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

`H_(2)O_(2) rarr H_(2)O + [O]`
`(2KI + H_(2)SO_(4) + [O] rarr K_(2)SO_(4) + H_(2)O + I_(2))/(2KI + H_(2)SO_(4) + H_(2)O_(2) rarr K_(2)SO_(4) + 2H_(2)O + I_(2))`
Liberated moles of `I_(2) = 0.508 xx 10^(3)/254 = 2`
Meq. of `I_(2) = 2 xx 2 = 4`
Meq. of liberated `I_(2)` = Meq. of used `H_(2)O_(2)`
Meq. of `H_(2)O_(2) = 4 `
Weight of `H_(2)O_(2) = "Meq." xx "Eq. wt. of" H_(2)O_(2) xx 10^(-3) = 4 xx 17 xx 10^(-3)g = 0.068 g`
Strength of `H_(2)O_(2) = 0.068 xx 1000/5 g//L = 13.6 g//L`
60.7 g `H_(2)O_(2)` is used for 20 volume `H_(2)O_(2)`
`13.6 g//L` strength of `H_(2)O_(2)` is used for = `20 xx 13.6/60.7 volume = 4.48 volume`