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A bottle of H2O2 is labelled as 10 vol H...

A bottle of `H_2O_2` is labelled as `10 vol H_2O_2. 112mL` of this solution of `H_2O_2` is titrated against `0.04M` acidified solution of `KMnO_4`. Calculate the volume of `KMnO_4` in terms of litre.

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The correct Answer is:
1
`5.6 vol of H_(2)O_(2)=1N=`
`10 vol of H_(2)O_(2)=10/5.6 N`
`mEq of H_(2)O_(2)=mEq of MnO_(4)^(ɵ)`
`10/5.6xx112-=0.04xx5` (`n`factor)`xxV`
`V-=1000 mL =1L`
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