Question

Vapour pressure of pure benzene is $119\mathrm{torr}$ and toluene is $37\mathrm{torr}$at the same temperature. 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:

Answer 1

Step 1: Calculation of Total pressure

The total pressure is given by

${\mathrm{P}}_{\mathrm{s}}={\mathrm{x}}_{\mathrm{T}}{\mathrm{P}}_{\mathrm{T}}^{°}+{\mathrm{x}}_{\mathrm{B}}{\mathrm{P}}_{\mathrm{B}}^{°}$

Where ${\mathrm{P}}_{\mathrm{s}}=$Total pressure , ${\mathrm{x}}_{\mathrm{T}}\mathrm{and}{\mathrm{x}}_{\mathrm{B}}$ are the mole fraction of Toluene and Benzene in equilibrium respectively and ${\mathrm{P}}_{\mathrm{T}}^{°}\mathrm{and}{\mathrm{P}}_{\mathrm{B}}^{°}$ are the vapor pressure of Toluene and Benzene respectively

${\mathrm{P}}_{\mathrm{s}}=\left(0.5\times 37\right)+\left(0.5\times 119\right)$

${\mathrm{P}}_{\mathrm{S}}=78\mathrm{torr}$

Step 2: Calculating Mole fraction of toluene in a vapor phase

Mole fraction in the vapor phase is given by

${\mathrm{y}}_{\mathrm{T}}=\frac{{\mathrm{x}}_{\mathrm{T}}{\mathrm{P}}_{\mathrm{T}}^{°}}{{\mathrm{P}}_{\mathrm{s}}}$

Where ${\mathrm{y}}_{\mathrm{T}}$ is the mole fraction of toluene in a vapor phase

${\mathrm{y}}_{\mathrm{T}}=\frac{0.5\times 37}{78}$

${\mathrm{y}}_{\mathrm{T}}=0.237$

Therefore, the Mole fraction of toluene in the vapor phase is $0.237$.