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QJEEAdv21Ch1b95: JEE Advanced 2021 Chemistry - Class 11 JEEAdv21Ch1b95 Solution: A sample (5.6 g) containing iron is completely dissolved in…

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 KMnO4 solution to reach the end point. Number of moles of Fe2+ present in 250 mL solution is x × 10−2 (consider complete dissolution of FeCl2). The amount of iron present in the sample is y% by weight.

(Assume : KMnO4 reacts only with Fe2+ in the solution

Use : Molar mass of iron as 56 g mol−1)

The value of y is ______.

Step-by-Step Verified Solution

Mass of sample = 5.6 g

Number of moles of Fe = 1.88 × 10−2 mol

Molar mass of Fe = 56 g/mol

Mass of Fe = Moles of Fe × Molar mass

= 1.88 × 10−2 × 56 = 1.05 g

% Fe = MassofFeMassofsample×100=1.055.6×100

⇒ y = 18.75%

The value of y is 18.75.

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