Vapour Pressure of Solutions, Henry's Law and Raoult's Law – the vapour pressure of solutions and Raoult's Law: NEET MCQs with Answers

14 free questions on Vapour Pressure of Solutions, Henry's Law and Raoult's Law – the vapour pressure of solutions and Raoult's Law for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: the vapour pressure of solutions and Raoult's Law - Ideal and non-ideal solutions, vapour pressure - composition, plots for ideal and non-ideal solutions;

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Question 1

Previous-year question
The mixture which shows positive deviation from Raoult's law is [NEET 2020]
  1. Benzene + Toluene
  2. Acetone + Chloroform
  3. Chloroethane + Bromoethane
  4. Ethanol + Acetone
Show answer
Answer: (D) Ethanol + Acetone
Explanation: Acetone breaks the hydrogen bonds between the ethanol molecules, so both components escape more readily and the observed vapour pressure exceeds that predicted by Raoult's law.

Question 2

Previous-year question
For an ideal solution, the correct option is [NEET 2019]
  1. Δ(mix)S = 0 at constant T and P
  2. Δ(mix)V ≠ 0 at constant T and P
  3. Δ(mix)H = 0 at constant T and P
  4. Δ(mix)G = 0 at constant T and P
Show answer
Answer: (C) Δ(mix)H = 0 at constant T and P
Explanation: An ideal solution forms with no change in enthalpy because the A–B interactions equal the A–A and B–B interactions. The volume change is also zero, while ΔS is positive and ΔG is negative.

Question 3

Previous-year question
Which of the following statements about the composition of the vapour over an ideal 1 : 1 molar mixture of benzene and toluene is correct? Assume that the temperature is constant at 25 °C. (Vapour pressure data at 25 °C: benzene = 12.8 kPa, toluene = 3.85 kPa) [NEET 2016]
  1. The vapour will contain a higher percentage of benzene
  2. The vapour will contain a higher percentage of toluene
  3. The vapour will contain equal amounts of benzene and toluene
  4. Not enough information is given to make a prediction
Show answer
Answer: (A) The vapour will contain a higher percentage of benzene
Explanation: The partial pressures are 6.4 kPa for benzene and 1.925 kPa for toluene, so benzene contributes about 77% of the total. The vapour is always richer in the more volatile component.

Question 4

A volatile liquid A has a vapour pressure of 240 mm Hg in the pure state at a certain temperature. It forms an ideal solution in which its mole fraction is 0.25. If the partial vapour pressure of A is written in the Henry's law form p_A = K_H x_A, what are the values of K_H and p_A?
  1. K_H = 960 mm Hg, p_A = 240 mm Hg
  2. K_H = 240 mm Hg, p_A = 180 mm Hg
  3. K_H = 60 mm Hg, p_A = 15 mm Hg
  4. K_H = 240 mm Hg, p_A = 60 mm Hg
Show answer
Answer: (D) K_H = 240 mm Hg, p_A = 60 mm Hg
Explanation: Raoult's law gives p_A = p°_A x_A and Henry's law gives p_A = K_H x_A, so both make the partial pressure directly proportional to the mole fraction. Raoult's law is thus a special case of Henry's law in which K_H equals p°_A, here 240 mm Hg. The partial pressure is 240 × 0.25 = 60 mm Hg.

Question 5

Phenol and aniline each show intermolecular hydrogen bonding in the pure liquid state. Which statement correctly describes a mixture of the two with respect to Raoult's law?
  1. No deviation, because both liquids are aromatic and their molecules are of similar size
  2. Positive deviation, because mixing breaks the hydrogen bonds between like molecules and nothing replaces them
  3. Negative deviation, because the hydrogen bond between the phenolic proton and the lone pair on nitrogen of aniline is stronger than those between like molecules
  4. Positive deviation, because the hydrogen bond between the phenolic proton and nitrogen increases the escaping tendency of both components
Show answer
Answer: (C) Negative deviation, because the hydrogen bond between the phenolic proton and the lone pair on nitrogen of aniline is stronger than those between like molecules
Explanation: When A–B attractions are stronger than A–A and B–B attractions, the escaping tendency of the molecules falls and the vapour pressure is lower than Raoult's law predicts. In the phenol–aniline mixture, the phenolic O–H proton forms a hydrogen bond with the lone pair on the nitrogen of aniline, and this bond is stronger than the hydrogen bonds between similar molecules. The mixture therefore shows negative deviation, like chloroform and acetone.

