To a `25 mL` of `H_(2)O_(2)` solution, excess of acidified solution of `KI` was added. The iodine liberated required `20 mL` of 0.3 N `Na_(2)S_(2)O_(3)` solution. Calculate the volume strength of `H_(2)O_2` solution.
Strategy : Volume strength of `H_(2)O_(2)` solution is related to its normality by the following relation
Volume strength `(V)=5.6xx"Normality" (N)`
where, Normality`=((meq)H_(2)O_(2))/V_(mL)`
According to the law of equivalence
`(meq)_(Na_(2)S_(2)O_(3))=(meq)_(I_(2))=(meq)_(H_(2)O_(2))`

Explanation

`(meq)_(Na_(2)S_(2)O_(3))=V_(mL)xxN`
`=(20)(0.3)=6`
`=(meq)_(H_(2)O_(2))`
Thus, normality of `H_(2)O_(2)=(meq)_(H_(2)O_(2))//V_(mL)`
`6/25eq L^(-1)`
Volume strength `(V)=5.6xx "Normality"`
`=(5.6)(6//25)`
`=1.344`