Question

The vapour pressure of pure liquid $A$ at $300K$ is $577Torr$ and that of pure liquid $B$ is $390Torr$. 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.

• A. 0.628
• B. 0.872
• C. 0.267
• D. 0.834

Answer 1

Answer: (C)

0.267

$P_A^o=577torr,P_B^o=390torr,{x_A}^{\prime }=0.35$

Mole fraction of $A$ in vapour phase= $\frac{P_A}{P}$

$P_A=x_A{P}_A^o,P=P_A+P_B=x_A{P}_A^o+(1-x_A)P_B^o$

$\therefore 0.35=\frac{x_A\times 577}{x_A\times 577+(1-x_A)390}$

$\therefore x_A=0.267$