Benzene and toluene form ideal solution over the entire range of composition.
Question:

Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.

Solution:

Molar mass of benzene $\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)=6 \times 12+6 \times 1$

$=78 \mathrm{~g} \mathrm{~mol}^{-1}$

Molar mass of toluene $\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}\right)=7 \times 12+8 \times 1$

$=92 \mathrm{~g} \mathrm{~mol}^{-1}$

Now, no. of moles present in $80 \mathrm{~g}$ of benzene $=\frac{80}{78} \mathrm{~mol}=1.026 \mathrm{~mol}$

And, no. of moles present in $100 \mathrm{~g}$ of toluene $=\frac{100}{92} \mathrm{~mol}=1.087 \mathrm{~mol}$

$\therefore$ Mole fraction of benzene, $x_{b}=\frac{1.026}{1.026+1.087}=0.486$

And, mole fraction of toluene, $x_{t}=1-0.486=0.514$

It is given that vapour pressure of pure benzene, $p_{b}^{0}=50.71 \mathrm{~mm} \mathrm{Hg}$

And, vapour pressure of pure toluene, $p_{t}^{0}=32.06 \mathrm{~mm} \mathrm{Hg}$

Therefore, partial vapour pressure of benzene, $p_{b}=x_{b} \times p_{b}$

$=0.486 \times 50.71$

$=24.645 \mathrm{~mm} \mathrm{Hg}$

And, partial vapour pressure of toluene, $p_{t}=x_{t} \times p_{t}$

$=0.514 \times 32.06$

$=16.479 \mathrm{~mm} \mathrm{Hg}$

Hence, mole fraction of benzene in vapour phase is given by:

$\frac{p_{b}}{p_{b}+p_{t}}$

$=\frac{24.645}{24.645+16.479}$

$=\frac{24.645}{41.124}$

$=0.599$

$=0.6$