Answer: (b)
In acidic medium the number of electrons involved are 5.
Mno4 + 8H+ + 5e– → Mn2+ + 4H2O
Eq. mass of KMnO4 = 158/ 5 =31.6
In basic medium the number of electrons involved are 3.
Mno4 + 2H2O+ + 3e– → MnO2 + 4OH-
Eq. mass of KMnO4 = 158/ 3 = 52.7
Equivalent mass of KMnO4 in acidic and alkaline medium is 31.6 and 52.7 respectively.