Chemistry · Solutions and Colligative Properties

JEE Main 2025 — 2 April, Morning Shift — Question 2

A solution is made by mixing one mole of volatile liquid $A$ with 3 moles of volatile liquid $B$ . The vapour pressure of pure $A$ is 200 mm Hg and that of the solution is 500 mm Hg . The vapour pressure of pure $B$ and the least volatile component of the solution, respectively, are

Option A:
$600 \mathrm{~mm} \mathrm{Hg}, \mathrm{B}$
Option B:
$600 \mathrm{~mm} \mathrm{Hg}, \mathrm{A}$
Correct
Option C:
$1400 \mathrm{~mm} \mathrm{Hg}, \mathrm{A}$
Option D:
$1400 \mathrm{~mm} \mathrm{Hg}, \mathrm{B}$

Answer: B

Step-by-step solution

$p_{\text {total }}=p_{A}^{0} x_{A}+p_{B}^{0} x_{B}$

$500=200\left(\frac{1}{4}\right)+\mathrm{p}_{\mathrm{B}}^{\mathrm{o}}\left(\frac{3}{4}\right)$

$2000=200+3 p_{B}^{0}$

$\mathrm{p}_{\mathrm{B}}^{0}=600$

$\mathrm{p}_{\mathrm{A}}^{0}=200$

A is least volatile

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 2 April, Morning Shift
Subject: Chemistry
Chapter: Solutions and Colligative Properties
Topic: Liquid in Liquid Solutions (Raoult's Law)