Q 1189167017.     `KMnO_4` reacts with oxalic acid as:


`MnO_4^− + C_2O_4^(2−) + H^+ → Mn^(2+) + CO_2 + H_2O`

Hence, `50` ml of `0.04` `M` `KMnO_4` in acidic medium is chemically equivalent to


A

`25` ml of `0.1` `M` `H_2C_2O_4`

B

`100` ml of `0.1` `M` `H_2C_2O_4`

C

`50` ml of `0.2` `M` `H_2C_2O_4`

D

`50` ml of `0.1` `M` `H_2C_2O_4`

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