In order to prepare one litre normal solution of ${\mathrm{KMnO}}_4,\mathrm{how}\mathrm{many}\mathrm{grams}\mathrm{of}{\mathrm{KMnO}}_4$ are required if the solution is used in acidic medium for oxidation

For One litre normal solution , Weight = Equivalent weight (Since , ) $\stackrel{+7}{\mathrm{Mn}}{\mathrm{O}}_4^{-}\stackrel{{\mathrm{H}}^{+}}{\to }{\mathrm{Mn}}^{+2} ; \mathrm{E}=\frac{\mathrm{M}}{\mathrm{Change} \mathrm{in} \mathrm{oxidation} \mathrm{state}}$

In Acidic medium ,  $\mathrm{N}=\frac{\mathrm{W}}{\mathrm{E}}\times \frac{1}{\mathrm{V}\left(\mathrm{l}\right)}\Rightarrow 1=\frac{\mathrm{W}}{\mathrm{E}}\times \frac{1}{1}$

$\mathrm{E}=\frac{\mathrm{M}}{5}=\frac{158}{5}=31.6\mathrm{g}.$