Question

Equivalent weight of $KMnO_{4}$ acting as an oxidant in acidic medium is equal to.

• A. Molecular weight of $KMnO_{4}$
• B. $\frac{1}{2}$× Molecular weight of $KMnO_{4}$
• C. $\frac{1}{3}$× Molecular weight of $KMnO_{4}$
• D. $\frac{1}{5}$× Molecular weight of $KMnO_{4}$

Answer 1

Answer: (D)

$\frac{1}{5}$× Molecular weight of $KMnO_{4}$

Answer 2

$\frac{1}{5} \times \text{ molecular weigth of }KMnO_{4}$

as transfer of $5e^{-}$ takes place when $KMnO_{4}$ acts as oxidant in acidic medium.

$2KMnO_{4} + 3H_{2}SO_{4} \rightarrow K_{2}SO_{4} + 2MnSO_{4} + 3H_{2}O + 5O$