Question

A solution is made by mixing one mole of volatile liquid A with 3 moles of volatile liquid B. The vapor pressure of pure A is 200 mm Hg and that of the solution is 500 mm Hg. The vapor pressure of pure B and the least volatile component of the solution, respectively, are:

• A. 1400 mm Hg, A
• B. 1400 mm Hg, B
• C. 600 mm Hg, A
• D. 600 mm Hg, B

Hint:

Raoult's Law is useful for determining the vapor pressure of components in an ideal solution. It states that the vapor pressure is proportional to the mole fraction of the component in the solution.

Answer 1

The Correct Option is D

The relation between vapor pressures is given by Raoult's law: \[ P_A = P_A^0 \cdot X_A, \quad P_B = P_B^0 \cdot X_B \] Given: \[ P_A + P_B = 500 \, \text{mm Hg}, \quad P_A^0 = 200 \, \text{mm Hg}, \quad P_B^0 = 600 \, \text{mm Hg} \] Using the mole fractions and Raoult's law: \[ P_A = 200 \times \frac{1}{4}, \quad P_B = 600 \times \frac{3}{4} \] Thus, the answer is \( P_A = 600 \, \text{mm Hg}, P_B = 600 \, \text{mm Hg} \).