Calculate the mass of sodium acetate $(CH_{3}COONa)$ required to make $500 \,mL$ of $0.375 \,M$ aqueous solution. Molar mass of sodium acetate is $82.0245 \,g \,mol^{-1}$.
Molarity $(M)$ is defined as the number of moles of solute per liter of solution.
$0.375 \,M$ aqueous solution means $0.375$ moles of sodium acetate are present in $1000 \,mL$ of solution.
Number of moles in $500 \,mL = \frac{0.375 \,mol}{1000 \,mL} \times 500 \,mL = 0.1875 \,mol$.
Mass of sodium acetate = $\text{Number of moles} \times \text{Molar mass}$.
Mass = $0.1875 \,mol \times 82.0245 \,g \,mol^{-1} = 15.3796 \,g \approx 15.38 \,g$.
$0.375 \,M$ aqueous solution means $0.375$ moles of sodium acetate are present in $1000 \,mL$ of solution.
Number of moles in $500 \,mL = \frac{0.375 \,mol}{1000 \,mL} \times 500 \,mL = 0.1875 \,mol$.
Mass of sodium acetate = $\text{Number of moles} \times \text{Molar mass}$.
Mass = $0.1875 \,mol \times 82.0245 \,g \,mol^{-1} = 15.3796 \,g \approx 15.38 \,g$.
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