Question

In alkaline medium,KMnO4 reacts as follows:2KMnO4+2KOH→2K2MnO4+H2O+O.Therefore,the equivalent weight of KMnO4,will be

Answer 1

Answer to your question is in the above image.

Answer 2

Answer : The equivalent weight of $KMnO_4$ will be 158 g/eq.

Explanation :

The given balanced reaction is,

$2KMnO_4+2KOH\rightarrow 2K_2MnO_4+H_2O+O$

The oxidation number of Mn in $KMnO_4$,

(+1) + (x) + 4(-2)=0

x = (+7)

The oxidation number of Mn in $K_2MnO_4$,

2(+1) + (x) + 4(-2)=0

x = (+6)

The change in oxidation number = 1

Formula used :

$\text{ Equivalent weight of }=\frac{\text{ Molecular weight }}{\text{ Change in oxidation number}}$

The molecular weight of $KMnO_4$ = 158 g/mole

$\text{ Equivalent weight of }KMnO_4=\frac{\text{ Molecular weight of }KMnO_4}{\text{ Change in oxidation number}}=\frac{158g/mole}{1eq/mole}=158g/eq$

Therefore, the equivalent weight of $KMnO_4$ will be 158 g/eq.