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` reacts with acidified Kl solution as :
`2Kl + H_2 SO_4 + underset(5 cm^3)underset(34g)(H_2 O_2) to K_2 SO_4 + underset(0.508 g)underset(254g)(I_2) + 2H_2 O`
254 g of `I_2` is liberated by `H_2 O_2` =34 g
0.508 g of `I_2 ` is liberated by `H_2 O_2 =(34)/(254)xx0.508`
`=0.068 g`
`5 cm^3` of `H_2 O_2` contains `=0.068 g`
` 1 cm^3 ` of `H_2 O_2` contains `=(0.068)/(5) =0.0136 g`
Now `underset(68g)underset(2H_2 O_2)to 2H_2 O + underset(22400 mL "at S.T.P.")(O_2)`
68 g of `H_2 O_2` give `O_2` at S.T.P. = 22400 mL
`1 cm^3` or 0.0136 g of `H_2 O_2` give `O_2` at S.T.P.
=`(22400)/(68) xx 0.0136g `
`=4.48 ` mL of S.T.P.
`1 cm^3` of `H_2 O_2 ` at S.T.P. gives 4.48 mL of `O_2`
`therefore ` Strength of `H_2 O_2 =4.48` volumes.