The mass of sodium acetate $\left(\mathrm{CH}_3 \mathrm{COONa}\right)$ required to prepare $250 \mathrm{~mL}$ of $0.35 \mathrm{~M}$ aqueous solution is ________ g. (Molar mass of $\mathrm{CH}_3 \mathrm{COONa}$ is $82.02 \mathrm{~g} \mathrm{~mol}^{-1}$)
Answer (integer)
7
Solution
<p>$$\begin{aligned}
& \text { Moles }=\text { Molarity } \times \text { Volume in litres } \\
& =0.35 \times 0.25 \\
& \text { Mass }=\text { moles } \times \text { molar mass } \\
& =0.35 \times 0.25 \times 82.02=7.18 \mathrm{~g}
\end{aligned}$$</p>
<p>Ans. 7</p>
About this question
Subject: Chemistry · Chapter: Some Basic Concepts of Chemistry · Topic: Mole Concept
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