Discussion
The mass of sodium acetate required to prepare 250 mL of 0.4 M aqueous solution is (Molar mass of sodium acetate = 82 g mol-1)
Answer: B
The mass required is 8.2 g.
Moles needed = molarity times volume in litres; mass = moles times molar mass:
\[ \begin{aligned} n &= 0.4 \times 0.250 = 0.1\ \text{mol} \\ m &= 0.1 \times 82 = 8.2\ \text{g} \end{aligned} \]
4.1 g and 16.4 g are half and double the right amount, as from using 125 mL or 500 mL. 0.82 g is ten times too small.
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