Chemistry · Solutions and Colligative Properties

JEE Main 2024 — 9 April, Shift 2 — Question 75

The vapour pressure of pure benzene and methyl benzene at $27^{\circ} \mathrm{C}$ is given as 80 Torr and 24 Torr, respectively. The mole fraction of methyl benzene in vapour phase, in equilibrium with an equimolar mixture of those two liquids (ideal solution) at the same temperature is $\qquad$ $\times 10^{-2}$ (nearest integer)

Answer: 23

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{X}_{\text {methylbenzene }}=0.5$

$\mathrm{Y}_{\text {methylbenzene }}=\frac{P_{\text {methylbenzene }}}{P_{\text {total }}}$

$\mathrm{Y}_{\text {methylbenzene }}=\frac{0.5 \times 24}{0.5 \times 80+0.5 \times 24}$

$=\frac{12}{40+12}=0.23=23 \times 10^{-2}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 9 April, Shift 2
Subject: Chemistry
Chapter: Solutions and Colligative Properties
Topic: Liquid in Liquid Solutions (Raoult's Law)