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)`
`therefore 0.508 g` of `I_(2)` will be liberated from
`H_(2)O_(2) = 34/254 xx 0.508 = 0.068` g
(b) The decomposition of `H_(2)O_(2)` occurs as
`underset(2 xx 34-= 68 g)(2H_(2)O_(2)) to underset(22400 cm^(3) "at NTP")(2H_(2)O) + O_(2)`
`O_(2) = 22400/68 xx 0.068 = 22.4 cm^(3)`
( c) Now `5.0 cm^(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^(3)` at STP
Thus, volume strength of given `H_(2)O_(2)` solution = 4.48.