Question

A 4 mol mixture of Mohr’s salt and $F{e}_2(S{O}_4{)}_3$ requires 400 ml of $0.5{\text{M KMnO}}_4$ for complete oxidation in an acidic medium. The percentage of Mohr’s salt in the mixture is:

• A. 40 %
• B. 5 %
• C. 50 %
• D. 25 %

Answer 1

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
$KMn{O}_4$
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 $KMn{O}_4$ : 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%.