JEE MAIN - Chemistry (2024 - 30th January Morning Shift - No. 22)
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
7
Explanation
$$\begin{aligned} & \text { Moles }=\text { Molarity } \times \text { Volume in litres } \\ & =0.35 \times 0.25 \\ & \text { Mass }=\text { moles } \times \text { molar mass } \\ & =0.35 \times 0.25 \times 82.02=7.18 \mathrm{~g} \end{aligned}$$
Ans. 7
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