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A bottle of `H_(2)O` is labelled as 10 vol `H_(2)O_(2)`. 112 " mL of " this solution of `H_(2)O_(2)` is titrated against 0.04 M acidified solution of `KMnO_(4)` the volume of `KMnO_(4)` in litre isA. 1 LB. 2 LC. 3 LD. 4 L |
Answer» Correct Answer - A `5.6 V H_(2)O_(2)=1N` `10 V H_(2)O_(2)=(10)/(5.6)N` `m" Eq of "H_(2)O_(2)=m" Eq of "MnO_(4)^(ɵ)` `(10)/(5.6)xx112-=0.04xx5` ("n-factor")`xxV` `V-=1000ml=1L` |
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