A 4 mol mixture of Mohr’s salt and $F e_{2} (S O_{4})_{3}$ requires 400 ml of $0.5 {M KMnO}_{4}$ for complete oxidation in an acidic medium. The percentage of Mohr’s salt in the mixture is:

40 %

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5 %

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50 %

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25 %

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Solution

The correct option is D 25 %
In this redox reaction, Iron(II) in Mohr's salt is oxidised to Iron(III) and Mn is reduced from +7 to +2.
We know,
gram equivalents of Mohr's salt = gram equivalents of
$K M n O_{4}$
As change in O.S of Fe is +1 the n-factor will be 1
and let x moles of Mohr's salt is present in the solution
Mn is changing from +7 to +2 hence n- factor will be +5
So gram-equivalents of $K M n O_{4}$ : molarity * volume
(in litres) * n-factor
= 0.5 * $\frac{400}{1000}$ * 5
=1
So from eq. 1 moles of Mohr's salt =1
So moles of $F e_{2} (S O_{4})_{3}$ is 3
hence mole fraction of Mohr's salt is $\frac{1}{1 + 3}$
= 0.25
and in percentage of Mohr's salt in mixture is 25%.

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A 4 mol mixture of Mohr’s salt and Fe2(SO4)3 requires 400 ml of 0.5 M KMnO4 for complete oxidation in an acidic medium. The percentage of Mohr’s salt in the mixture is:

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