`100mL` of `H_2O_2` is oxidised by `100mL` of `0.01M KMnO_4` in acidic medium `(MnO_4^(ɵ)` reduced to `Mn^(2+)`). `100mL` of the same `H_2O_2` is oxidised by `VmL` of `0.01M KMnO_4` in basic medium. Hence `V` is

100 mL of H_(2)O_(2) is oxidized by 100 mL of 1M KMnO_(4) in acidic medium ( MnO_(4)^(-) reduced to Mn^(+2) ) 100 mL of same H_(2)O_(2) is oxidized by v mL of 1M KMnO_(4) in basic medium ( MnO_(4)^(-) reduced to MnO_(2) ). Find the value of v:

100 mL of H_(2)O_(2) is oxidized by 100 mL of 1M KMnO_(4) in acidic medium ( MnO_(4)^(-) reduced to Mn^(+2) ) 100 mL of same H_(2)O_(2) is oxidized by v mL of 1M KMnO_(4) in basic medium ( MnO_(4)^(-) reduced to MnO_(2) ). Find the value of v:

100mL of 1M KMnO_(4) oxidised 100mL of H_(2)O_(2) in acidic medium ( when MnO_(4)^(-) is reduced to Mn^(2+) ) , voluem of same KMnO_(4) required to oxidise 100mL of H_(2)O_(2) in basic medium ( when MnO_(4)^(-) is reduced to MnO_(2) ) will be :

10 g sample of H_(2)O_(2) just decolorised 100 ml of 0.1 M KMnO_(4) in acidic medium % by mass of H_(2)O_(2) in the sample is

10 g sample of H_(2)O_(2) just decolorised 100 ml of 0.1 M KMnO_(4) in acidic medium % by mass of H_(2)O_(2) in the sample is

100 mL of NaHC_(2)O_(4) required 50 mL of 0.1M KMnO_(4) solution in acidic medium. Volume of 0.1M NaOH required by 100 mL of NaHC_(2)O_(4) is :

100 mL of NaHC_(2)O_(4) required 50 mL of 0.1M KMnO_(4) solution in acidic medium. Volume of 0.1M NaOH required by 100 mL of NaHC_(2)O_(4) is :

Moles of KHC_(2)O_(4) (potassium acid oxalate) required to reduce 100 mL of 0.02M KMnO_(4) in acidic medium (to Mn^(2+) ) is

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

Moles of KHC_(2)O_(4) (potassium acid oxalate) required to reduce 100ml of 0.02M KMnO_(4) in acidic medium (to Mn^(2+) ) is :