Question

$KMnO_{4}$ (mol. wt. = 158) oxidizes oxalic acid in acidic medium to $CO_{2}$ and water as follows

$5C_{2}O_{4}^{2 -}+2MnO_{4}^{-}+16H^{+} \rightarrow 10CO_{2}+2Mn^{2 +}+8H_{2}O$

What is the equivalent weight of $KMnO_{4}$ ?

• A. 158
• B. 31.6
• C. 39.5
• D. 79

Answer 1

Answer: (B)

$MnO_{4}^{-}+8H^{+}+5e^{-} \rightarrow Mn^{2 +}+4H_{2}O$

Equivalent weight of $KMnO_{4}$ in acidic medium $=\frac{m o l e c u l a r w e i g h t}{5}=\frac{158}{5}=31.6$