Question

In an alkaline medium, $KMn{O}_4$ reacts as follows -

$2KMn{O}_4+2KOH\to 2K_{2}Mn{O}_4+H_{2}O+O$

Equivalent weight of $KMn{O}_4$ will be:

• A. 31.6
• B. 52.7
• C. 7
• D. 158

Answer 1

Answer: (D)

158

Equivalent mass of $KMn{O}_4=\frac{Molecularweight}{No.ofelectronsgainedbyMn{O}_4}$

$Mn{O}_4^{-}$ $+$ $e^{-}$ $\to$ $Mn{O}_4^{2-}$

The number of electrons gained per molecule of $KMn{O}_4$ is $1$.

So, the equivalent mass is $158$.