Question

A `5.0 mL` of solution of `H_(2)O_(2)` liberates `0.508 g` of iodine from acidified `KI` solution. Calculate the strength of `H_(2)O_(2)` solution in terms of volume strength at `STP`.

Answer 1

`mEq of I_(2)=0.508/127xx1000`
`mEq of KI=0.508/127xx1000`
`mEq of H_(2)O_(2)=0.508/127xx1000`
Normality of `H_(2)O_(2)=0.503/127xx1000/5`
For `H_(2)O_(2)`
`5.6xx` Normality =Volume strength
`:.` Volume strength =`5.6xx(0.508xx1000)/(127xx5)`
`=4.48 V`