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Vapour pressure and Raoult’s law: NEET MCQs with answers

65 questions on this topic of Solutions (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. At 300 K, the vapour pressures of pure benzene (C₆H₆) and pure toluene (C₇H₈) are 50 mm Hg and 20 mm Hg, respectively. What is the total vapour pressure of an equimolar solution of these two volatile liquids?

  1. 70 mm Hg
  2. 25 mm Hg
  3. 30 mm Hg
  4. 35 mm Hg
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Answer: (D) 35 mm Hg

For an equimolar solution, x_benzene = x_toluene = 0.5. Total pressure P_total = p_benzene + p_toluene = (0.5 × 50) + (0.5 × 20) = 25 + 10 = 35 mm Hg.

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2. The vapour pressure of pure water at a certain temperature is 40 mm Hg. For an ideal solution in which the mole fraction of a non-volatile solute is 0.1, the vapour pressure is:

  1. 44 mm Hg
  2. 36 mm Hg
  3. 4 mm Hg
  4. 40 mm Hg
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Answer: (B) 36 mm Hg

Raoult's law: p = x_solvent p° = (1 − 0.1) × 40 = 36 mm Hg.

hard

3. An ideal solution is formed by mixing two volatile liquids C and D. At a certain temperature, the vapour pressure of pure C is 200 mm Hg and that of pure D is 500 mm Hg. If the vapour phase contains 40 mol% of C, what is the mole fraction of C in the liquid phase?

  1. 0.250
  2. 0.500
  3. 0.400
  4. 0.625
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Answer: (D) 0.625

Given y_C = 0.40, so y_D = 0.60. Using (y_C / y_D) = (p_C° x_C) / (p_D° x_D): 0.40 / 0.60 = (200 / 500) × (x_C / (1 - x_C)). This simplifies to 2 / 3 = (2 / 5) × [x_C / (1 - x_C)], so 5 / 3 = x_C / (1 - x_C). Solving gives 5(1 - x_C) = 3 x_C, leading to 8 x_C = 5, which means x_C = 5 / 8 = 0.625.

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4. At a given temperature, an ideal solution of two volatile components X and Y has mole fraction of X in the liquid phase equal to 0.4. If the vapour pressures of pure X and pure Y are 200 mm Hg and 100 mm Hg respectively, what is the mole fraction of X in the vapour phase (y_X) in equilibrium with this solution?

  1. 0.40
  2. 0.57
  3. 0.72
  4. 0.28
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Answer: (B) 0.57

p_X = x_X · p°_X = 0.4 × 200 = 80 mm Hg. p_Y = x_Y · p°_Y = 0.6 × 100 = 60 mm Hg. P_total = 80 + 60 = 140 mm Hg. The mole fraction in the vapour phase is y_X = p_X / P_total = 80 / 140 ≈ 0.571.

easy

5. At 298 K, a solution is prepared by mixing 2 moles of heptane (vapour pressure of pure heptane = 105 kPa) and 3 moles of octane (vapour pressure of pure octane = 45 kPa). Assuming an ideal solution, the total vapour pressure of the mixture is:

  1. 69 kPa
  2. 75 kPa
  3. 60 kPa
  4. 81 kPa
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Answer: (A) 69 kPa

Total moles = 2 + 3 = 5. Mole fraction of heptane x_H = 2/5 = 0.4; mole fraction of octane x_O = 3/5 = 0.6. P_total = (0.4 × 105) + (0.6 × 45) = 42 + 27 = 69 kPa.

hard

6. Two volatile liquids A and B form an ideal solution. If the mole fraction of A in the vapour phase is denoted by y_A and in the liquid phase by x_A, which of the following expressions is strictly correct?

  1. 1 / y_A = (p°_B / p°_A) · (1 / x_A - 1) + 1
  2. 1 / y_A = (p°_A / p°_B) · (1 / x_A) + 1
  3. y_A / (1 - y_A) = (p°_B / p°_A) · [x_A / (1 - x_A)]
  4. 1 / P_total = x_A / p°_A + (1 - x_A) / p°_B
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Answer: (A) 1 / y_A = (p°_B / p°_A) · (1 / x_A - 1) + 1

We know y_A = p_A / P_total = (x_A · p°_A) / [x_A · p°_A + (1 - x_A)p°_B]. Taking the reciprocal: 1 / y_A = 1 + [(1 - x_A)p°_B] / [x_A · p°_A] = 1 + (p°_B / p°_A)(1/x_A - 1). This is a standard linear relation between 1/y_A and 1/x_A.

