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

In the reactant of KMnO_(4) with an oxalate in acidic medium. MnO_(4)^

1.	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
Answer» 100 ML of 0.05 M `H_(2)C_(2)O_(4)` 50 ml of 0.05 M `H_(2)C_(2)O_(4)` 25 ml of 0.2 M `H_(2)C_(2)O_(4)` 50 ml of 0.10 M `H_(2)C_(2)O_(4)` Solution :EQ. `KMnO_(4)` = eq `C_(2)O_(4)^(2-)` `50xx0.02xx5=MxxVxx2` `implies MxxV=2.5=50.0.05`