolved in 180 g of water. Calculate the molality of the solution and mole fraction of glucose. (Molar mass of glucose = 180 g mol⁻¹)

Answer:

Moles of glucose:

n₁ = 18/180 = 0.1 mol

Moles of water:

n₂ = 180/18 = 10 mol

Mass of water = 180 g = 0.18 kg

Molality:

m = 0.1/0.18

m = 0.556 mol kg⁻¹

Mole fraction of glucose:

xglucose = 0.1/(0.1 + 10)

xglucose = 0.1/10.1

xglucose ≈ 0.0099

Answer: Molality = 0.556 mol kg⁻¹; Mole fraction ≈ 0.0099

12. Explain why the addition of a non-volatile solute causes lowering of vapour pressure, elevation of boiling point and depression of freezing point.

Answer:
When a non-volatile solute is added to a solvent:

The solute particles reduce the number of solvent molecules escaping into the vapour phase. Hence, the vapour pressure decreases.
Since the vapour pressure is lower, a higher temperature is required for the vapour pressure to become equal to atmospheric pressure. Hence, the boiling point increases.
The presence of solute lowers the escaping tendency of solvent molecules from the liquid phase. Therefore, the solution must be cooled to a lower temperature for freezing. Hence, the freezing point decreases.

Thus:

Vapour pressure ↓ → Boiling point ↑ → Freezing point ↓

13. Explain the effect of temperature and pressure on the solubility of gases in liquids.

Answer:

Effect of pressure:

According to Henry’s law:

p = Kₕx

At constant temperature, increasing the pressure of a gas above the solution increases its solubility in the liquid.

Effect of temperature:

The dissolution of most gases in liquids is exothermic. Therefore, according to Le Chatelier’s principle, increasing temperature decreases the solubility of gases.

Thus:

Pressure ↑ → Gas solubility ↑

Temperature ↑ → Gas solubility ↓

14. A solution is prepared by dissolving 4 g of a non-volatile solute in 100 g of water. The freezing point of the solution is −0.744°C. Calculate the molar mass of the solute. (K𝒇 for water = 1.86 K kg mol⁻¹)

Answer:

Given:

ΔT𝒇 = 0.744 K

K𝒇 = 1.86 K kg mol⁻¹

Mass of solute = 4 g

Mass of solvent = 100 g = 0.1 kg

Using:

ΔT𝒇 = K𝒇m

Therefore,

m = 0.744/1.86 = 0.4 mol kg⁻¹

Now,

m = (1000w₂)/(M₂w₁)

Therefore,

M₂ = (1000 × 4)/(0.4 × 100)

M₂ = 100 g mol⁻¹

Answer: Molar mass = 100 g mol⁻¹

15. Explain the important applications of colligative properties in everyday life.

Answer:
Colligative properties have several practical applications:

Antifreeze: Ethylene glycol is added to automobile radiators to lower the freezing point of water and increase its boiling point.
Ice cream: Salt is added to ice around the container to lower the freezing point and produce a colder mixture.
Desalination: Reverse osmosis is used to remove dissolved salts from seawater.
Molar mass determination: Osmotic pressure and other colligative properties can be used to determine the molar masses of solutes.
Medical applications: Osmotic pressure is important in maintaining the proper balance of fluids in biological systems.

Solutions — 25 MCQs for Board Exams

1. Which concentration term is independent of temperature?

A. Molarity
B. Molality
C. Normality
D. Volume percentage

Answer: B. Molality

2. The number of moles of solute present in one litre of solution is called:

A. Molality
B. Mole fraction
C. Molarity
D. Mass percentage

Answer: C. Molarity

3. Which equation represents Henry’s law?

A. π = CRT
B. ΔTᵦ = Kᵦm
C. p = Kₕx
D. ΔT𝒇 = K𝒇m

Answer: C. p = Kₕx

4. According to Raoult’s law, the partial vapour pressure of a component is:

A. pᵢ = pᵢ°/xᵢ
B. pᵢ = xᵢpᵢ°
C. pᵢ = xᵢ/pᵢ°
D. pᵢ = Kₕxᵢ

Answer: B. pᵢ = xᵢpᵢ°

5. An ideal solution obeys Raoult’s law:

A. Only at low concentration
B. Only at high concentration
C. At all concentrations
D. Only at boiling point

Answer: C. At all concentrations

6. For an ideal solution, the enthalpy of mixing is:

A. Positive
B. Negative
C. Zero
D. Infinite

Answer: C. Zero

7. Which pair forms an approximately ideal solution?

A. Ethanol + water
B. Benzene + toluene
C. Chloroform + acetone
D. HNO₃ + water

Answer: B. Benzene + toluene

8. Addition of a non-volatile solute to a solvent generally:

A. Increases vapour pressure
B. Decreases vapour pressure
C. Does not affect vapour pressure
D. Makes vapour pressure zero

