Question

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

• A. $0.449$
• B. $0.26$
• C. $0.73$
• D. $0.551$

Answer 1

The correct option is C $0.73$

$\text{Volume of}1mol\text{of liquid benzene}=\frac{78}{0.877}=88.94mL$

$\text{Volume of}1mol\text{of benzene vapour}=88.94\times 2750=244.585L$

Assuming the vapour to behave like an ideal gas,

$P_{\text{Benzene}}^{\circ }=\frac{RT}{V}\left(\text{For 1 mol}\right)$
$P_{\text{Benzene}}^{\circ }=\frac{0.082\times 293}{244.585}atm=0.098atm$

$\text{Volume of}1mol\text{of liquid toluene}=\frac{92}{0.867}=106.11mL$

$\text{Volume of}1mol\text{of toluene vapour}=106.11\times 7720=819.169L$

Assuming the vapour to behave like an ideal gas,

$P_{Toluene}^{\circ }=\frac{0.082\times 293}{819.169}=0.029atm$

$P_{Benzene}=X_{Benzene}\times P_{Benzene}^{\circ }=X_{Benzene}\times 0.098atm$

$P_{Toluene}=X_{Toluene}\times 0.029atm$

$P_{Toluene}=(1-X_{Benzene})\times 0.029atm[\because X_{Toluene}=1-X_{Benzene}]$

$P_{total}=P_{Benzene}+P_{Toluene}$

$46torr=X_{Benzene}\times 0.098+(1-X_B)\times 0.029$

$\frac{46}{760}atm=0.069X_{Benzene}+0.029$

$0.06=0.069X_{Benzene}+0.029$

$X_{Benzene}=\frac{0.06-0.029}{0.069}=\frac{0.031}{0.069}=0.449$

$\text{Thus, mole fraction of benzene in vapour phase}=\frac{0.449\times 0.098}{0.06}=0.73$