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

Answer: (C)

$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}4H_{2}O$
Gain electrons $=5$
Molecular weight $=M$
Equivalent weight $=\frac{\text{molecular weight}}{\text{gain electron}}$
$=\frac{M}{5}$