A `5.0cm^(3)` solutions of `H_(2)O_(2)` liberates of 0.508g of iodine from acidified KI solution. Calculate the volume strength of `H_(2)O_(2)` at N.T.P.

Explanation

Step I. Calcination of mass of `H_(2)O_(2)`.
`2KI+H_(2)SO_(4)+H_(2)O_(2) to K_(2)SO_(4)+2H_(2)+I_(2)`
`H_(2)O_(2)(34g)-=I_(2)(254g)`
`254"g of" I_(2) "is liberated from" H_(2)O_(2)=34g`
`therefore "0.508g of" I_(2) "is liberated from" H_(2)O =(34)/(245)xx0.508g=0.68g`
Step-II Calcination of the volume of oxygen at N.T.P
`underset(=68g)underset(2xx34)(2H_(2)O_(2)) to underset(at N.T.P)underset(22400cm^(3))(2H_(2)O+O_(2))`
`68"g of" H_(2)O_(2) "at N.T.P evolve oxygen"=22400cm^(3)`
`0.068"g of" H_(2)O_(2) "at N.T.P evolve oxygen"=(22400)/(68)xx0.068=22.4cm^(3)`
Step III. Calculation of volume strength of `H_(2)O_(2)`.
`5cm^(3) ` of `H_(2)O_(2)` solution gives `= 22.4 cm^(3)` of oxyen at N.T.P.
`1cm^(3)` of `H_(2)O_(2)` solution will give `= (22.4)/(5) = 4.48 cm^(3)` of oxygen at N.T.P.
`:.` Volume strength of `H_(2)O_(2) = 4.48`