Answer: B. Decreases vapour pressure

9. Which of the following is NOT a colligative property?

A. Osmotic pressure
B. Elevation in boiling point
C. Depression in freezing point
D. Viscosity

Answer: D. Viscosity

10. The elevation in boiling point is given by:

A. ΔTᵦ = Kᵦm
B. ΔTᵦ = K𝒇m
C. ΔTᵦ = CRT
D. ΔTᵦ = Kₕx

Answer: A. ΔTᵦ = Kᵦm

11. The depression in freezing point is expressed as:

A. ΔT𝒇 = Kᵦm
B. ΔT𝒇 = K𝒇m
C. ΔT𝒇 = CRT
D. ΔT𝒇 = xKₕ

Answer: B. ΔT𝒇 = K𝒇m

12. Osmotic pressure of a dilute solution is given by:

A. π = K𝒇m
B. π = Kᵦm
C. π = CRT
D. π = Kₕx

Answer: C. π = CRT

13. During osmosis, solvent molecules move from:

A. Higher solute concentration to lower solute concentration
B. Lower solute concentration to higher solute concentration
C. Higher pressure to lower pressure only
D. Solute to solvent

Answer: B. Lower solute concentration to higher solute concentration

14. A semipermeable membrane allows the passage of:

A. Solute particles only
B. Solvent molecules only
C. Both solute and solvent
D. Neither solute nor solvent

Answer: B. Solvent molecules only

15. Which colligative property is most suitable for determining the molar mass of proteins?

A. Elevation in boiling point
B. Depression in freezing point
C. Osmotic pressure
D. Relative lowering of vapour pressure

Answer: C. Osmotic pressure

16. If a solute undergoes dissociation in solution, the Van’t Hoff factor is generally:

A. i = 0
B. i < 1
C. i = 1
D. i > 1

Answer: D. i > 1

17. If solute molecules associate in solution, the Van’t Hoff factor is generally:

A. Greater than 1
B. Less than 1
C. Equal to 2
D. Equal to infinity

Answer: B. Less than 1

18. Which factor increases the solubility of a gas in a liquid?

A. Increase in temperature
B. Decrease in pressure
C. Increase in pressure
D. Addition of a non-volatile solute

Answer: C. Increase in pressure

19. The solubility of most gases in liquids generally:

A. Increases with increase in temperature
B. Decreases with increase in temperature
C. Remains constant
D. Becomes zero at high temperature

Answer: B. Decreases with increase in temperature

20. Positive deviation from Raoult’s law occurs when:

A. A–B interactions are stronger
B. A–B interactions are weaker
C. A–B interactions are equal to A–A interactions
D. No intermolecular forces exist

Answer: B. A–B interactions are weaker

21. Negative deviation from Raoult’s law generally results from:

A. Weaker A–B interactions
B. Stronger A–B interactions
C. Complete dissociation
D. Increase in temperature

Answer: B. Stronger A–B interactions

22. One mole of solute is dissolved in 9 moles of solvent. The mole fraction of solute is:

A. 0.01
B. 0.10
C. 0.90
D. 1.00

Answer: B. 0.10

23. If the Van’t Hoff factor of a solute is 2, it indicates:

A. Association of particles
B. Dissociation producing more particles
C. No change in particles
D. Complete evaporation

Answer: B. Dissociation producing more particles

24. Which quantity is used in the formula ΔTᵦ = Kᵦm?

A. Molarity
B. Molality
C. Mole fraction
D. Normality

Answer: B. Molality

25. The addition of a non-volatile solute to a solvent causes its freezing point to:

A. Increase
B. Decrease
C. Remain unchanged
D. Become equal to its boiling point

Answer: B. Decrease

Solutions — 20 True or False Questions

1. Molality is independent of temperature.
Answer: True

2. Molarity is expressed in mol kg⁻¹.
Answer: False

3. Mole fraction is a dimensionless quantity.
Answer: True

4. Henry’s law relates gas solubility to pressure.
Answer: True

5. Adding a non-volatile solute increases the vapour pressure of a solvent.
Answer: False

6. An ideal solution obeys Raoult’s law at all concentrations.
Answer: True

7. For an ideal solution, ΔHmix is zero.
Answer: True

8. Positive deviation occurs when A–B interactions are stronger than A–A and B–B interactions.
Answer: False

9. Colligative properties depend on the number of solute particles.
Answer: True

10. Osmotic pressure is a colligative property.
Answer: True

11. Addition of a non-volatile solute raises the freezing point of a solvent.
Answer: False

12. The boiling point of a solution containing a non-volatile solute is higher than that of the pure solvent.
Answer: True

