Question

If molecular weight of $KMn{O}_4$ is $M$ then its equivalent weight in acidic medium would be :

• A. $M$
• B. $\frac{M}{2}$
• C. $\frac{M}{5}$
• D. $\frac{M}{3}$

Answer 1

The correct option is C $\frac{M}{5}$

$Mn{O}_4^{-}+8H^{+}+5e^{-}⟶M{n}^{2+}+4H_{2}O$
Gained electrons = 5
Moleclar weight = M
Equivalent weight
$=\frac{\text{Molecular weight}}{\text{gained electrons}}=\frac{M}{5}$