Question 6

The vapour pressure of a solvent falls when a non-volatile solute is dissolved in it at the same temperature. What is the reason for this at the molecular level?
  1. Solute molecules escape into the vapour along with the solvent and condense the solvent vapour
  2. The solute raises the temperature of the solvent, which lowers its tendency to vaporise
  3. The solute changes the chemical identity of the solvent molecules, which then cannot vaporise
  4. Solute molecules at the surface reduce the fraction of the surface covered by solvent molecules, so fewer solvent molecules escape
Show answer
Answer: (D) Solute molecules at the surface reduce the fraction of the surface covered by solvent molecules, so fewer solvent molecules escape
Explanation: In a pure liquid the whole surface is occupied by solvent molecules. In a solution the surface holds both solute and solvent molecules, so the fraction of it covered by solvent is reduced. Fewer solvent molecules can then escape into the vapour, and since a non-volatile solute does not itself contribute to the vapour, the vapour pressure is lower.

Question 7

Pure liquids P and Q have vapour pressures of 120 torr and 200 torr at a certain temperature. A mixture of 3 mol P and 1 mol Q has a measured vapour pressure of 160 torr at this temperature. Which conclusion is correct?
  1. The mixture is ideal, with Δ_mix H = 0
  2. It shows positive deviation, and heat is released on mixing
  3. It shows negative deviation, and P–Q attractions are stronger than the average of P–P and Q–Q attractions
  4. It shows positive deviation, and P–Q attractions are weaker than the average of P–P and Q–Q attractions
Show answer
Answer: (D) It shows positive deviation, and P–Q attractions are weaker than the average of P–P and Q–Q attractions
Explanation: Raoult's law predicts 0.75 × 120 + 0.25 × 200 = 140 torr for this mixture. The measured 160 torr is higher, so the deviation is positive. More molecules escape into the vapour because the P–Q attractions are weaker than those in the pure liquids. Such mixing absorbs heat, so Δ_mix H is positive.
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Question 8

Henry's law constant of oxygen in water is taken as 35 kbar at 20 °C and 47 kbar at 30 °C. For water in contact with oxygen at a partial pressure of 0.21 bar, what happens to the mole fraction of dissolved oxygen on warming from 20 °C to 30 °C?
  1. It falls from about 6.0 × 10⁻³ to about 4.5 × 10⁻³, so warm water holds less oxygen for aquatic animals
  2. It rises from about 4.5 × 10⁻⁶ to about 6.0 × 10⁻⁶, so warm water supports aquatic animals better
  3. It falls from about 6.0 × 10⁻⁶ to about 4.5 × 10⁻⁶, so warm water holds less oxygen for aquatic animals
  4. It stays at about 6.0 × 10⁻⁶, because the partial pressure of oxygen is unchanged
Show answer
Answer: (C) It falls from about 6.0 × 10⁻⁶ to about 4.5 × 10⁻⁶, so warm water holds less oxygen for aquatic animals
Explanation: From p = K_H x, the mole fraction is x = p/K_H. At 20 °C, x = 0.21/35000 = 6.0 × 10⁻⁶, and at 30 °C, x = 0.21/47000 ≈ 4.5 × 10⁻⁶. K_H rises with temperature, so gas solubility falls, which is why aquatic species are more comfortable in cold water.

Question 9

A dilute solution of ethanol in water obtained by fermentation is fractionally distilled. The ethanol content of the distillate cannot be raised beyond about 95% by volume. What is the reason?
  1. This mixture is a minimum boiling azeotrope formed through large positive deviation from Raoult's law, so its liquid and vapour have the same composition
  2. This mixture is a maximum boiling azeotrope formed through negative deviation from Raoult's law, so ethanol stays behind in the flask
  3. Ethanol and water form an ideal solution, so their vapour has exactly the composition of the liquid at every stage
  4. Ethanol decomposes on heating once its concentration in the distillate passes this value
Show answer
Answer: (A) This mixture is a minimum boiling azeotrope formed through large positive deviation from Raoult's law, so its liquid and vapour have the same composition
Explanation: Ethanol and water show large positive deviation from Raoult's law and form a minimum boiling azeotrope at about 95% ethanol by volume. At this composition the liquid and the vapour in equilibrium have the same composition. Further fractional distillation therefore gives no enrichment.

