The balanced chemical equation for the reaction between ferrous sulfate ($Fe^{2+}$) and potassium permanganate ($KMnO_4$) in an acidic medium is: $$8H^+ + 5Fe^{2+} + MnO_4^- \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O$$ From the stoichiometry of the reaction, 1 mole of $MnO_4^-$ reacts with 5 moles of $Fe^{2+}$. Given: Volume of $KMnO_4$ ($V_1$) = 6 mL = 0.006 L Molarity of $KMnO_4$ ($M_1$) = 0.02 M Number of moles of $KMnO_4$ used: $$ \begin{aligned} n_{KMnO_4} &= M_1 \times V_1 \\ &= 0.02 \times 0.006 \\ &= 0.00012 \text{ mol} \end{aligned} $$ Since 1 mole of $MnO_4^-$ reacts with 5 moles of $Fe^{2+}$, the moles of $Fe^{2+}$ present in the sample are: $$ \begin{aligned} n_{Fe^{2+}} &= 5 \times n_{KMnO_4} \\ &= 5 \times 0.00012 \\ &= 0.0006 \text{ mol} \end{aligned} $$ The molar mass of $FeSO_4 \cdot 7H_2O$ is approximately $55.85 + 32.06 + 4 \times 16 + 7 \times 18 = 277.91 \text{ g/mol}$. However, using the standard value of 278 g/mol or the calculation provided in the context (151.85 + 126 = 277.85 g/mol): Mass of pure $FeSO_4 \cdot 7H_2O$: $$ \begin{aligned} m &= n_{Fe^{2+}} \times 277.85 \\ &= 0.0006 \times 277.85 \\ &= 0.16671 \text{ g} \end{aligned} $$ Thus, the amount of pure ferrous sulfate in the sample is 0.16671 g.