Calculate the equivalent weight of KMnO4 and MnO2. And explain it.

Molecular mass of KMnO₄ = 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^- ---> MnO₄^2-
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

For, MnO₂ (molar mass = 87) , here  n = 3
So, equivalent weight = molar mass/ 3 = 87/3 = 29 gm/equivalent.