In an oxidation - reduction reaction, ${\mathrm{MnO}}_4^{-}$ ion is converted to ${\mathrm{Mn}}^{+2}$. What is the number of equivalents of ${\mathrm{KMnO}}_4$ (Mol. wt - 158) present in 250 ml. of 0.04 N ${\mathrm{KMnO}}_4$ ?  

The chemical reaction for the provided question is:

${\mathrm{MnO}}_4{}^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}{\mathrm{Mn}}^{2+}+4{\mathrm{H}}_2\mathrm{O}$

Change in oxidation state is +7-(+2)=5n

Let the mass be $\mathrm{m}$

Molarity is $\frac{n}{ \mathrm{V}}$ where
$\mathrm{n}=$ no. of moles $=\frac{\mathrm{m}}{158}$
$\mathrm{V}=$ volume of solution in litre $=\frac{250}{1000}=0.25$

Molarity $=0.04$
$⟹\mathrm{m}=1.58 \mathrm{g}$
Normality is $\frac{n_{\mathrm{e}}}{\mathrm{V}}$ where
 ${\mathrm{n}}_{\mathrm{e}}$ is no of equivalents
 $\mathrm{V}$ is volume of solution in liter
 

Normality $=$ molarity $\times \mathrm{n}=5\times 0.04=0.2$
No of equivalents is $0.2\times 0.25=0.05$

Therefore, option A is the correct answer.