Solutions Question 9
Question 9 - 2024 (30 Jan Shift 1)
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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Answer (7)
Solution
Moles $=$ Molarity $\times$ Volume in litres
$=0.35 \times 0.25$
Mass $=$ moles $\times$ molar mass
$=0.35 \times 0.25 \times 82.02=7.18 \mathrm{~g}$
Ans. 7
