Chemistry · Solutions and Colligative Properties

JEE Main 2026 — 24 January, Morning Shift — Question 51

A solution is prepared by dissolving 0.3 g of a nonvolatile non-electrolyte solute ' A ' of molar mass $60 \mathrm{~g} \mathrm{~mol}^{-1}$ and 0.9 g of a non-volatile nonelectrolyte solute ' B ' of molar mass $180 \mathrm{~g} \mathrm{~mol}^{-1}$ in $100 \mathrm{~mL} \mathrm{H}_{2} \mathrm{O}$ at $27^{\circ} \mathrm{C}$ . Osmotic pressure of the solution will be[0pt] [Given : $\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ ]

Option A:
1.23 atm
Option B:
2.46 atm
Correct
Option C:
0.82 atm
Option D:
1.47 atm

Answer: B

Step-by-step solution

Mass of solute ' A ' $=0.3 \mathrm{~g}$ Moles of solute ${ }^{\prime} \mathrm{A}^{\prime}=\frac{0.3 \mathrm{~g}}{60 \mathrm{~g} / \mathrm{mol}}=\frac{1}{200} \mathrm{~mol}$ Mass of solute ${ }^{\prime} \mathrm{B}^{\prime}=0.9 \mathrm{~g}$

Moles of solute ${ }^{\prime} \mathrm{B}^{\prime}=\frac{0.9 \mathrm{gm}}{180 \mathrm{~g} / \mathrm{mol}}=\frac{1}{200} \mathrm{~mol}$ Total molarity of all solutes

\begin{aligned} & =\frac{2 / 200}{100} \times 1000=\frac{1}{10} \mathrm{M} \therefore \pi & =\frac{1}{10} \times 0.082 \times 300 \pi & =2.46 \mathrm{~atm} . \end{aligned}

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Exam: JEE Main 2026
Paper: 24 January, Morning Shift
Subject: Chemistry
Chapter: Solutions and Colligative Properties
Topic: Solid in Liquid Solutions (Colligative Properties)

W ' g of a non-volatile electrolyte solid solute of molar mass ' M ' g mol^-1 when dissolved in 100 mL water, decreases vapour pressure of…

Answer: B

Step-by-step solution

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