1)

160 g of non-volatile solute ‘A’ is dissolved in 54 mL of water at 373 K. What is the vapour pressure of aqueous solution of A? (Given, molecular weight of A = 160 g mol⁻¹)
a) 760 Torr
b) 720 Torr
c) 570 Torr
d) 450 Torr
Answer: c

2)

Assertion: Standard boiling point of a liquid is slightly higher than the normal boiling point.
Reason: 1 bar pressure is slightly less than 1 atm pressure.
a) Assertion and Reason are correct statements and reason is the correct explanation for Assertion.
b) Assertion and Reason are correct statements and Reason is not the explanation for Assertion.
c) Assertion is correct, Reason is incorrect.
d) Assertion is incorrect, Reason is correct.
Answer: d

3)

Match Column-I (method) and Column-II (colligative property)
a) (A-II), (B-IV), (C-III), (D-I)
b) (A-I), (B-IV), (C-II), (D-III)
c) (A-II), (B-III), (C-IV), (D-I)
d) (A-III), (B-IV), (C-I), (D-II)
Answer: d

4)

What is the density of water vapour at boiling point of water?
a) 1 × 10⁻⁴ g cm⁻³
b) 1 g cm⁻³
c) 6 × 10⁻⁴ g cm⁻³
d) 4 × 10⁻⁴ g cm⁻³
Answer: c

5)

A non-volatile solute ‘A’ tetramerises in water to the extent of 80%. 2.5 g of ‘A’ in 100 g of water lowers freezing point by 0.3°C. Molar mass of ‘A’ is (Kf = 1.86)
a) 62
b) 221
c) 155
d) 354
Answer: a

6)

Pure vapour pressures: A = 400 mmHg, B = 600 mmHg at 298 K. Ideal mixing with XB(liquid)=0.5. Vapour pressure and yA,yB are:
a) 480 mmHg, 0.4, 0.6
b) 500 mmHg, 0.5, 0.5
c) 450 mmHg, 0.3, 0.5
d) 500 mmHg, 0.4, 0.6
Answer: d

7)

For 1 m glucose, ΔTb = 2 K. For 2 m glucose, ΔTf = 2 K. Relation between Kb and Kf is:
a) Kb = 1.5 Kf
b) Kb = 0.5 Kf
c) Kb = Kf
d) Kb = 2Kf
Answer: d

8)

Freezing point of 4% aqueous solution of X equals freezing point of 12% aqueous solution of Y. If molar mass of X is A, molar mass of Y is:
a) 3A
b) A
c) 2A
d) 4A
Answer: a

9)

1 g solute dissolved in 100 g of solvents A and B. Ebullioscopic constants ratio KA:KB = 1:5. Ratio ΔTb(A)/ΔTb(B) is:
a) 5:1
b) 10:1
c) 1:5
d) 1:0.2
Answer: c

10)

Freezing point of diluted milk is -0.2°C, pure milk should be -0.5°C. How much water is added?
a) 2 cups water : 3 cups pure milk
b) 1 cup water : 3 cups pure milk
c) 3 cups water : 2 cups pure milk
d) 1 cup water : 2 cups pure milk
Answer: c

11)

Benzoic acid dimerises in benzene. w g in 30 g benzene gives ΔTf = 2 K. % association = 80. Find w (Kf=5, M=122).
a) 1.8 g
b) 1.0 g
c) 2.4 g
d) 1.5 g
Answer: c

12)

If molarity = 0.05 and elevation in boiling point = 0.16 K, molal elevation constant is:
a) 3.2
b) 1.6
c) 2.2
d) 2.3
Answer: a

13)

Depression in freezing point of 0.1 M solutions of HCl, CuSO₄ and K₂SO₄ are in ratio:
a) 1 : 1 : 1
b) 1 : 2 : 3
c) 1 : 1 : 1.5
d) 2 : 4 : 3
Answer: c

14)

w₂ g of solute (M₂) in w₁ g solvent gives depression ΔTf. Expression is:
a) ΔTf = (Kf × w₂ × 1000)/(M₂ × w₁)
b) ΔTf = (Kf × w₂ × 100)/(M₂ × w₁)
c) ΔTf = (Kf × w₁ × 1000)/(M₂ × w₂)
d) ΔTf = (Kf × w₁ × 100)/(M₂ × w₂)
Answer: a

