Question

The equivalent weight of KMnO₄ (in acid medium) is (At. wt. of K = 39, Mn = 55)

(a) 158

(b) 15.8

(c) 31.6

(d) 3.16

Answer 1

Correct option is (c) 31.6

The oxidation half reaction of KMnO₄ in acidic medium: 

MnO₄^- + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O

Hence, Eq. mass of KMnO₄ = Mol. mass / 5

= \(\frac{158}5\)

= 31.6