Question

In $KMn{O}_4$ titration in acidic meadium, equivalent mass of $KMn{O}_4$ is:

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

Answer 1

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

In acidic medium $KMn{O}_4$ undergoes five electron change.
$\stackrel{+7}{Mn}{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$

Equivalent Mass =$\frac{\text{Molecular Mass}}{\text{No. of electron gain or lost}}=\frac{M}{5}$