1. Use Raoult’s law for an ideal solution

For components $A$ and $B$:

$p_A = x_A P_A^0, \qquad p_B = x_B P_B^0$

Given:

$P_A^0 = 90\ \text{mm Hg}, \qquad P_B^0 = 15\ \text{mm Hg}$

Liquid-phase mole fraction of $A$:

$x_A = 0.6$

So,

$x_B = 1 - 0.6 = 0.4$

2. Calculate partial pressures

$p_A = 0.6 \times 90 = 54\ \text{mm Hg}$

$p_B = 0.4 \times 15 = 6\ \text{mm Hg}$

3. Calculate total vapour pressure

$P_{\text{total}} = p_A + p_B = 54 + 6 = 60\ \text{mm Hg}$

4. Find mole fraction of $B$ in vapour phase

If $y_B$ is the mole fraction of $B$ in vapour,

$y_B = \frac{p_B}{P_{\text{total}}} = \frac{6}{60} = 0.1$

Given that mole fraction of $B$ in vapour phase is

$x \times 10^{-1}$

So,

$0.1 = 1 \times 10^{-1}$

Hence,

$x = 1$

5. Comparison with stored answer

Stored correct answer = $1$

Our derived answer = $1$

They agree.