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

4.48

Explanation

`2Kl + H_(2)SO_(4)+ H_(2)O_(2) to K_(2)SO_(4)+ 2H_(2)O + I_(2)`
From the above equation
`H_(2)O_(2) -=I_(2)` or 34g of `I_(2)`
`therefore 0.508 g` of `I_(2)` will be liberated from `H_(2)O_(2)`
`=(34)/(254)xx0.508`
`=0.068` g
The decomposition of `H_(2)O_(2)` occurs as
`underset(2xx34=68 g )(2H_(2)O_(2))to underset(22400 cm^(3) " at " NTP)(2H_(2)O) + O_(2)`
`therefore 0.068 of `H_(2)O_(2)` upon decomposition will give `O_(2)=(22400)/(68)xx0.068`
`=22.4 cm^(3)`
Now `5.0cm^(3)` of `H_(2)O_(2)` solution gives `O_(2)`
`=22.4 cm^(3)` at STP
`therefore 1.0 cm^(3) ` of `H_(2)O_(2)` solution will give `O_(2)`
`=(22.4)/(5)=4.48 cm^(8)` at STP