The mass of sodium acetate $\left(\mathrm{CH}_3 \mathrm{COONa}\right)$ required to prepare 250 mL of 0.35 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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Solution
Moles = Molarity × Volume in litres
=0.35×0.25
Mass = moles × molar mass
=0.35×0.25×82.02=7.18" " g
Ans. 7
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