Two liquids A and B form an ideal solution at temperature T K. At T K, the vapour pressures of pure A and B are 55 and 15 kNm -2 respectively. What is the mole fraction of A in solution of A and B in equilibrium with a vapour in which the mole fraction of A is 0.8 ?

Chemistry · Solutions and Colligative Properties

JEE Main 2026 — 24 January, Evening Shift — Question 51

Two liquids A and B form an ideal solution at temperature T K. At T K, the vapour pressures of pure A and B are 55 and $15 \mathrm{kNm}^{-2}$ respectively. What is the mole fraction of A in solution of A and $B$ in equilibrium with a vapour in which the mole fraction of A is 0.8 ?

Option A:
0.5217
Correct
Option B:
0.48
Option C:
0.663
Option D:
0.34

Answer: A

Step-by-step solution

$\frac{Y_{A}}{Y_{B}}=\frac{P_{A}^{o}}{P_{B}^{0}} \cdot \frac{X_{A}}{X_{B}}$ $\frac{0.8}{0.2}=\frac{55}{15} \times \frac{\mathrm{X}_{\mathrm{A}}}{\mathrm{X}_{\mathrm{B}}}$ $\frac{\mathrm{X}_{\mathrm{A}}}{\mathrm{X}_{\mathrm{B}}}=\frac{60}{55}=\frac{12}{11}$ $\mathrm{X}_{\mathrm{A}}=\frac{12}{23}=0.5217$

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Exam: JEE Main 2026
Paper: 24 January, Evening Shift
Subject: Chemistry
Chapter: Solutions and Colligative Properties
Topic: Liquid in Liquid Solutions (Raoult's Law)

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