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`.

Explanation

Meq.of `H_(2)O_(2)="Meq.of" I_(2)`
`(w//17)xx1000=[0.508//(254//2)]xx1000`
`therefore w=0.068g`
`H_(2)O_(2)rarrH_(2)O+1//2O_(2)`
`because 34 g H_(2)O_(2)` gives `11.2 "litre"O_(2)`,
`therefore 0.068g` gives`(11.2xx0.068)//34=22.4 mL O_(2)`
`therefore` Volume strength of `H_(2)O_(2)=22.4//5=4.48%`