The vapour pressures of A and B at 25$$^\circ$$C are 90 mmHg and 15 mm Hg respectively. If A and B are mixed such that the mole fraction of A in the mixture is 0.6, then the mole fraction of B in the vapour phase is $$x \times 10^{-1}$$. The value of x is ___. (Nearest integer)

For an ideal solution, the partial vapour pressures are given by Raoult's law. Given $$P_A^* = 90$$ mmHg, $$P_B^* = 15$$ mmHg, $$x_A = 0.6$$, and $$x_B = 0.4$$:

Partial pressure of A: $$p_A = x_A \cdot P_A^* = 0.6 \times 90 = 54$$ mmHg

Partial pressure of B: $$p_B = x_B \cdot P_B^* = 0.4 \times 15 = 6$$ mmHg

Total vapour pressure: $$P_{\text{total}} = p_A + p_B = 54 + 6 = 60$$ mmHg

The mole fraction of B in the vapour phase is:

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

Since $$y_B = x \times 10^{-1}$$, we have $$0.1 = 1 \times 10^{-1}$$, so $$x = \mathbf{1}$$.