The equivalent weight of `KMnO_(4)` in (a) neutral medium, (b) acidic medium and (c ) alkaline medium is `M//..` ( where `M` is mol.wt. of `KMnO_(4)`)

Options

A. `158.0`
B. `79.0`
C. 52.7
D. 31.6

Answer

D

Explanation

In acidic medium, `MnO_4^(-) + 8H^(+) + 5e^(-) to Mn^(2+) + 4H_2O`
`therefore` Equivalent weight of `KMnO_4 = M/5=158/5`=31.6