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 of \(\mathbf{y \%}\) 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 \(\mathbf{x}\) is ____.

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Practical and Analytical Chemistry

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 KMnO_{4}

solution to reach the end point. Number of moles of

Fe^{2 +}

present in 250 mL solution is

x \times 1 0^{- 2}

(consider complete dissolution of

FeCl_{2}

). The amount of iron present in the sample of

y%

by weight.
(Assume :

KMnO_{4}

reacts only with

Fe^{2 +}

in the solution
Use : Molar mass of iron as

56 g mol^{- 1}

)

The value of

x

is ____.

Answer: 1.87

Solution

In acidic medium,

MnO_{4}^{-} + 8 H^{+} + 5 e^{-} \to Mn^{2 +} + 4 H_{2} O

(n-factor = 5).

Fe^{2 +} \to Fe^{3 +} + e^{-}

(n-factor = 1). Using the equivalence: moles of

Fe^{2 +}

in 25 mL × 1 = moles of

KMnO_{4}

× 5. Moles of

KMnO_{4}

= 0.03 × 12.5/1000 = 3.75 × 10⁻⁴. Moles of

Fe^{2 +}

in 25 mL = 5 × 3.75 × 10⁻⁴ = 1.875 × 10⁻³. Moles in 250 mL = 10 × 1.875 × 10⁻³ = 1.875 × 10⁻². So

x = 1.875

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Numerical answer

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JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Practical and Analytical Chemistry

Solution

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