A student dissolves 1.96 g of Mohr's salt, FeSO₄(NH₄)₂SO₄· 6H₂O (392 g/mol), in water with a little…

Step 1. $n = \dfrac{1.96\ \text{g}}{392\ \text{g/mol}} = 5.00\ \text{mmol}$ in $100.0\ \text{mL}$, so $c = \dfrac{5.00\ \text{mmol}}{100.0\ \text{mL}} = 0.0500\ \text{mol/L}$.

Step 2. The portion holds $0.0500\ \text{mol/L} \times 0.0200\ \text{L} = 1.00\ \text{mmol}$ of iron(II).

Step 3. $n(\text{MnO}_4^-) = \dfrac{1.00}{5} = 0.200\ \text{mmol}$.

Step 4. $V = \dfrac{0.200\ \text{mmol}}{0.0200\ \text{mol/L}} = 10.0\ \text{mL}$.