# 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.

$\begin{array}{1 1}0.60\\0.80\\0.50\\0.40\end{array}$

Molar mass of benzene $(C_6H_6)=78gmol^{-1}$
Molar mass of toluene $(C_6H_5CH_3)=92gmol^{-1}$
$\therefore$ No of moles in 80g of benzene =$\large\frac{80g}{78g mol^{-1}}$
$\Rightarrow 1.026$ mole
Number of moles in 100g of toluene =$\large\frac{100g}{92g mol^{-1}}$
$\Rightarrow 1.087$ mole
In the solution ,mole fraction of benzene =$\large\frac{1.026}{1.026+1.087}$
$\Rightarrow 0.486$
Mole fraction of toluene =1-0.486=0.514
$P^{\large\circ}_{Benzene}=50.71mm$
$P^{\large\circ}_{Toluene}=32.06mm$
Applying Raoult's law
$P_{Benzene}=x_{Benzene}\times P^0_{Benzene}$
$\qquad\quad=0.486\times$ 50.71 mm=24.65 mm
$P_{Toluene}=x_{Toluene}\times P^0_{Toluene}$
$\qquad\quad=0.514\times 32.06$ mm =16.48 mm
Mole fraction of Benzene in vapour phase =$\large\frac{P_{Benzene}}{P_{Benzene}+P_{Toluene}}$
$\Rightarrow \large\frac{24.65}{24.65+16.48}$
$\Rightarrow \large\frac{24.65}{41.13}$
$\Rightarrow 0.60$