Chemistry · Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

JEE Main 2024 — 6 April, Shift 1 — Question 72

The density of $x\ \mathrm{M}$ solution of $\mathrm{NaOH}$ is $1.12\ \mathrm{g\ mL^{-1}}$ while in molality the concentration of the solution is $3\ \mathrm{m}$ . Then $x$ is $\_\_\_\_$ .

[Given: Molar mass of $\mathrm{NaOH} = 40\ \mathrm{g\ mol^{-1}}$ ]

Option A:
3.5
Option B:
3
Correct
Option C:
3.8
Option D:
2.8

Answer: B

Step-by-step solution

Take $1\ kg$ solvent.

Molality $= \mathrm{3\ m = 3\ mol\ solute}$

Mass of solute $= \mathrm{3 \times 40 = 120\ g}$

Mass of solution $= \mathrm{1000 + 120 = 1120\ g}$

Volume $= \mathrm{\frac{1120}{1.12} = 1000\ mL = 1\ L}$

Molarity, $\mathrm{x = 3\ M}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 6 April, Shift 1
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry
Topic: Concentration Terms and Their Interconversion