Question

$KMnO_4$ (mol. wt $= 158$) oxidizes oxalic acid in acidic medium to $CO_2$and water as follows.
$5C_2O_4^{2-} + 2MnO_4^- + 16H^+ \to 10CO_2 + 2Mn^{2+} + 8H_2O$
What is the equivalent weight of $KMnO_4$?

• A. 158
• B. 31.6
• C. 39.5
• D. 79

Answer 1

Answer: (B)

$MnO _{4}^{-}+8 H ^{+}+5 e ^{-} \longrightarrow Mn ^{2+} +4 H _{2} O $
Eq. wt of $KMnO _{4} $ in acidic medium $=\frac{ \text{Mol . wt }}{5} $
$=\frac{158}{5}=31.6 $