Equivalent mass of KMnO4 in acidic, basic and neutral are in the ratio of: (1) 3:5:15 (2) 5:3:1 (3) 5:1:3 (4) 3:15:5

Answer: (4)

Equivalent weight can be calculated by using the formula

Equivalent weight = Molar mass/number of electrons lost or gained (n)

The molecular mass of KMno4 = 158.04g

In acidic medium: MnO_4- + 8H +5e^– –> Mn₂ + 4H₂O

Here n = 5

Eq wt = 158.04/5

Eq wt = 31.61g/equivqlent

In basic medium: MnO₄⁻+ e⁻ → MnO₄²⁻

Here n = 1

Eq wt = 158.04/1

Eq wt = 158.04g/Equivalent

In neutral medium: Mno₄– + 4H⁺ + 3e⁻ –> MnO₂+ 2H₂O

Here n = 3

Eq wt = 158.04/3

Eq wt = 52.68g/Equivalent

Ratio of equivalent mass = 3:15:5