`KMnO_(4)` (m.w.=158) oxidises oxalic acid in acid medium to `CO_(2)` and water as follows :
`5C_(2)O_(4)^(2-)+2MnO_(4)^(-)+16H^(+) to 10 CO_(2)+2Mn^(2+)+8H_(2)O`
What is the equivalent weigth of `KMnO_(4)` ?

Options

A. 158
B. 31.6
C. 39.5
D. 79

Answer

B

Explanation

`{:(MnO_(4)^(-), to , Mn^(+)),("Oxidation number", , "Oxidation number"),(of Mn=+7,,of Mn=+2):}`
Equivalent mass of `KMnO_(4)=("Molecular mass")/("Change in oxidation number")`
`=(158)/(5)=31.6`