A student titrates 25.00 mL of oxalic acid of concentration 0.0500 mol/L with potassium…

Step 1. $n(\text{oxalate}) = 0.0500\ \text{mol/L} \times 0.02500\ \text{L} = 1.25\ \text{mmol}$.

Step 2. $n(\text{MnO}_4^-) = 1.25 \times \dfrac{2}{5} = 0.500\ \text{mmol}$.

Step 3. $c = \dfrac{0.500\ \text{mmol}}{20.00\ \text{mL}} = 0.0250\ \text{mol/L}$.

Step 4. $M(\text{KMnO}_4) = 158\ \text{g/mol}$, so the mass is $0.0250 \times 158 = 3.95\ \text{g/L}$.