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In the reactant of KMnO(4) with an oxala...

In the reactant of `KMnO_(4)` with an oxalate in acidic medium. `MnO_(4)^(-)` is reduced to `Mn^(2+)` and `C_(2)O_(4)^(2-)` is oxidised to `CO_(2)`. Hence, 50 ml of 0.02 M `KMnO_(4)` is equivalent to

A

100 ml of 0.05 M `H_(2)C_(2)O_(4)`

B

50 ml of 0.05 M `H_(2)C_(2)O_(4)`

C

25 ml of 0.2 M `H_(2)C_(2)O_(4)`

D

50 ml of 0.10 M `H_(2)C_(2)O_(4)`

Text Solution

Verified by Experts

The correct Answer is:
B
eq. `KMnO_(4)` = eq `C_(2)O_(4)^(2-)`
`50xx0.02xx5=MxxVxx2`
`implies MxxV=2.5=50.0.05`
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