Liquids A and B form an ideal solution. The plot of (Y-axis) versus (X-axis) (where and are the mole — Solutions and Colligative Properties Chemistry Question
Question
Liquids A and B form an ideal solution. The plot of $\frac{1}{X_A}$ (Y-axis) versus $\frac{1}{Y_A}$ (X-axis) (where $X_A$ and $Y_A$ are the mole fractions of A in liquid and vapour phases at equilibrium, respectively) is linear whose slope and intercept, respectively, are given as
💡 Solution & Explanation
By Raoult's and Dalton's laws, $Y_A = P_A^\circ X_A / (P_A^\circ X_A + P_B^\circ (1 - X_A))$. Rearranging gives $\frac{1}{Y_A} = \frac{P_A^\circ X_A + P_B^\circ - P_B^\circ X_A}{P_A^\circ X_A} = 1 + \frac{P_B^\circ}{P_A^\circ X_A} - \frac{P_B^\circ}{P_A^\circ}$. Rearranging for $\frac{1}{X_A}$ as the y-variable: $\frac{P_B^\circ}{P_A^\circ} \frac{1}{X_A} = \frac{1}{Y_A} - \frac{P_A^\circ - P_B^\circ}{P_A^\circ} \Rightarrow \frac{1}{X_A} = \frac{P_A^\circ}{P_B^\circ} \frac{1}{Y_A} + \frac{P_B^\circ - P_A^\circ}{P_B^\circ}$. The slope is $P_A^\circ / P_B^\circ$ and intercept is $(P_B^\circ - P_A^\circ) / P_B^\circ$. Therefore, correct answer is B.