Question

Vapour pressure of pure benzene is $119\text{torr}$ and of toluene is $37.0\text{torr}$ at the same temperature. Then mole fraction of toluene in vapour phase which is in equilibrium with a solution of benzene and toluene having a mole fraction of toluene $0.50,$ will be:

• A. $0.137$
• B. $0.205$
• C. $0.237$
• D. $0.435$

Answer 1

The correct option is C $0.237$

Given,
$\text{Vapour pressure of pure benzene,}{P}_A^0=119\text{torr}$
$\text{Vapour pressure of pure toluene,}{P}_B^0=37\text{torr}$
$\text{Then, partial pressure of benzene,}{P}_A={\chi }_A{P}_A^o=Y_A{P}_T$
$\text{Where,}{\chi }_A=\text{mole fraction of benzene}=1-0.5=0.5$
$Y_A=\text{Mole fraction of benzene in vapour phase}$
$P_T=\text{total pressure}$
So, $P_A=0.5\times 119=Y_A{P}_T...\left(1\right)$
Similarly,
$P_B=0.5\times 37=Y_B{P}_T...\left(2\right)$
Where, $Y_B=\text{Mole fraction of toluene in vapour phase}$
So, dividing $\left(1\right)$ by $\left(2\right)$ we get,
$\frac{0.5\times 119}{0.5\times 37}=\frac{(1-Y_B)P_T}{Y_B{P}_T}$
$\Rightarrow 119Y_B=37-37Y_B$
$\Rightarrow 156Y_B=37$
$\Rightarrow Y_B=\frac{37}{156}\approx 0.237$