what is the equivalent weight of kmn04 in acidic, basic,neutral medium

Molecular mass of KMnO

4

= 158 gm
In acidic medium, no. of electrons involved n is:
 ​MnO4-+ 8H++5e- ---> Mn+2 + 4H2O

So, equivalent mass = molar mass/ electrons involved= 158/ 5 = 31.6 gm/ equivalent

In strongly basic medium, 
​MnO4- + 1e- ---> MnO42-

So, n = 1 , equivalent weight = 158/1 = 158gm/ qeuivalent

In neutral medium
MnO4- + 4H2O +3e- ---> MnO2 + 4OH-

Here, n = 3 so, equivalent weight = 158/3 = 52.66 gm/equivalent

Regards

Arun (askIITian forum expert)