The mass of sodium acetate (CH 3COONa) required to prepare 250\ mL of 0.35\ M aqueous solution is \ \ \ \ \ g . [Given: molar mass of CH 3COONa = 82.02\ g\ mol -1 ]

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

JEE Main 2024 — 30 January, Shift 1 — Question 77

The mass of sodium acetate $\mathrm{(CH_3COONa)}$ required to prepare $250\ \mathrm{mL}$ of $0.35\ \mathrm{M}$ aqueous solution is $\_\_\_\_$ $\ \mathrm{g}$ .

[Given: molar mass of $\mathrm{CH_3COONa} = 82.02\ \mathrm{g\ mol^{-1}}$ ]

Answer: 7.18

Numerical answer — enter this value.

Step-by-step solution

Molarity relation, $\mathrm{M = \frac{n}{V}}$

\mathrm{n = M \times V}

Volume $= \mathrm{250\ mL = 0.25\ L}$

Moles required, $\mathrm{n = 0.35 \times 0.25 = 0.0875\ mol}$

Mass relation, $\mathrm{mass = n \times molar\ mass}$

\mathrm{mass = 0.0875 \times 82.02 \approx 7.18\ g}

Thus, the required mass is $\mathrm{7.18\ g}$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 30 January, Shift 1
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry
Topic: Concentration Terms and Their Interconversion