Question

Equal moles of benzene and toluene are mixed the V.P. of benzene and toluene in pure state are $700$ and $600mmHg$ 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 $\chi =\frac{p_{\text{benzene }}}{p_{\text{total }}}$
${\chi }_{\text{berzene }}=\frac{\frac{1}{2}}{\frac{1}{2}+\frac{1}{2}}=\frac{1}{2}$
${\chi }_{\text{berzene }(g)}=\frac{p_{\text{benzene }}^{\circ }{x}_{\text{benzene }}}{{\left(P^{\circ }x\right)}_{\text{benzene }}+{\left(p^{\circ }x\right)}_{\text{toluene }}}$
${\chi }_{\text{toluene }}=\frac{\frac{1}{2}}{\frac{1}{2}+\frac{1}{2}}=\frac{1}{2}$
$=\frac{700\times \frac{1}{2}}{700\times \frac{1}{2}+600\times \frac{1}{2}}=\frac{350}{350+300}=0.53$