13. Reverse osmosis is used for water purification.
Answer: True

14. The Van’t Hoff factor is always equal to one.
Answer: False

15. Dissociation of a solute generally gives a Van’t Hoff factor greater than one.
Answer: True

16. Association of solute particles can make the Van’t Hoff factor less than one.
Answer: True

17. Increasing pressure generally decreases the solubility of a gas in a liquid.
Answer: False

18. The solubility of most gases in liquids decreases with increase in temperature.
Answer: True

19. Osmosis involves the movement of solute particles through a semipermeable membrane.
Answer: False

20. The equation for osmotic pressure of a dilute solution is π = CRT.
Answer: True

Solutions — 10 Fill in the Blanks Questions

1. The component present in larger amount in a solution is called the ________.

Answer: Solvent

2. The number of moles of solute present in one kilogram of solvent is called ________.

Answer: Molality

3. The equation p = Kₕx represents ________ law.

Answer: Henry’s

4. The vapour pressure of a solution containing a non-volatile solute is ________ than that of the pure solvent.

Answer: Lower

5. Properties depending on the number of solute particles are called ________ properties.

Answer: Colligative

6. The pressure required to stop osmosis is called ________ pressure.

Answer: Osmotic

7. The formula for elevation in boiling point is ________.

Answer: ΔTᵦ = Kᵦm

8. The formula for depression in freezing point is ________.

Answer: ΔT𝒇 = K𝒇m

9. The Van’t Hoff factor is represented by the symbol ________.

Answer: i

10. Reverse osmosis is commonly used for the ________ of water.

Answer: Purification

Important Points – Solutions

Molarity depends on temperature; molality does not.
Mole fraction is dimensionless and the sum of all mole fractions is 1.
Henry’s law: p = Kₕx
Increasing pressure generally increases gas solubility in liquids.
Increasing temperature generally decreases gas solubility.
Raoult’s law: pᵢ = xᵢpᵢ°
An ideal solution obeys Raoult’s law at all concentrations.
For an ideal solution: ΔHmix = 0 and ΔVmix = 0.
Positive deviation: vapour pressure is higher than expected.
Negative deviation: vapour pressure is lower than expected.
Colligative properties depend on the number of solute particles.
Four colligative properties: relative lowering of vapour pressure, elevation in boiling point, depression in freezing point and osmotic pressure.
Elevation in boiling point: ΔTᵦ = Kᵦm
Depression in freezing point: ΔT𝒇 = K𝒇m
Osmotic pressure: π = CRT
Reverse osmosis is used for water purification and desalination.
Van’t Hoff factor (i) accounts for association and dissociation.
Dissociation → i > 1
Association → i < 1
For no association or dissociation, i = 1.
Always pay attention to units, molar mass and conversion of grams into kilograms in numerical problems.

Solutions – Complete Formula Sheet

1. Concentration of Solutions

Mass Percentage

$\text{Mass \%}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times100$

If mass of solvent = $w_1$ g and mass of solute = $w_2$ g: $\text{Mass \%}=\frac{w_2}{w_1+w_2}\times100$

Volume Percentage

$\text{Volume \%}=\frac{\text{Volume of solute}}{\text{Volume of solution}}\times100$

Mass by Volume Percentage

$\text{Mass by Volume \%}= \frac{\text{Mass of solute in g}}{\text{Volume of solution in mL}}\times100$

Parts per Million (ppm)

$\text{ppm}= \frac{\text{Number of parts of component}} {\text{Total number of parts of solution}} \times10^6$

2. Mole Fraction

For a binary solution containing components 1 and 2: $x_1=\frac{n_1}{n_1+n_2}$ $x_2=\frac{n_2}{n_1+n_2}$

Therefore, $x_1+x_2=1$

For any component $i$ : $x_i=\frac{n_i}{\sum n_i}$

3. Molarity

Molarity is defined as the number of moles of solute present in one litre of solution. $M=\frac{n_2}{V}$

Since, $n_2=\frac{w_2}{M_2}$

Therefore, $M=\frac{w_2}{M_2V}$

If volume is given in mL: $M=\frac{w_2\times1000}{M_2V_{\mathrm{mL}}}$

4. Molality

Molality is defined as the number of moles of solute present in one kilogram of solvent. $m=\frac{n_2}{w_1(\mathrm{kg})}$

Since, $n_2=\frac{w_2}{M_2}$

and solvent mass in kg is: $w_1(\mathrm{kg})=\frac{w_1}{1000}$

Therefore, $m=\frac{w_2\times1000}{M_2w_1}$

5. Raoult’s Law

For a solution containing volatile components 1 and 2: $p_1=x_1p_1^\circ$ $p_2=x_2p_2^\circ$

