Question

Equal moles of benzene and toluene are mixed the V.P. of benzene and toluene in pure state are $700$ and $600 \,mm \,Hg$ respectively. The mole fraction of benzene in vapour state is

• A. 0.7
• B. 0.47
• C. 0.50
• D. 0.54

Answer 1

Answer: (D)

Total moles of benzene and toluene let $1$ mole
So, benzene $=\frac{1}{2}$, Toluene $=\frac{1}{2}$
Total $=\frac{1}{2}+\frac{1}{2}=1$
$p_{\text {benzene }}^{\circ}=700$
$p_{\text {Toluene }}^{\circ}=600$
Mole fraction of benzene in vapours $\chi=\frac{p_{\text {benzene }}}{p_{\text {total }}}$
$\chi_{\text {berzene }}=\frac{\frac{1}{2}}{\frac{1}{2}+\frac{1}{2}}=\frac{1}{2}$
$\chi_{\text {berzene }(g)}=\frac{p_{\text {benzene }}^{\circ} x_{\text {benzene }}}{\left(P^{\circ} x\right)_{\text {benzene }}+\left(p^{\circ} x\right)_{\text {toluene }}}$
$\chi_{\text {toluene }}=\frac{\frac{1}{2}}{\frac{1}{2}+\frac{1}{2}}=\frac{1}{2}$
$=\frac{700 \times \frac{1}{2}}{700 \times \frac{1}{2}+600 \times \frac{1}{2}}=\frac{350}{350+300}=0.53$