JEE Advanced 2021 · Paper 2, Chemistry question 12

Question Stem for Question Nos. 11 and 12

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 M $\ce{KMnO4}$ solution to reach the end point. Number of moles of $\ce{Fe^{2+}}$ present in 250 mL solution is $\mathbf x \times 10^{-2}$ (consider complete dissolution of $\ce{FeCl2}$). The amount of iron present in the sample is $\mathbf y$% by weight.

(Assume: $\ce{KMnO4}$ reacts only with $\ce{Fe^{2+}}$ in the solution

Use: Molar mass of iron as 56 g mol$^{-1}$)