The total vapour pressure is: $P=p_1+p_2$

Therefore, $P=x_1p_1^\circ+x_2p_2^\circ$

Since, $x_2=1-x_1$

we get: $P=p_2^\circ+(p_1^\circ-p_2^\circ)x_1$

6. Solution Containing a Non-Volatile Solute

For a solution containing a non-volatile solute: $p_1=x_1p_1^\circ$

Since, $x_1=1-x_2$

Therefore, $p_1=(1-x_2)p_1^\circ$

Hence, $p_1^\circ-p_1=x_2p_1^\circ$

Therefore, $\frac{p_1^\circ-p_1}{p_1^\circ}=x_2$

7. Relative Lowering of Vapour Pressure

The relative lowering of vapour pressure is: $\frac{p_1^\circ-p_1}{p_1^\circ}=x_2$

For a dilute solution: $x_2\approx\frac{n_2}{n_1}$

Therefore, $\frac{p_1^\circ-p_1}{p_1^\circ} \approx\frac{n_2}{n_1}$

Since, $n_1=\frac{w_1}{M_1}$

and $n_2=\frac{w_2}{M_2}$

we get: $\frac{p_1^\circ-p_1}{p_1^\circ} \approx \frac{w_2M_1}{w_1M_2}$

Hence, molar mass of solute: $M_2= \frac{w_2M_1p_1^\circ} {w_1(p_1^\circ-p_1)}$

8. Elevation in Boiling Point

The elevation in boiling point is: $\Delta T_b=T_b-T_b^\circ$

For a dilute solution: $\Delta T_b=K_bm$

Since, $m=\frac{w_2\times1000}{M_2w_1}$

Therefore, $\Delta T_b= K_b\frac{w_2\times1000}{M_2w_1}$

Hence, $M_2= \frac{K_bw_2\times1000} {\Delta T_bw_1}$

9. Depression in Freezing Point

The depression in freezing point is: $\Delta T_f=T_f^\circ-T_f$

For a dilute solution: $\Delta T_f=K_fm$

Since, $m=\frac{w_2\times1000}{M_2w_1}$

Therefore, $\Delta T_f= K_f\frac{w_2\times1000}{M_2w_1}$

Hence, $M_2= \frac{K_fw_2\times1000} {\Delta T_fw_1}$

10. Osmotic Pressure

Osmotic pressure is represented by $\pi$ .

For a dilute solution: $\pi=CRT$

Since, $C=\frac{n_2}{V}$

Therefore, $\pi=\frac{n_2RT}{V}$

Since, $n_2=\frac{w_2}{M_2}$

Therefore, $\pi=\frac{w_2RT}{M_2V}$

Hence, $M_2=\frac{w_2RT}{\pi V}$

11. Van’t Hoff Factor

The Van’t Hoff factor is represented by $i$ . $i= \frac{\text{Observed colligative property}} {\text{Calculated colligative property}}$

It can also be written as: $i= \frac{\text{Normal molar mass}} {\text{Observed molar mass}}$

Therefore, $\text{Observed molar mass} = \frac{\text{Normal molar mass}}{i}$

12. Modified Colligative Property Equations

For abnormal molar masses: $\pi=iCRT$ $\Delta T_b=iK_bm$ $\Delta T_f=iK_fm$

For relative lowering of vapour pressure: $\frac{p_1^\circ-p_1}{p_1^\circ}=ix_2$

13. Van’t Hoff Factor for Dissociation

Suppose one molecule dissociates into $n$ particles. $AB\rightarrow A+B$

If the degree of dissociation is $\alpha$ , then: $i=1+(n-1)\alpha$

Therefore, $\alpha=\frac{i-1}{n-1}$

For complete dissociation: $\alpha=1$

Therefore, $i=n$

14. Van’t Hoff Factor for Association

Suppose $n$ molecules associate to form one molecule: $nA\rightarrow A_n$

If the degree of association is $\alpha$ : $i=1-\alpha+\frac{\alpha}{n}$

or, $i=1-\left(1-\frac{1}{n}\right)\alpha$

Therefore, $\alpha= \frac{1-i}{1-\frac{1}{n}}$

15. Important Relations for Board Numericals

Molarity

$M=\frac{w_2\times1000}{M_2V_{\mathrm{mL}}}$

Molality

$m=\frac{w_2\times1000}{M_2w_1}$

Elevation in Boiling Point

$\Delta T_b=iK_bm$

Depression in Freezing Point

$\Delta T_f=iK_fm$

Osmotic Pressure

$\pi=iCRT$

Molar Mass from Osmotic Pressure

$M_2=\frac{w_2RT}{\pi V}$

Molar Mass from Elevation in Boiling Point

$M_2= \frac{iK_bw_2\times1000} {\Delta T_bw_1}$

Molar Mass from Depression in Freezing Point

$M_2= \frac{iK_fw_2\times1000} {\Delta T_fw_1}$