1. Given data

For an ideal solution, Raoult’s law applies: $p_A = x_A P_A^0, \qquad p_B = x_B P_B^0$

Given:

• $P_A^0 = 7 \times 10^3\,\text{Pa}$
• $P_B^0 = 12 \times 10^3\,\text{Pa}$
• Liquid solution has $40$ mole percent of $A$

So in the liquid phase: $x_A = 0.40, \qquad x_B = 0.60$

2. Calculate partial pressures

Using Raoult’s law: $p_A = x_A P_A^0 = 0.40 \times 7\times 10^3 = 2.8 \times 10^3\,\text{Pa}$ $p_B = x_B P_B^0 = 0.60 \times 12\times 10^3 = 7.2 \times 10^3\,\text{Pa}$

3. Calculate total vapor pressure

$P_{\text{total}} = p_A + p_B = 2.8\times 10^3 + 7.2\times 10^3 = 10.0\times 10^3\,\text{Pa}$

4. Find composition of vapor phase

Mole fraction in vapor phase is: $y_A = \frac{p_A}{P_{\text{total}}} = \frac{2.8\times 10^3}{10.0\times 10^3} = 0.28$ $y_B = \frac{p_B}{P_{\text{total}}} = \frac{7.2\times 10^3}{10.0\times 10^3} = 0.72$

Thus, vapor composition is: $y_A = 0.28, \qquad y_B = 0.72$

5. Match with options

Option B: $x_A = 0.28, x_B = 0.72$

This matches the vapor-phase composition (though the option uses $x$, it should technically be $y$ for vapor mole fraction).

6. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So the answer agrees.