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7. The vapour pressure of pure water at a certain temperature is 40 mm Hg. For an ideal solution in which the mole fraction of a non-volatile solute is 0.05, the vapour pressure is:

  1. 2 mm Hg
  2. 38 mm Hg
  3. 42 mm Hg
  4. 40 mm Hg
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Answer: (B) 38 mm Hg

Raoult's law: p = x_solvent p° = (1 − 0.05) × 40 = 38 mm Hg.

easy

8. At a given temperature, 1 mole of liquid A (p°_A = 400 mm Hg) and 3 moles of liquid B (p°_B = 200 mm Hg) are mixed to form an ideal solution. What is the mole fraction of component A in the vapour phase in equilibrium with this solution?

  1. 0.40
  2. 0.25
  3. 0.50
  4. 0.67
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Answer: (A) 0.40

Mole fraction of A in liquid phase is x_A = 1/(1+3) = 0.25, and x_B = 0.75. Partial pressures are p_A = 0.25 × 400 = 100 mm Hg and p_B = 0.75 × 200 = 150 mm Hg. The total pressure is P_total = 100 + 150 = 250 mm Hg. Therefore, y_A = p_A / P_total = 100 / 250 = 0.40.

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9. The vapour pressure of pure water at a certain temperature is 31.8 mm Hg. For an ideal solution in which the mole fraction of a non-volatile solute is 0.05, the vapour pressure is:

  1. 1.59 mm Hg
  2. 33.4 mm Hg
  3. 30.2 mm Hg
  4. 31.8 mm Hg
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Answer: (C) 30.2 mm Hg

Raoult's law: p = x_solvent p° = (1 − 0.05) × 31.8 = 30.2 mm Hg.

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10. The vapour pressure of pure water at a certain temperature is 100 mm Hg. For an ideal solution in which the mole fraction of a non-volatile solute is 0.02, the vapour pressure is:

  1. 2 mm Hg
  2. 102 mm Hg
  3. 98 mm Hg
  4. 100 mm Hg
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Answer: (C) 98 mm Hg

Raoult's law: p = x_solvent p° = (1 − 0.02) × 100 = 98 mm Hg.

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11. At 300 K, a binary liquid solution containing 1 mole of component A and 4 moles of component B has a total vapour pressure of 560 mm Hg. At the same temperature, the vapour pressure of pure component B is 500 mm Hg. Assuming ideal behavior, the vapour pressure of pure component A (p°_A) is:

  1. 700 mm Hg
  2. 600 mm Hg
  3. 800 mm Hg
  4. 750 mm Hg
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Answer: (C) 800 mm Hg

Mole fraction of A, x_A = 1/(1 + 4) = 0.20, and x_B = 0.80. P_total = x_A p°_A + x_B p°_B. Therefore, 560 = 0.20 × p°_A + 0.80 × 500 = 0.20 p°_A + 400. This gives 0.20 p°_A = 160, so p°_A = 160 / 0.20 = 800 mm Hg.

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12. An equimolar liquid mixture of two volatile liquids X and Y is prepared. At a given temperature, the vapour pressure of pure X is 240 mm Hg and that of pure Y is 80 mm Hg. Assuming the solution behaves ideally, what is the mole fraction of X in the vapour phase at equilibrium?

  1. 0.75
  2. 0.50
  3. 0.67
  4. 0.25
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Answer: (A) 0.75

Since the liquid mixture is equimolar, x_X = 0.5 and x_Y = 0.5. The partial vapour pressures are P_X = 240 × 0.5 = 120 mm Hg and P_Y = 80 × 0.5 = 40 mm Hg. The total vapour pressure P_total = 120 + 40 = 160 mm Hg. The mole fraction of X in the vapour phase is y_X = P_X / P_total = 120 / 160 = 0.75.

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