/CBSE Class 12 Chemistry (RT) Question Paper - Chapter: Solutions

Time: 1 Hour 30 Minutes

Maximum Marks: 50

Section A: Multiple Choice Questions (1 × 11 = 11 Marks)

Q1. Which of the following is a colligative property?

a) Surface tension
b) Vapour pressure
c) Boiling point
d) Osmotic pressure
Answer: d) Osmotic pressure

Q2. What is the molality of a solution containing 0.5 mol of solute in 1 kg of water?
a) 0.05 m
b) 0.5 m
c) 1.0 m
d) 5.0 m
Answer: b) 0.5 m

Q3. According to Raoult’s law, the relative lowering of vapour pressure is equal to the:
a) Mole fraction of solute
b) Mole fraction of solvent
c) Mass fraction of solute
d) Mass fraction of solvent
Answer: a) Mole fraction of solute

Q4. Which of the following will show the highest depression in freezing point for the same
concentration?
a) Glucose
b) NaCl
c) K₂SO₄
d) Urea
Answer: c) K₂SO₄

Q5. Which of the following is not a colligative property?

a) Relative lowering of vapour pressure
b) Elevation of boiling point
c) Depression of freezing point
d) Heat of vaporization
Answer: d) Heat of vaporization
Q6. The van’t Hoff factor (i) for NaCl in water is:
a) 1
b) 2
c) 3
d) 4
Answer: b) 2

Q7. A solution obeying Raoult’s law is called:

a) Dilute solution
b) Ideal solution
c) Saturated solution
d) Non-ideal solution
Answer: b) Ideal solution

Q8. What is the S.I. unit of molality?

a) mol/kg
b) mol/L
c) mol/m³
d) mol
Answer: a) mol/kg

Q9. Which of the following solutions will have the highest boiling point?
a) 1 M glucose
b) 1 M NaCl
c) 1 M KCl
d) 1 M MgCl₂
Answer: d) 1 M MgCl₂

Q10. The boiling point of a solution is always:

a) Equal to pure solvent
b) Lower than pure solvent
c) Higher than pure solvent
d) Same as freezing point
Answer: c) Higher than pure solvent

Q11. The depression in freezing point is directly proportional to:

a) Molality of the solution
b) Molarity of the solute
c) Volume of solvent
d) Temperature of the solution
Answer: a) Molality of the solution
 Section B: Assertion-Reason (1 × 2 = 2 Marks)

Q12.
Assertion (A): Addition of a non-volatile solute lowers the vapour pressure of the solution.
Reason (R): The solute particles occupy the surface of liquid and reduce the escaping
tendency of solvent molecules.
a) Both A and R are true, and R is the correct explanation.
b) Both A and R are true, but R is not the correct explanation.
c) A is true, R is false.
d) A is false, R is true.
Answer: a) Both A and R are true, and R is the correct explanation.

Q13.
Assertion (A): In highly concentrated solutions, van’t Hoff factor deviates from expected
value.
Reason (R): Electrolytes associate at higher concentration.
a) Both A and R are true, and R is the correct explanation.
b) Both A and R are true, but R is not the correct explanation.
c) A is true, R is false.
d) A is false, R is true.
Answer: c) A is true, R is false.

Section C: Case-Based Questions (2 × 4 = 8 Marks)

Q14.

A student prepares a 0.1 molal solution of urea in water. He observes that the freezing point
of the solution is lower than that of pure water. This behavior is attributed to colligative
properties, which depend on the number of solute particles present, not their identity. Urea,
being a non-electrolyte, does not dissociate in water.

(a) What is the colligative property responsible for the observed decrease in freezing point?
(1 mark)
Answer: Depression in freezing point

(b) Write the formula used to calculate the depression in freezing point and define each
term. (2 marks)
Answer: ΔTf=i⋅Kf⋅m\Delta T_f = i \cdot K_f \cdot m, where ΔTf\Delta T_f = freezing point
depression, ii = van’t Hoff factor, KfK_f = cryoscopic constant, mm = molality

(c) Why does urea, a non-electrolyte, have a lesser effect on freezing point compared to an
electrolyte at the same molality? (1 mark)
Answer: Urea does not dissociate; hence, it contributes fewer particles to the solution.
Q15.

