Equivalent weight of `MnO_(4)^(ɵ)` in acidic neutral and basic media are in ratio of:

Answer

0.6

Explanation

`MnO_(4)^(-) + 8H^(+) + 5e^(-) to Mn^(2+) + 4H_(2)O`
If M is mol. Wt of `KMnO_(4)` then its Eq. wt. in acidic, basic are
`M/5: M/3` or `3:5=0.6`