Volume of 3\ M NaOH (formula weight 40\ g\ mol -1 ) which can be prepared from 84\ g of NaOH is \ \ \ × 10 -1 \ dm 3 .

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

JEE Main 2024 — 27 January, Shift 2 — Question 80

Volume of $3\ \mathrm{M}$ $\mathrm{NaOH}$ (formula weight $40\ \mathrm{g\ mol^{-1}}$ ) which can be prepared from $84\ \mathrm{g}$ of $\mathrm{NaOH}$ is $\_\_\_ \times 10^{-1}\ \mathrm{dm^3}$ .

Answer: 7

Numerical answer — enter this value.

Step-by-step solution

Moles of $\mathrm{NaOH}$ $\mathrm{= \dfrac{84}{40} = 2.1\ mol}$

Using molarity, $\mathrm{M = \dfrac{n}{V}}$

Volume of solution, $\mathrm{V = \dfrac{2.1}{3} = 0.7\ dm^3}$

Expressing in required form, we have

\mathrm{0.7\ dm^3 = 7 \times 10^{-1}\ dm^3}

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 27 January, Shift 2
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry
Topic: Concentration Terms and Their Interconversion

Answer: 7

Step-by-step solution

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