Question

A cylinder contains $0.9$ moles of benzene vapour and $0.1$ moles of toluene vapour at $50{}^{o}C$ . The pure vapour pressure of benzene liquid and the pure vapour pressure of toluene liquid at $50{}^o$C are 600 torr and 200 torr respectively.
What is the mole fraction of liquid benzene in the solution?

• A. $0.25$
• B. $0.75$
• C. $0.50$
• D. $0.90$

Answer 1

The correct option is B $0.75$

$\Rightarrow$ The mole fraction of benzene in vapour phase,
$Y_{benzene}=0.9=\frac{X_{benzene}\times P_{benzene}^0}{P_T}\left(1\right)$
$\Rightarrow$ The mole fraction of toluene in vapour phase,
$Y_{toluene}=0.1=\frac{X_{toluene}\times P_{toluene}^0}{P_T}\left(2\right)$
$\Rightarrow$ $Eqn\left(1\right)÷Eqn\left(2\right)$
$9=\frac{X_{benzene}\times 600}{X_{toluene}\times 200}$
$9=\frac{X_{benzene}\times 600}{(1-X_{benzene})\times 200}(X_A+X_B=1)$
$X_{benzene}=0.75$