The vapour pressure of two pure liquids A and B which form an ideal solution are 1000 and 1600 torr respectively at 400 K. A liquid solution of A and B for which the mole fraction of A is 0.60 is contained in a cylinder by a piston on which the pressure can be varied. The solution use slowly vapourised at 400 K by decreasing the applied pressure. What is the composition of last droplet of liquid remaining in equilibrium with vapour?

Engineering

Chemistry

Relative Lowering in Vapour Pressure

Question

X_A = 0.30, X_B = 0.70

X_A = 0.40, X_B = 0.60

X_A = 0.70, X_B = 0.30

X_A = 0.50, X_B = 0.50

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

Given, P_A° = 1000 torr and P_B° = 1600 torr [T = 400 K]

Here,

Y_{A} = \frac{P_{A}^{o} X_{A}}{P_{A}^{o} X_{A} + P_{B}^{o} (1 - X_{A})} \Rightarrow 0.6 = \frac{1000 X_{A}}{1000 X_{A} + 1600 (1 - X_{A})}

∴ X_A = 0.70, X_B = 1 – X_A = 0.30.

Please subscribe our Youtube channel to unlock this solution.