Chemistry · Solutions and Colligative Properties

JEE Main 2026 — 4 April, Morning Shift — Question 67

A non-volatile, non-electrolyte solid solute when dissolved in 40 g of a solvent, the vapour pressure of the solvent decreased from 760 mm Hg to 750 mm Hg . If the same solution boils at 320 K , then the number of moles of the solvent present in the solution is $\_\_\_\_$ . (Nearest integer)[0pt] [Given : boiling point of the pure solvent $=319.5 \mathrm{~K}, \mathrm{~K}_{\mathrm{b}}$ of the solvent $=0.3 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ ]

Answer: 5

Numerical answer — enter this value.

Step-by-step solution

$\quad \frac{\mathrm{P}^{\circ}-\mathrm{Ps}}{\mathrm{Ps}}=\mathrm{i}$ . molality $\times \frac{(\text { M.solvent })}{1000}$ $\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{i} . \mathrm{K}_{\mathrm{b}}$ . molality $\Rightarrow$ molality $=\frac{0.5}{0.3}$ $($ Molecular Mass $)=\frac{600}{75} \mathrm{~g}$ Moles $=\frac{40}{600 / 75}=5$

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Exam: JEE Main 2026
Paper: 4 April, Morning Shift
Subject: Chemistry
Chapter: Solutions and Colligative Properties
Topic: Solid in Liquid Solutions (Colligative Properties)