Raoult’s law explains how the addition of a non-volatile solute to a solvent results in
lowering of the vapour pressure of the solvent. This happens because fewer solvent
molecules are present at the surface to escape into the vapour phase. The extent of vapour
pressure lowering is proportional to the mole fraction of the solute.

(a) State Raoult’s law for a solution containing a non-volatile solute. (1 mark)
Answer: Vapour pressure of solution = mole fraction of solvent × vapour pressure of pure
solvent

(b) Derive the relation between relative lowering of vapour pressure and mole fraction of
the solute. (2 marks)
Answer: P0−PP0=Xsolute\frac{P^0 - P}{P^0} = X_{solute}

(c) Why does addition of a volatile solute not always lower the total vapour pressure? (1
mark)
Answer: A volatile solute adds its own vapour pressure to the solution, possibly increasing
the total pressure.

Section D: Short Answer Questions – Type I (3 × 2 = 6 Marks)

Q16. What is the effect of temperature on molarity? Explain briefly.

Answer: Molarity decreases with increase in temperature because volume expands with
temperature.

Q17. Define ideal and non-ideal solutions with one example each.
Answer: Ideal: Obeys Raoult’s law (e.g., benzene + toluene). Non-ideal: Deviates from
Raoult’s law (e.g., acetone + chloroform).

Q18. Calculate the mass of NaCl (i = 2) required to be dissolved in 100 g of water to lower
the freezing point by 3.72°C. (Kf = 1.86 K kg/mol)
Answer:
3.72=2×1.86×w58.5×0.1⇒w=10.9 g3.72 = 2 \times 1.86 \times \frac{w}{58.5 \times 0.1}
\Rightarrow w = 10.9\text{ g}

Section E: Short Answer Questions – Type II (2 × 3 = 6 Marks)

Q19. Define abnormal molar mass. How is van’t Hoff factor used to correct it?
Answer: Abnormal molar mass arises due to association/dissociation. Van’t Hoff factor
i=Observed propertyCalculated propertyi = \frac{\text{Observed property}}{\text{Calculated
property}}, and M=Mnormal/iM = M_{normal}/i

Q20. Calculate the boiling point of a solution containing 1.8 g glucose in 100 g of water. (Kb =
0.52 K kg/mol; M = 180 g/mol)
Answer:
Moles = 1.8/180 = 0.01 mol; molality = 0.1 m; ΔTb=0.052\Delta T_b = 0.052 K;
Boiling point = 100 + 0.052 = 100.05°C

Q21. Differentiate between molarity and molality. Why is molality preferred over molarity?
Answer: Molarity = mol/L, temp. dependent; Molality = mol/kg, temp. independent →
molality is preferred.

Section F: Long Answer Questions (2 × 6 = 12 Marks)

Q22.
(a) What are azeotropes? Explain minimum and maximum boiling azeotropes with examples.
(3 marks)
Answer: Azeotropes: Constant boiling mixtures. Minimum: Ethanol-water. Maximum: HNO₃-
water.

(b) 5.85 g NaCl in 100 g water. Find relative lowering in vapour pressure (P₀ = 23.8 mm Hg).
(3 marks)
Answer:
Moles NaCl = 0.1 mol; Moles water = 5.56 mol;
Mole fraction = 0.0177; ΔP=0.0177×23.8=0.421\Delta P = 0.0177 \times 23.8 = 0.421 mm Hg

Q23.
(a) Explain how elevation in boiling point is a colligative property. Derive the formula. (3
marks)
Answer: Depends on number of solute particles: ΔTb=iKbm\Delta T_b = iK_bm

(b) 2 g benzoic acid in 25 g benzene. ΔTf = 1.62 K, Kf = 4.9. Find molar mass. (3 marks)
Answer:
Molality = 0.33; Moles = 0.00825; Molar mass ≈ 242.4 g/mol (higher than actual ⇒
association)

End of Question Paper