15)

Assertion: Ideal solutions obey Raoult’s law.
Reason: ΔHmix and ΔVmix for ideal solution are less than zero.
a) Both correct and reason explains
b) Both correct but reason doesn’t explain
c) Assertion correct, Reason incorrect
d) Both incorrect
Answer: c

16)

Vapour pressure of CCl₄ at 25°C is 143 mmHg. 0.5 g solute (M=65) in 100 mL CCl₄. Find vapour pressure. (Density CCl₄=1.58 g/cm³)
a) 141.93 mm Hg
b) 94.39 mm Hg
c) 199.34 mm Hg
d) 143.99 mm Hg
Answer: a

17)

0.5 m aqueous weak acid HX is 20% ionized. Kf = 1.86. Lowering in freezing point is:
a) -1.12 K
b) 0.56 K
c) 1.12 K
d) -0.56 K
Answer: c

18)

Which solution shows maximum elevation in boiling point?
a) 0.1 m NaCl
b) 0.1 m FeCl₃
c) None of these
d) 0.1 m CaCl₂
Answer: b

19)

Correct relation between elevation in boiling point and molar mass of solute is:
a) M₂ = (Kb · w₂)/(ΔTb · w₁)
b) M₂ = (Kb · w₁)/(ΔTb · w₂)
c) M₂ = (Kb · ΔTb)/(w₂ · w₁)
d) M₂ = (w₁ · ΔTb)/(Kb · w₂)
Answer: a

20)

For 1 molal aqueous solution, which compound shows the highest freezing point?
a) [Co(H₂O)₆]Cl₃
b) [Co(H₂O)₄Cl₂]Cl·2H₂O
c) [Co(H₂O)₃Cl₃]·3H₂O
d) [Co(H₂O)₅]Cl₂·H₂O
Answer: c

21)

Ethylene glycol used as antifreeze to reduce freezing point to -2.4°C. Mass required for 2 L water? (Kf=1.86)
a) 16 kg
b) 160 g
c) 1.60 kg
d) 16 g
Answer: b

22)

Freezing point of 0.4 m weak monoprotic acid solution is -0.1°C. van’t Hoff factor i? (Kf=1.86)
a) 1.6
b) 1.2
c) 1.34
d) 1.86
Answer: c

23)

45 g solute in 600 g water lowers freezing point by 2.2 K. Find molar mass (Kf=1.86).
a) 63.4 g/mol
b) 80 g/mol
c) 90 g/mol
d) 21 g/mol
Answer: a

24)

Assertion: Adding non-volatile solute decreases freezing point.
Reason: Vapour pressure decreases, so equality of VP(solid) and VP(liquid) occurs at lower T.
a) Both correct and reason explains
b) Both correct but reason doesn’t explain
c) Assertion correct, reason incorrect
d) Both incorrect
Answer: a

25)

If molality of a dilute solution is doubled, Kf will be:
a) halved
b) tripled
c) unchanged
d) doubled
Answer: c

26)

Freezing point of benzene decreases by 0.45°C when 0.2 g acetic acid is added to 20 g benzene. If acetic acid dimerizes, % association is (Kf=5.12):
a) 64.6%
b) 80.4%
c) 74.6%
d) 94.6%
Answer: d

27)

Colligative properties of dilute solution depend upon:
a) nature of solute
b) diffusion of solvent
c) number of solute particles
d) number of solvent particles
Answer: c

28)

When pure solvent diffuses out of solution through a semipermeable membrane, the process is called:
a) sorption
b) dialysis
c) reverse osmosis
d) osmosis
Answer: c

29)

Freezing point of equimolal solution will be highest for:
a) C₆H₅NH₃Cl
b) AgNO₃
c) Ca(NO₃)₂
d) D-fructose
Answer: d

30)

Raoult’s law becomes a special case of Henry’s law when:
a) Kh = P₁⁰
b) Kh > P₁⁰
c) Kh < P₁⁰
d) Kh ≥ P₁⁰
Answer: a

Conclusion on Benzoic Acid Assertion Reason Questions