Question

In a binary mixture of two ideal volatile liquids A and B, calculate the mole fraction of component A in vapour phase if:
Vapour pressure of A in pure state is $500mmHg$
Vapour pressure of B in pure state is$700mmHg$
Total pressure $\left(P_T\right)$ of the solution is $600mmHg$

• A. 0.25
• B. 0.42
• C. 0.75
• D. 0.58

Answer 1

The correct option is B 0.42

${\chi }_A=\frac{y_A\times P_T}{p_A^0}$
${\chi }_A$ is the mole fraction of component A in liquid phase

${\chi }_B=\frac{y_B\times P_T}{p_B^0}$
${\chi }_B$ is the mole fraction of component B in liquid phase

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

So, $\frac{1}{P_T}=\frac{y_A}{p_A^0}+\frac{y_B}{p_B^0}\frac{1}{P_T}=\frac{y_A}{p_A^0}+\frac{1-y_A}{p_B^0}$
Solving,

$P_T=\frac{p_A^0\times p_B^0}{p_A^0+(p_B^0-p_A^0)\times y_A}$
Here ,
$P_T=300mmHg{p}_A^0=400mmHg{p}_B^0=700mmHg$
$600=\frac{500\times 700}{500+(700-500)\times y_A}6=\frac{35}{5+2y_A}30+12y_A=35y_A=\frac{5}{12}=0.416\approx 0.42$