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A bottle of H(2)O is labelled as 10 vol ...

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 is

A

1 L

B

2 L

C

3 L

D

4 L

Text Solution

Verified by Experts

The correct Answer is:
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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