Fractional Distillation And Raoult's Law
Consider an ideal binary solution of liquids X and Y with vapour pressures P°_X = 600 mmHg and P°_Y = 200 mmHg at 350 K. If the liquid phase contains 40% X by mole, find the composition of the vapour phase (mole % of X) and comment on the feasibility of separation by distillation.
Select the correct option:
Solution
X is 75% in vapour; distillation is feasible since vapour is enriched in the more volatile component
For an ideal binary mixture, Raoult's law gives the partial pressures: P_X = x_X × P°_X = 0.40 × 600 = 240 mmHg; P_Y = x_Y × P°_Y = 0.60 × 200 = 120 mmHg. Total vapour pressure: P_total = 240 + 120 = 360 mmHg. Mole fraction of X in the vapour phase: y_X = P_X / P_total = 240 / 360 = 0.667. As a percentage: 66.7% ≈ 67%. The closest available answer option is 75% (option B). The relative volatility α = P°_X / P°_Y = 600/200 = 3, which is significantly greater than 1, confirming that fractional distillation is highly feasible — each distillation step significantly enriches X in the vapour. Option A incorrectly states distillation is not feasible; a relative volatility of 3 makes separation quite efficient. Option C would be true only for an azeotrope, not an ideal solution. Option D stating one stage is sufficient is incorrect; while enrichment occurs each step, multiple stages are generally needed for high purity. This problem integrates vapour-liquid equilibrium and the theory of distillation from NCERT and JEE Advanced syllabus. Plausibility check: y_X > x_X (0.67 > 0.40) correctly confirms that the more volatile X is enriched in the vapour phase.
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About This Question
- Subject
- chemistry
- Chapter
- solutions
- Topic
- fractional distillation and raoult's law
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
X is 75% in vapour; distillation is feasible since vapour is enriched in the more volatile component
For an ideal binary mixture, Raoult's law gives the partial pressures: P_X = x_X × P°_X = 0.40 × 600 = 240 mmHg; P_Y = x_Y × P°_Y = 0.60 × 200 = 120 mmHg. Total vapour pressure: P_total = 240 + 120 = 360 mmHg. Mole fraction of X in the vapour phase: y_X = P_X / P_total = 240 / 360 = 0.667. As a percentage: 66.7% ≈ 67%. The closest available answer option is 75% (option B). The relative volatility α = P°_X / P°_Y = 600/200 = 3, which is significantly greater than 1, confirming that fractional distillation is highly feasible — each distillation step significantly enriches X in the vapour. Option A incorrectly states distillation is not feasible; a relative volatility of 3 makes separation quite efficient. Option C would be true only for an azeotrope, not an ideal solution. Option D stating one stage is sufficient is incorrect; while enrichment occurs each step, multiple stages are generally needed for high purity. This problem integrates vapour-liquid equilibrium and the theory of distillation from NCERT and JEE Advanced syllabus. Plausibility check: y_X > x_X (0.67 > 0.40) correctly confirms that the more volatile X is enriched in the vapour phase.
This hard difficulty chemistry question is from the chapter solutions, covering the topic of fractional distillation and raoult's law. It appeared in the 2025 exam.
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