A solution at $20^{\circ} \mathrm{C}$ is composed of $1.5 \mathrm{~mol}$ of benzene and $3.5 \mathrm{~mol}$…

A solution at $20^{\circ} \mathrm{C}$ is composed of $1.5 \mathrm{~mol}$ of benzene and $3.5 \mathrm{~mol}$ of toluene. If the vapour pressure of pure benzene and pure toluene at this temperature are $74.7$ torr and $22.3$ torr, respectively, then the total vapour pressure of the solution and the benzene mole fraction in equilibrium with it will be, respectively :
  1. $35.8$ torr and $0.280$
  2. $38.0$ torr and $0.589$
  3. $30.5$ torr and $0.389$
  4. $30.5$ torr and $0.480$

Solution

$\quad$ Total V.P. of solution $=P_{A}^{\circ} x_{A}+P_{B}^{\circ} x_{B}$
Given, $P_{A}^{\circ}=74.7$ torr, $P_{B}^{\circ}=22.3$ torr
$\mathrm{n}_{\text {benzene }}=1.5 \mathrm{~mol}, \mathrm{n}_{\text {toluene }}=3.5 \mathrm{~mol}$
$\mathrm{n}_{\text {solution }}=1.5+3.5=5 \mathrm{~mol}$
$\mathrm{x}_{\mathrm{A}}=\frac{n_{\text {benzene }}}{n_{\text {solution }}}=\frac{1.5}{5}$
Total V.P. of solution
$=\left(\frac{1.5}{5} \times 74.7+\frac{3.5}{5} \times 22.3ight)$ torr
$=(22.4+15.6)$ torr $=38$ torr
Mole fraction of benzene in vapour form $=$
$\frac{22.4}{38}=0.589$ ,

Asked in: JEE-TOPICTESTS-CHEMISTRY

Practice more SOLUTIONS questions on Aicharya