The equilibrium vapour pressures of two liquids A and B in their pure states are in the ratio of . A — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

Let $P_A^0 = P$ and $P_B^0 = 2P$. The ratio of liquid moles is $n_A : n_B = 1 : 2$, meaning $X_A = 1/3$ and $X_B = 2/3$. Using Raoult's law, the partial pressures are $P_A = X_A P_A^0 = \frac{P}{3}$ and $P_B = X_B P_B^0 = \frac{2}{3}(2P) = \frac{4P}{3}$. Total pressure $P_{total} = P_A + P_B = \frac{5P}{3}$. By Dalton's law, the mole fraction in vapour is $Y_A = \frac{P_A}{P_{total}} = \frac{P/3}{5P/3} = \frac{1}{5} = 0.20$.