Test Index

JEE Advanced 2021 Paper 2

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A sample

(5.6\,\mathrm{g})

containing iron is completely dissolved in cold dilute

\mathrm{HCl}

to prepare a

250\,\mathrm{mL}

of solution. Titration of

25.0\,\mathrm{mL}

of this solution requires

12.5\,\mathrm{mL}

0.03\,\mathrm{M}\,\mathrm{KMnO}_4

solution to reach the end point. Number of moles of

\mathrm{Fe}^{2+}

present in

250\,\mathrm{mL}

solution is

x \times 10^{-2}

(consider complete dissolution of

\mathrm{FeCl}_2

). The amount of iron present in the sample of

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}

)

Section: Chemistry

Question : 30 of 57

Marks: +1, -0

The value of

x

Solution:

5\,\mathrm{Fe}^{2+} + \mathrm{MnO}_4^{-} + 8\,\mathrm{H}^{+} \rightarrow 5\,\mathrm{Fe}^{3+} + \mathrm{Mn}^{2+} + 4\,\mathrm{H}_{2}\mathrm{O}

For

25\,\text{mL}

meq of

\mathrm{Fe}^{2+}=

meq of

\mathrm{MnO}_4^{-}=12.5 \times 0.03 \times 5

For

250\,\text{mL}

\mathrm{M}

moles of

\mathrm{Fe}^{2+} = \frac{12.5 \times 0.03 \times 5 \times 250}{25}

Moles of

\mathrm{Fe}^{2+} = \frac{18.75}{1000}\,\text{mol} = 18.75 \times 10^{-3} = 1.875 \times 10^{-3}

x = 1.875

Weight of

\mathrm{Fe}^{2+} = 1.875 \times 10^{-2} \times 56 = 1.05\,\text{g}

\%

Purity of

\mathrm{Fe}^{2+} = \frac{1.05}{5.6} \times 100 = 18.75\%

y = 18.75

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