The mass of sodium acetate ( CH_3 COONa) required to prepare 250 mL of 0.35 M aqueous solution is ________ g. (Molar mass of CH_3 COONa is 82.02 g mol^-1)

Moles = Molarity x Volume in litres \ =0.35 x 0.25 \ Mass = moles x molar mass \ =0.35 x 0.25 x 82.02=7.18 g Ans. 7

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Chemistry

Exam

Chemistry

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 $82.02 \mathrm{~g} \mathrm{~mol}^{-1}$ )

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Explanation

$\begin{aligned} & \text { Moles }=\text { Molarity } \times \text { Volume in litres } \\ & =0.35 \times 0.25 \\ & \text { Mass }=\text { moles } \times \text { molar mass } \\ & =0.35 \times 0.25 \times 82.02=7.18 \mathrm{~g} \end{aligned}$