Multiple choice

The vapour pressure of pure liquid $A$ at $300\ K$ is $577\ Torr$ and that of pure liquid $B$ is $390\ Torr$. These two compounds form ideal liquid and gaseous mixtures. Consider the equilibrium composition of a mixture in which the mole fraction of $A$ in the vapour is $0.35$. Find the mole % of $A$ in liquid.

  1. 0.628

  2. 0.872

  3. 0.267

  4. 0.834
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Raoult's Law: P_A = x_A * P_A_pure. Total pressure P = x_A * P_A_pure + (1-x_A) * P_B_pure. Mole fraction in vapor y_A = (x_A * P_A_pure) / P. Given y_A = 0.35, P_A_pure = 577, P_B_pure = 390. 0.35 = (x_A * 577) / (x_A * 577 + (1-x_A) * 390). 0.35 = 577x_A / (187x_A + 390). 0.35(187x_A + 390) = 577x_A. 65.45x_A + 136.5 = 577x_A. 136.5 = 511.55x_A. x_A = 0.2668.

AI explanation

By Dalton's law, the total pressure of the mixture is the partial pressure of A divided by its vapour mole fraction, which is 0.35 times 577 divided by 0.35, but setting up the equilibrium with Raoult's law gives the total pressure as 577x + 390(1 - x) where x is the liquid mole fraction of A. The partial pressure of A is 0.35 times the total pressure, so 577x = 0.35 multiplied by (577x + 390(1 - x)). Solving 577x = 202.0x + 136.5 - 136.5x gives 511.5x = 136.5, so x equals 0.267. The mole percent of A in the liquid is 0.267.

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