Question

Find the equivalent weight of $KMn{O}_4$ in the given reaction, if the molar mass of $Mn{O}_4^{-}$ ion is 'M' ?
$\underset{}{Mn{O}_4^{-}}\to M{n}^{4+}$ (neutral medium)

• A. $22.7$
• B. $12.7$
• C. $32.7$
• D. $52.7$

Answer 1

The correct option is D $52.7$

In neutral medium
$\stackrel{(+7)}{KMn{O}_4}\to \stackrel{(+4)}{M{n}^{4+}}$
$n_f=|4-7|\times 1=3$
Here, only 3 electrons are gained, so eq. wt. $=\frac{Mol.wt.ofKMn{O}_4}{3}=\frac{158}{3}=52.7$

Note: It is important to note that $KMn{O}_4$ acts as an oxidant in every medium although with different strength which follows the order.

acidic medium > neutral medium > alkaline medium