Consider an ideal binary liquid mixture obeying Raoult's law. Let be the mole fraction of A in the l — Solutions and Colligative Properties Chemistry Question
Question
Consider an ideal binary liquid mixture obeying Raoult's law. Let $X_A$ be the mole fraction of A in the liquid phase and $Y_A$ be its mole fraction in the vapour phase at equilibrium. If pure A has a strictly higher equilibrium vapour pressure than pure B ($P_A^0 > P_B^0$), which of the following statements are thermodynamically correct?
💡 Solution & Explanation
Konowaloff's rule states that the vapour phase is always richer in the more volatile component, so $Y_A > X_A$ (A). $P_{total} = P_A^0 + (P_B^0 - P_A^0)X_B$, which mathematically defines a straight line (B). Since $Y_A = \frac{P_A}{P_{total}}$ and $Y_B = \frac{P_B}{P_{total}}$, dividing them yields $\frac{Y_A}{Y_B} = \frac{P_A}{P_B} = \frac{P_A^0 X_A}{P_B^0 X_B}$ (D). C is false; $P_A$ or $P_B$ values depend entirely on the specific mole fractions chosen for the mixture.