Question 10

Volatile liquid A obeys Raoult's law over the whole composition range in its solution with liquid B. The vapour pressure of pure A is 180 torr. If the partial pressure of A is written in the Henry's law form p = K_H x, what are the value of K_H for A and the partial pressure of A at x = 0.25?
  1. K_H = 180 torr; p = 45 torr
  2. K_H = 720 torr; p = 180 torr
  3. K_H = 180 torr; p = 135 torr
  4. K_H = 45 torr; p = 11.25 torr
Show answer
Answer: (A) K_H = 180 torr; p = 45 torr
Explanation: Raoult's law, p = p°x, has the same form as Henry's law, p = K_H x, so it is a special case in which K_H equals the vapour pressure of the pure component. Here K_H = 180 torr. The partial pressure at x = 0.25 is 180 × 0.25 = 45 torr.

Question 11

Fractional distillation of a dilute aqueous solution of ethanol cannot give ethanol stronger than about 95% by volume, however efficient the column. This is because ethanol and water
  1. react to form a stable compound at this composition
  2. form an ideal solution, so the vapour always has the composition of the liquid
  3. show positive deviation from Raoult's law and form a minimum-boiling azeotrope
  4. show negative deviation from Raoult's law and form a maximum-boiling azeotrope
Show answer
Answer: (C) show positive deviation from Raoult's law and form a minimum-boiling azeotrope
Explanation: Ethanol–water mixtures show large positive deviation because hydrogen bonds between water molecules are disrupted by ethanol. At about 95% ethanol by volume the mixture boils below both pure liquids as an azeotrope. Its vapour has the same composition as the liquid, so further distillation brings no enrichment. Azeotropes are physical mixtures, not compounds, and their composition shifts with pressure.

Question 12

Liquids A and B form an ideal solution. At a certain temperature the vapour pressures of pure A and pure B are 90 torr and 30 torr. What must be the mole fraction of A in the liquid so that the vapour in equilibrium with it contains A and B in equal mole fractions?
  1. 0.33
  2. 0.75
  3. 0.25
  4. 0.50
Show answer
Answer: (C) 0.25
Explanation: Equal mole fractions in the vapour require equal partial pressures. So 90 x(A) = 30 (1 − x(A)), which gives 120 x(A) = 30 and x(A) = 0.25. The more volatile liquid need only be the minor component of the liquid to match the other in the vapour.

Question 13

One mole of liquid X (pure vapour pressure 300 torr) and 3 mol of liquid Y (pure vapour pressure 100 torr) are mixed at a fixed temperature. The measured vapour pressure of the mixture is 180 torr. Which statement is correct?
  1. The mixture shows positive deviation from Raoult's law, X–Y attractions are stronger than X–X and Y–Y attractions, and ΔHmix < 0
  2. The mixture shows negative deviation from Raoult's law, X–Y attractions are stronger than X–X and Y–Y attractions, and ΔHmix < 0
  3. The mixture is ideal, with X–Y attractions similar to X–X and Y–Y attractions, and ΔHmix = 0
  4. The mixture shows positive deviation from Raoult's law, X–Y attractions are weaker than X–X and Y–Y attractions, and ΔHmix > 0
Show answer
Answer: (D) The mixture shows positive deviation from Raoult's law, X–Y attractions are weaker than X–X and Y–Y attractions, and ΔHmix > 0
Explanation: Raoult's law predicts 0.25 × 300 + 0.75 × 100 = 150 torr. The observed 180 torr is higher, which is a positive deviation. Molecules escape more easily because X–Y attractions are weaker than those in the pure liquids. Energy is absorbed on mixing, so ΔHmix is positive.

Question 14

A rigid flask contains dry nitrogen at 300 torr. A little of an ideal liquid mixture of A and B with mole fraction of A = 0.40 is introduced at constant temperature, and some liquid remains at equilibrium. Pure-component vapour pressures are 200 torr for A and 100 torr for B. Taking the liquid composition and gas volume as unchanged, the total pressure in the flask is
  1. 440 torr
  2. 460 torr
  3. 600 torr
  4. 140 torr
Show answer
Answer: (A) 440 torr
Explanation: By Raoult's law the vapour pressure of the mixture is 0.40 × 200 + 0.60 × 100 = 140 torr. Nitrogen in the same fixed volume and temperature keeps its 300 torr. By Dalton's law the total pressure is 300 + 140 = 440 torr.

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