Question

Consider a binary mixture of two ideal volatile liquids A and B having vapour pressure of $300mmHg$ and $600mmHg$ in their pure state.
What will be the mole fraction of component A in vapour phase $\left(y_A\right)$ if the total pressure $\left(P_T\right)$ of the solution is $400mmHg$?

• A. $0.2$
• B. $0.5$
• C. $0.8$
• D. $0.9$

Answer 1

The correct option is B $0.5$

${\chi }_A=\frac{y_A\times P_T}{p_A^{\circ }}$
${\chi }_A$ is the mole fraction of component A in liquid phase

${\chi }_B=\frac{y_B\times P_T}{p_B^{\circ }}$
${\chi }_B$ is the mole fraction of component B in liquid phase

Also ${\chi }_A+{\chi }_B=1$

So, $\frac{1}{P_T}=\frac{y_A}{p_A^{\circ }}+\frac{y_B}{p_B^{\circ }}\frac{1}{P_T}=\frac{y_A}{p_A^{\circ }}+\frac{1-y_A}{p_B^{\circ }}$
Solving,

$P_T=\frac{p_A^{\circ }\times p_B^{\circ }}{p_A^{\circ }+(p_B^{\circ }-p_A^{\circ })\times y_A}$
Here ,
$P_T=400mmHg{p}_A^{\circ }=300mmHg{p}_B^{\circ }=600mmHg$
$400=\frac{300\times 600}{300+(600-300)\times y_A}4=\frac{6}{1+y_A}4+4y_A=6y_A=0.5$
(b) is the correct option