Chemistry · Solutions and Colligative Properties

JEE Main 2026 — 6 April, Morning Shift — Question 50

When 0.25 moles of a non-volatile, non-ionizable solute was dissolved in 1 mole of a solvent the vapor pressure of solution was $\mathrm{x} \%$ of vapor pressure of pure solvent. What is $\mathrm{x} \%$ ?

Option A:
$50 %$
Option B:
$60 %$
Option C:
$70 %$
Option D:
$80 %$
Correct

Answer: D

Step-by-step solution

$\mathrm{P}_{\mathrm{s}}=\mathrm{P}^{0} \cdot \mathrm{X}_{\mathrm{A}}$ $\frac{\mathrm{P}_{\mathrm{S}}}{\mathrm{P}^{0}}=\mathrm{X}_{\mathrm{A}}=\frac{1}{1+0.25}=0.80$ $\%=0.8 \times 100=80$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 6 April, Morning Shift
Subject: Chemistry
Chapter: Solutions and Colligative Properties
Topic: Liquid in Liquid Solutions (Raoult's Law)