Question bank/JEE Advanced/Chemistry

A sample $(5.6 \mathrm{~g})$ containing iron is completely dissolved in cold dilute $\mathrm{HCl}$ to prepare a $250 \mathrm{~mL}$ of solution. Titration of $25.0 \mathrm{~mL}$ of this solution requires $12.5 \mathrm{~mL}$ of $0.03 \mathrm{M} \mathrm{KMnO}_4$ solution to reach the end point. Number of moles of $\mathrm{Fe}^{2+}$ present in $250 \mathrm{~mL}$ solution is $\mathbf{x} \times 10^{-2}$ (consider complete dissolution of $\mathrm{FeCl}_2$ ). The amount of iron present in the sample is $\mathbf{y} \%$ by weight.
(Assume: $\mathrm{KMnO}_4$ reacts only with $\mathrm{Fe}^{2+}$ in the solution
Use: Molar mass of iron as $56 \mathrm{~g} \mathrm{~mol}^{-1}$ ). What is the value of $\mathbf{y}$?

Source: JEEBench (IIT Delhi) / JEE Adv 2021 Paper 2