3.92 g of ferrous ammonium sulphate are dissolved in 100 ml water. 20ml of this solution requires 18ml of potassium permanganate during titration for complete oxidation. The weight of KMnO₄ present in one liter of the solution is

Molecular mass of Ferrous ammonium sulphate ((NH₄​)₂​Fe(SO_4​)2​.6H_2​O)=392g 

No. of moles of Ferrous ammonium sulphate ((NH4​)2​Fe(SO4​)2​.6H2​O) =$\frac{3.92}{392}=0.01\mathrm{moles}$ 

Normality of Ferrous ammonium sulphate ((NH₄​)₂​Fe(SO_4​)2​.6H_2​O) solution=$\frac{0.01\times 1000}{100}=0.1\mathrm{M}$

${\mathrm{MnO}}_4^{-}+{\mathrm{Fe}}^{+2}\stackrel{{\mathrm{H}}^{+}}{\to }{\mathrm{Mn}}^{+2}+{\mathrm{Fe}}^{+3}$

Normality of Ferrous ammonium sulphate=M(n-factor)=0.1(1)=0.1N

20 ml of Ferrous ammonium sulphate ((NH₄​)₂​Fe(SO_4​)2​.6H_2​O)  neutralizes 18 ml of KMnO₄​.

$$\begin{gathered} {\mathrm{N}}_1{\mathrm{V}}_1={\mathrm{N}}_2{\mathrm{V}}_2 \\ 20(0.01)=18{\mathrm{N}}_{{\mathrm{KMnO}}_4} \\ {\mathrm{N}}_{{\mathrm{KMnO}}_4}=\frac{20(0.1)}{18}=\frac{1}{9}\mathrm{N} \\ \mathrm{N}=\frac{\mathrm{W}}{\mathrm{E}}\times \frac{1}{\mathrm{V}\left(\mathrm{l}\right)} \\ \frac{1}{9}=\frac{\mathrm{W}}{31.6}\times \frac{1}{1} \left(\because {\mathrm{E}}_{{\mathrm{KMnO}}_4}=\frac{\mathrm{Mol}.\mathrm{wt}.}{\mathrm{n}-\mathrm{factor}}=\frac{158}{5}\right) \\ \mathrm{Wt}. \mathrm{of} {\mathrm{KMnO}}_4=\frac{31.6}{9}=3.476\mathrm{g} \end{gathered}$$