Vapour Pressure Of Binary Liquid Mixtures

Question

Two volatile liquids A (vapour pressure 400 mmHg) and B (vapour pressure 100 mmHg) are mixed in equal molar proportions to form an ideal solution. What is the total vapour pressure of the mixture and the mole fraction of A in the vapour phase?

RankGuru · Accepted Answer

For an ideal solution of two volatile components, Raoult's law applies to both. With equal molar proportions: x_A = x_B = 0.5. By Raoult's law: P_A = x_A × P°_A = 0.5 × 400 = 200 mmHg; P_B = x_B × P°_B = 0.5 × 100 = 50 mmHg. Total vapour pressure: P_total = P_A + P_B = 200 + 50 = 250 mmHg. The mole fraction of A in the vapour phase: y_A = P_A / P_total = 200 / 250 = 0.80. This demonstrates that the more volatile component is enriched in the vapour phase — y_A = 0.80 > x_A = 0.50. Option B incorrectly adds the pure vapour pressures without accounting for mole fractions. Option C correctly gives the total pressure but assigns equal vapour mole fractions, ignoring volatility difference. Option D invents an incorrect total pressure. The enrichment of the more volatile component in the vapour phase is the basis of fractional distillation. Plausibility check: total pressure (250 mmHg) lies between P°_A and P°_B, as required by Raoult's law, and y_A > x_A confirms enrichment of the more volatile component.

Vapour Pressure Of Binary Liquid Mixtures

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