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`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` isA. `100/3 mL`B. `500/3 mL`C. `300/5 mL`D. None |
Answer» Correct Answer - B `{:(MnO_(4)^(ɵ)+5e^(ɵ)toMn^(2+),("acidic")),(MnO_(4)^(ɵ)+3e^(ɵ)toMnO_(2),("basic")):}` `100 mL H_(2)O_(2)=100xx5N MnO_(4)^(ɵ)-=Vxx3N MnO_(2)` `N=500/3 mL` |
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