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Chemistry · Solutions and Colligative Properties

JEE Main 2026 — 6 April, Evening Shift — Question 49

Given below are two statements :

Chamber 1Chamber 2
18 g glucose in 100 mL aqueous solutionSemi permeable membrane30 g glucose in 250 mL aqueous solution

Statement I: H2O\mathrm{H}_{2} \mathrm{O} molecules move from the chamber 1 to chamber 2.

Statement II: The osmotic pressure of a solution prepared by dissolving 50 mg of potassium sulphate (molar mass =174 g/mol=174 \mathrm{~g} / \mathrm{mol} ) in 2 L of water (at 27∘C27^{\circ} \mathrm{C} ) is 0.0107 bar. (Given : R=0.083dm3\mathrm{R}=0.083 \mathrm{dm}^{3} bar K−1 mol−1\mathrm{K}^{-1} \mathrm{~mol}^{-1} and assume complete dissociation of electrolyte)

In the light of the above statements, choose the correct answer from the options given below :

  1. Option A:

    Both Statement I and Statement II are true

  2. Option B:

    Both Statement I and Statement II are false

  3. Option C:

    Statement I is true but Statement II is false

  4. Option D:

    Statement I is false but Statement II is true

    Correct

Answer: D

Step-by-step solution

On side (I) C1=18/1800.1=1M\mathrm{C}_{1}=\frac{18 / 180}{0.1}=1 \mathrm{M} On side (II) C2=30/180250/1000=23M;{π1>π2}\mathrm{C}_{2}=\frac{30 / 180}{250 / 1000}=\frac{2}{3} \mathrm{M} \quad ; \quad\left\{\pi_{1}>\pi_{2}\right\} Hence water will flow from chamber-II to Chamber-I. π=iCRT;i=3\pi=\mathrm{iCRT} \quad ; \mathrm{i}=3 for K2SO4\mathrm{K}_{2} \mathrm{SO}_{4} And (π=3×50×10−3/1742)×0.083×300\left(\pi=3 \times \frac{50 \times 10^{-3} / 174}{2}\right) \times 0.083 \times 300 =0.0107=0.0107 bar.

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
6 April, Evening Shift
Subject
Chemistry
Chapter
Solutions and Colligative Properties
Topic
Solid in Liquid Solutions (Colligative Properties)
Given below are two statements : Chamber 1 Chamber 2 --- --- --- 18 g… | JEE Main 2026 PYQ with Solution · DhiX AI