Question

At \( T(K) \), two liquids A and B form an ideal solution. The vapour pressures of pure liquids A and B at that temperature are 400 and 600 mm Hg respectively. If the mole fraction of liquid B is 0.3 in the mixture, the mole fractions of A and B in vapour phase respectively are:

• A. 0.391, 0.609
• B. 0.509, 0.491
• C. 0.609, 0.391
• D. 0.491, 0.509

Hint:

Raoult's law applies to ideal solutions where the vapor pressure is directly proportional to the mole fraction of the component in the liquid phase.

Answer 1

The Correct Option is C

The total pressure is given by Raoult's law: \[ P_{\text{total}} = P_A + P_B \] Where the individual pressures are calculated by: \[ P_A = x_A \cdot P_A^0 \quad \text{and} \quad P_B = x_B \cdot P_B^0 \] Substitute \( x_B = 0.3 \), so \( x_A = 1 - x_B = 0.7 \). The partial pressures are: \[ P_A = 0.7 \times 400 = 280 \, \text{mm Hg} \] \[ P_B = 0.3 \times 600 = 180 \, \text{mm Hg} \] Now calculate the mole fractions in vapour phase: \[ y_A = \frac{P_A}{P_{\text{total}}} = \frac{280}{280 + 180} = 0.609 \] \[ y_B = \frac{P_B}{P_{\text{total}}} = \frac{180}{280 + 180} = 0.391 \]