Question

The following reaction occurs in acidic medium
$KMn{O}_4+8H^{+}+5e^{-}\to K^{+}+M{n}^{2+}+4H_{2}O$
What is the equivalent weight of $KMn{O}_4$? (Molecular weight of $KMn{O}_4=158$)

• A. $79.0$
• B. $31.6$
• C. $158.0$
• D. $39.5$

Answer 1

Answer: (B)

$31.6$

For $KMn{O}_4$ In acidic medium

$E.W=\frac{M}{5}=\frac{158}{5}=31.6$