`KMnO_(4)` react with oxalic acid according to the equation, `2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+) rarr 2Mn^(2+)+ 10 CO_(2)+8H_(2)O`, here `20 ml` of `0.1 M KMnO_(4)` is equivalemt toA. 20 ml of 0.5 M `H_(2)C_(2)O_(4)`B. 50 ml of 0.1 M `H_(2)C_(2)O_(4)`C. 50 ml of 0.5 M `H_(2)C_(2)O_(4)`D. 20 ml of 0.1 M `H_(2)C_(2)O_(4)`

`KMnO_(4)` react with oxalic acid according to the equation, `2MnO_(4)

1.	`KMnO_(4)` react with oxalic acid according to the equation, `2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+) rarr 2Mn^(2+)+ 10 CO_(2)+8H_(2)O`, here `20 ml` of `0.1 M KMnO_(4)` is equivalemt toA. 20 ml of 0.5 M `H_(2)C_(2)O_(4)`B. 50 ml of 0.1 M `H_(2)C_(2)O_(4)`C. 50 ml of 0.5 M `H_(2)C_(2)O_(4)`D. 20 ml of 0.1 M `H_(2)C_(2)O_(4)`
Answer» Correct Answer - B `KMnO_(4)` Oxialic acid `(M_(1)V_(1))/n_(1)=(M_(2)V_(2))/n_(2), (20xx0.1)/2=(M_(2)V_(2))/5, M_(2)V_(2)=5` for (b) `M_(1)V_(1)=50xx0.1=5`

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