Question

An equimolar mixture of 'A' and 'B' is prepared. The total vapour pressure of this mixture as a function of mole fraction of B is found to be $P_T=300+600X_B$.

Calculate the composition of vapour of this mixture (Assume that the number of moles going into the vapour phase is negligible in comparison to the number of moles present in the liquid phase).

• A. $Y_A=40\%,Y_B=60\%$
• B. $Y_A=30\%,Y_B=70\%$
• C. $Y_A=10\%,Y_B=90\%$
• D. $Y_A=25\%,Y_B=75\%$

Answer 1

The correct option is D $Y_A=25\%,Y_B=75\%$

$\text{Total vapour pressure,}{P}_T=300+600X_B$
where, $X_B$ is mole faction of B.
When $X_B=1$,
$\text{Pure vapour pressure of B,}{p}_B^{\circ }=300+600=900$
When $X_B=0$,
$\text{Pure vapour pressure of A,}{p}_A^{\circ }=300$
We know, it is a equimolar solution
$\therefore$
$\text{Mole fraction of A,}{X}_A=\frac{1}{2}\text{Mole fraction of B,}{X}_B=\frac{1}{2}\text{Total vapour pressure,}{P}_T=300+600X_B{P}_T=300+600\times \frac{1}{2}{P}_T=600$
By raoults law,
$p_A=X_A\times p_A^{\circ }=Y_A\times P_T$
$\Rightarrow Y_A=\frac{X_A\times p_A^{\circ }}{P_T}$
where,
$Y_A\Rightarrow \text{Composition of A at vapour phase}\text{Also,}{Y}_B\Rightarrow \text{Composition of B at vapour phase}$
$Y_A=\frac{\frac{1}{2}\times 300}{600}{Y}_A=\frac{1}{4}=25\%$
Similary,
$Y_B=\frac{\frac{1}{2}\times 900}{600}{Y}_B=\frac{3}{4}=75\%$