Question

The molar volume of liquid benzene ($density=0.877gm{l}^{-1}$) increases by a factor of $2750$ as it vaporizes at ${20}^{\circ }$C and that of liquid toluene ($density=0.867gm{l}^{-1}$)increases by a factor of $7720$ at ${20}^{\circ }$C. A solution of benzene and toluene has a vapour pressure of $46.0$ torr. Find the mole fraction of benzene in the vapour above the solution:

• A. $0.38$
• B. $0.47$
• C. $0.53$
• D. $0.62$

Answer 1

The correct option is D $0.62$

Let the mole fraction of benzene be $x$.
Let the total number of moles of the solution be 1.
Hence, the number of moles of benzene are $x$.

The molar mass of benzene is $78$ g/mol.
The mass of benzene is $xmol\times 78g/mol=78xg$. The density of benzene is $0.87g/ml$.

Hence, the volume of liquid benzene is $\frac{78xg}{0.877g/ml}=88.9ml=0.0889L$.

In the vapor phase, the volume of benzene will be $2750\times 0.0889L=244.5L$. The vapor pressure of benzene is

$P=\frac{nRT}{V}=\frac{xmol\times 0.0821L\cdot atm/mol/K\times 293K}{244.5xL}=0.0984atm$

Convert the unit of pressure from atm to torr using the relation $1atm=760torr$.

$0.0984atm\times \frac{760torr}{1atm}=74.8torr$.

The mole fraction of benzene in the vapor phase is $\frac{46.0torr}{74.8torr}=0.62$.