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

7

Explanation

`H_(2)O_(2) + KI` shows the following change
`2I^(-) to I_(2) + 2e`
`O_(2)^(-) + 2e to 2O^(2-)`
or `H_(2)O_(2) + H_(2)SO_(4) + 2Kl to K_(2)SO_(4) + I_(2) + 2H_(2)O`
Now meq. Of `H_(2)O_(2)`=meq. Of `I_(2)`
`Nxx3.2=(0.508 g)/(254//2)xx1000 :. N_(H_(2)O_(2)) =4/3.2 =1.25 eq//L`
`:. W_(H_(2)O_(2))=(1.25xx34)/2=21.25 g//L =0.02125 g//mL`
`:. 68 g H_(2)O_(2)` gives `O_(2)` at NTP =22400 mL
`:. ` 0.02125 g or 1mL `H_(2)O_(2)`
gives `O_(2) =(22400xx0.02125)/68 =7mL`
`:. ` volume strength of `H_(2)O_(2) =7mL`