Question

If the molecular weight of ${KMnO}_4$ is taken as M, its equivalent weight in the reaction where it is converted to ${K_{2}MnO}_4$ is :

• A. M/2
• B. M/5
• C. M/4
• D. M

Answer 1

The correct option is D M

In $KMn{O}_4$,
Oxidation state of Mn is +7
$+1+x-8=0$
$x=+7$

In $K_{2}Mn{O}_4$,
Oxidation state of Mn is +6
$+2+x-8=0$
$x=6$

$\text{So valency factor}=7-6=1$

So equivalent weight $=$ molecular weight

$=M$

Hence, option D is correct.