At what pressure, 20% of the total moles of vapours will liquefy? () — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

Total moles = 5.0. 20% liquefied means $n_L = 1.0$ and $n_V = 4.0$. Mass balance for A: $1.0(X_A) + 4.0(Y_A) = 2.0$. We know $Y_A = \frac{0.4 X_A}{0.6 - 0.2 X_A} = \frac{2 X_A}{3 - X_A}$. Substituting gives $X_A + \frac{8 X_A}{3 - X_A} = 2$. Rearranging yields $X_A^2 - 13X_A + 6 = 0$. Solving for $X_A = \frac{13 - \sqrt{145}}{2} = \frac{13 - 12.04}{2} = 0.48$. Pressure $P = P_B^\circ + (P_A^\circ - P_B^\circ)X_A = 0.6 - 0.2(0.48) = 0.504 \text{ bar}$. Therefore, correct answer is B.