Question

Pure Benzene has a vapour pressure of $70\mathrm{torr}$, while methylbenzene has a vapour pressure of $20\mathrm{torr}$. Find the mole fraction of Benzene in the vapour phase if the two components are mixed in an equimolar ratio.

Answer 1

Given:

$\begin{array}{rcl}{\mathrm{P}}_{{\mathrm{C}}_6{\mathrm{H}}_6}^{°}& =& 70\mathrm{torr}\\ {\mathrm{P}}_{{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_3}^{°}& =& 20\mathrm{torr}\end{array}$

Pure Benzene and Methylbenzene are equimolar and represented by $\mathrm{n}$.

Dalton's law is used for the mole fraction of benzene in the vapour phase.

${\mathrm{P}}_{\mathrm{A}}={\mathrm{X}}_{\mathrm{A}}\times {\mathrm{P}}_{\mathrm{Total}}\cdots \left(\mathrm{i}\right)$

${\mathrm{P}}_{\mathrm{A}}=$Partial pressure of Benzene

${\mathrm{P}}_{\mathrm{B}}=$Partial pressure of Methylbenzene

${\mathrm{P}}_{\mathrm{Total}}=$Total pressure of solution

Now, According to Roult's Law:

$\begin{array}{rcl}{\mathrm{P}}_{\mathrm{Total}}& =& {\mathrm{P}}_{\mathrm{A}}+{\mathrm{P}}_{\mathrm{B}}\\ & =& {\mathrm{X}}_{\mathrm{A}}\times {\mathrm{P}}_{\mathrm{A}}^{°}+{\mathrm{X}}_{\mathrm{B}}\times {\mathrm{P}}_{\mathrm{B}}^{°}\\ & =& \frac{\mathrm{n}}{\mathrm{n}+\mathrm{n}}\times 70+\frac{\mathrm{n}}{\mathrm{n}+\mathrm{n}}\times 20\\ & =& 0.5\times 70+0.5\times 20\\ & =& 45\mathrm{torr}\\ & & \end{array}$

And,

$\begin{array}{rcl}{\mathrm{P}}_{\mathrm{A}}& =& {\mathrm{X}}_{\mathrm{A}}\times {\mathrm{P}}_{\mathrm{A}}^{°}\\ & =& \frac{\mathrm{n}}{\mathrm{n}+\mathrm{n}}\times 70\\ & =& 0.5\times 70\\ & =& 35\mathrm{torr}\end{array}$

Now from equation $\left(\mathrm{i}\right)$

$\begin{array}{rcl}{\mathrm{P}}_{\mathrm{A}}& =& {\mathrm{X}}_{\mathrm{A}}\times {\mathrm{P}}_{\mathrm{Total}}\\ {\mathrm{X}}_{\mathrm{A}}& =& \frac{{\mathrm{P}}_{\mathrm{A}}}{{\mathrm{P}}_{\mathrm{Total}}}\\ & =& \frac{35}{45}\\ & =& 0.778\mathrm{torr}\end{array}$

The mole fraction of Benzene in the vapour phase if the two components are mixed in an equimolar ratio is $0.778\mathrm{torr}$.