Pressure over an ideal binary liquid solution containing 10 moles each of liquid A and B is graduall — Solutions and Colligative Properties Chemistry Question
Question
Pressure over an ideal binary liquid solution containing 10 moles each of liquid A and B is gradually decreased isothermally. At what pressure, half of the total amount of liquid will get converted into vapour? ($P_A^\circ = 200 \text{ torr}, P_B^\circ = 100 \text{ torr}$)
💡 Solution & Explanation
Total moles = 20. Half evaporated = 10 moles liquid, 10 moles vapour. Let $X_A$ be mole fraction in liquid, $Y_A$ in vapour. Moles of A in liquid $= 10 X_A$, in vapour $= 10 Y_A$. Total A = 10. So $10 X_A + 10 Y_A = 10 \Rightarrow X_A + Y_A = 1$. Since $Y_A + Y_B = 1$, we get $Y_A = 1 - X_A$. Also $Y_A = P_A^\circ X_A / P = 200 X_A / P$. Thus $1 - X_A = 200 X_A / P \Rightarrow P = 200 X_A / (1 - X_A)$. Similarly for B, $P = 100(1 - X_A) / X_A$. Equating gives $2 X_A^2 = (1 - X_A)^2 \Rightarrow \sqrt{2} X_A = 1 - X_A \Rightarrow X_A = \sqrt{2} - 1$. Substituting $X_A$, $P = 100 \sqrt{2} \approx 141.4 \text{ torr}$. Therefore, correct answer is D.