Question

If two liquids \(A\) and \(B\) have \[ P_A^\circ:P_B^\circ=1:2 \] and have mole fraction in solution as \(1:2\), then mole fraction of \(B\) in vapour phase is

• A. \(0.2\)
• B. \(0.8\)
• C. \(0.4\)
• D. \(0.6\)

Hint:

For ideal solutions, \[ \boxed{ P_i=x_iP_i^\circ } \] and \[ \boxed{ y_i=\frac{P_i}{P_{\text{total}}}. } \]

Answer 1

The Correct Option is B

Step 1: Find the mole fractions in the liquid phase. Given, \[ x_A:x_B=1:2. \] Hence, \[ x_A=\frac13, \qquad x_B=\frac23. \] Also, \[ P_A^\circ:P_B^\circ=1:2. \] Let \[ P_A^\circ=P, \qquad P_B^\circ=2P. \]

Step 2: Apply Raoult's law. Partial pressures are \[ P_A=x_AP_A^\circ=\frac13P, \] \[ P_B=x_BP_B^\circ=\frac23(2P)=\frac43P. \] Total pressure, \[ P_{\text{total}} = \frac13P+\frac43P = \frac53P. \]

Step 3: Calculate the mole fraction in vapour phase. \[ y_B = \frac{P_B}{P_{\text{total}}} = \frac{\frac43P}{\frac53P} = \frac45 = 0.8. \] Hence, \[ \boxed{0.8}. \] Thus, \[ \boxed{(B)} \] is the correct answer.