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JEE Advance2021
Paper-2
Question
A sample $(5.6 \mathrm{~g})$ 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 $\times \times 10^{-2}$ (consider complete dissolution of $\mathrm{FeCl}_2$ ). The amount of iron present in the sample is $\mathrm{y} \%$ 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 $x$ is $\_\_\_\_$ .
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JEE Advance
Chemistry
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