JEE Advanced 2021 — Paper 2 — Question 39

A sample ( 5.6 g ) containing iron is completely dissolved in cold dilute HCl to prepare a 250 mL of solution. Titration of 25.0 mL of this solution requires 12.5 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 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 $56 \mathrm{~g} \mathrm{~mol}^{-1}$ )

The value of $\mathbf{y}$ is $\qquad$ .