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KMnO(4) (m.w.=158) oxidises oxalic acid ...

`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)` ?

Text Solution

Verified by Experts

The correct Answer is:
31.6
`MnO_(4)^(-) + 8H^(+) +5e^(-) to Mn^(2+) + 4H_(2)O`
Equivalent weight of `KMnO_(4)` in acidic medium
` = ("Molecular weight")/(5)`
` = (158)/(5) = 31.6`
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