The vapour pressures of two volatile liquids A and B at $$25°$$C are $$50$$ Torr and $$100$$ Torr, respectively. If the liquid mixture contains $$0.3$$ mole fraction of A, then the mole fraction of liquid B in the vapour phase is $$\frac{x}{17}$$. The value of $$x$$ is ______

We need to find the value of $$x$$ given that the mole fraction of liquid B in the vapour phase is $$\frac{x}{17}$$. The vapour pressures of pure A and B are $$P_A^0 = 50$$ Torr and $$P_B^0 = 100$$ Torr respectively, and in the liquid phase the mole fractions are $$x_A = 0.3$$ and $$x_B = 0.7$$.

By Raoult's law, the partial pressures are calculated as $$P_A = x_A \times P_A^0 = 0.3 \times 50 = 15 \text{ Torr}$$ and $$P_B = x_B \times P_B^0 = 0.7 \times 100 = 70 \text{ Torr}$$. The total vapour pressure is then $$P_{total} = P_A + P_B = 15 + 70 = 85 \text{ Torr}$$.

Applying Dalton's law, the mole fraction of B in vapour is $$y_B = \frac{P_B}{P_{total}} = \frac{70}{85}$$, which simplifies to $$y_B = \frac{14}{17}$$. Since this equals $$\frac{x}{17}$$, we have $$\frac{x}{17} = \frac{14}{17}$$, giving $$x = 14$$.