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Chemistry · Solutions and Colligative Properties

JEE Main 2026 — 21 January, Evening Shift — Question 66

A substance ' X ' ( 1.5 g ) dissolved in 150 g of a solvent ' Y ' (molar mass =300 g mol−1=300 \mathrm{~g} \mathrm{~mol}^{-1} ) led to an elevation of the boiling point by 0.5 K . The relative lowering in the vapour pressure of the solvent ' Y ' is ____\_\_\_\_ ×10−2\times 10^{-2}. (Nearest integer)[0pt] [Given : Kb\mathrm{K}_{\mathrm{b}} of the solvent =5.0 K kg mol−1=5.0 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} ] Assume the solution to be dilute and no association or dissociation of X takes place in solution.

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

ΔTb=i×Kb×m \Delta \mathrm{T}_{\mathrm{b}}=\mathrm{i} \times \mathrm{K}_{\mathrm{b}} \times \mathrm{m} 0.5=i×m×50.5=\mathrm{i} \times \mathrm{m} \times 5 i×m=0.55=0.1\mathrm{i} \times \mathrm{m}=\frac{0.5}{5}=0.1 i×a=151000\mathrm{i} \times \mathrm{a}=\frac{15}{1000} (where a=\mathrm{a}= moles of solute) Now,

Po−PSPo=iXsolute =i×aa+150300=i×a1/2=15/10001/2=301000=3×10−2=3\begin{aligned} & \frac{\mathrm{P}_{o}-\mathrm{P}_{\mathrm{S}}}{\mathrm{P}^{o}}=\mathrm{i} \mathrm{X}_{\text {solute }}=\mathrm{i} \times \frac{\mathrm{a}}{\mathrm{a}+\frac{150}{300}} & =\mathrm{i} \times \frac{\mathrm{a}}{1 / 2}=\frac{15 / 1000}{1 / 2}=\frac{30}{1000}=3 \times 10^{-2}=3 \end{aligned}

Answer key and solution verified before publishing.

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

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