Question

The equivalent weight of $KMn{O}_4$ in acidic medium is:

• A. 158
• B. 52.67
• C. 31.6
• D. 49

Answer 1

Answer: (C)

In acidic medium $KMn{O}_4$ reacts as
$2KMn{O}_2+2H_{2}S{O}_4⟶K_{2}S{O}_4+2MnS{O}_4$
$+3H_{2}O+5O$
or $Mn{O}_4^{-}⟶M{n}^{2+}$
Oxidation number of $Mn+7+2$
$\therefore$ change in oxidation number
$=7-2=5$
$\therefore$ equivalent weight of
$KMn{O}_2=\frac{\text{ formula weight of }KMn{O}_4}{4}$
$=\frac{158}{5}=31.6$