Question

Mole fraction of toluene in the vapour phase which is in equilibrium with a solution of benzene $\left({\text{p}}^{\circ }=120\text{ton}\right)$ and toluene $\left({\text{p}}^{\circ }=80\text{ton}\right)$ having 3 mole of each is:

• A. 0.25
• B. 0.4
• C. 0.5
• D. 0.6

Answer 1

Answer: (B)

0.4

Given ${\text{P}}_{\text{Benzene}}^{\text{O}}=120\text{ton}$, ${\text{P}}_{\text{toluene}}^{\text{O}}=80\text{ton}$
${\text{n}}_{\text{Benzene}}=3\text{moles}$, ${\text{n}}_{\text{toluene}}=3\text{moles}$
${\text{X}}_{\text{Benzene}}=\frac{3}{3+3}=\frac{1}{2}$
${\text{X}}_{\text{toluene}}=\frac{3}{3+3}=\frac{1}{2}$
${\text{P}}_{\text{toluene}}={\text{X}}_{\text{toluene}}\times {\text{P}}_{\text{toluene}}^{\text{o}}$
$=\frac{1}{2}\times 80=40\text{ton}$
${\text{P}}_{\text{Benzene}}={\text{X}}_{\text{Benzene}}\times {\text{P}}_{\text{Benzene}}^{\text{o}}=\frac{1}{2}\times 120=60\text{ton}$
${\text{P}}_{\text{total}}={\text{P}}_{\text{toluene}}\times {\text{P}}_{\text{Benzene}}=40+60=100\text{ton}$
Now,
${\text{P}}_{\text{toluene}}={\left({\text{X}}_{\text{toluene}}\right)}_{\text{vapour}}\times {\text{P}}_{\text{total}}$
$40={\left(X_{\text{tol}}\right)}_{\text{vapour}}\times 100$
${\left(X_{\text{tol}}\right)}_{\text{vapour}}=\frac{2}{5}=0.40$
Hence, option (B) is correct.