Chemistry · Solutions and Colligative Properties

JEE Main 2025 — 23 January, Evening Shift — Question 33

When a non-volatile solute is added to the solvent, the vapour pressure of the solvent decreases by 10 mm of Hg . The mole fraction of the solute in the solution is 0.2 . What would be the mole fraction of the solvent if decrease in vapour pressure is 20 mm of Hg ?

Option A:
0.6
Correct
Option B:
0.4
Option C:
0.2
Option D:
0.8

Answer: A

Step-by-step solution

$\because \mathrm{P}^{\circ}-\mathrm{P} \propto \mathrm{X}_{\text {solute }}$

and $\because 10 \propto 0.2$

$\therefore 20 \propto 0.4$

$\therefore \mathrm{X}_{\text {solvent }}=1-\mathrm{X}_{\text {solute }}$

$=1-0.4$

$=0.6$

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Exam: JEE Main 2025
Paper: 23 January, Evening Shift
Subject: Chemistry
Chapter: Solutions and Colligative Properties
Topic: Liquid in Liquid Solutions (Raoult's Law)

Consider a binary solution of two volatile liquid components 1 and 2 x 1 and y 1 are the mole fractions of component 1 in liquid and vapour…

Answer: A

Step-by-step solution

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