Calculate the equivalent weight of `KMnO_(4)` 

Explanation

`KMnO_(4)` acts as oxidant in acidic, basic and also in neutral medium.
In acidic medium :
`KMnO_(4)+8H^(+)+5e^(-)+MN^(2+)+4H_(2)O`
One molecule of `KMnO_(4)`gains five electrons. Hence, the equivalent weight of `KMnO_(4)` :
`"Eq. wt"=("Mol.wt.of "KMnO_(4))/(5)=(158.04)/(5)=31.608`
In neutral as well as weakly basic medium:
`KMnO_(4)+2H_(2)O+3e^(-)rarrK^(+)+4OH^(-)+MnO_(2)`
  One molecule of `KMnO_(4)` gains three electrons.
`"Eq. wt."=("Mol. Wt. of "KMnO_(4))/(3)=(158.04)/(3)=52.68`
In strongly alkaline medium : `MnO_(4)^(-)+e^(-)rarrMnO_(4)^(2-)`
One molecule of `KMnO_(4)` gains one electron.
`Eq. wt.=("Mol.wt.of" KMnO_(4))/(1)=158.04`