50 mL of an aqueous solution of `H_(2)O_(2)` was treated with an excess of KI solution and dilute `H_(2)SO_(4)`. The liberated iodine required 20 mL 0.1 N `Na_(2)S_(2)O_(3)` solution for complete interaction. Calculate the concentration of `H_(2)O_(2)` in `g//L`.

Answer

0.68 g

Explanation

`H_(2)O_(2)^(-1)KI+H_(2)SO_(4)rarrK_(2)SO_(4)+I_(2)+2H_(2)O^(-2)`
`Na_(2)S_(2)O_(3)+I_(2)rarrNa_(2)S_(4)O_(6)+2` Nal
Total change in O.N of O=2(-1)-2(-2)=2 `therefore` Eq of `H_(2)O_(2)=(2+2xx16)/(2)=17`
Let `N_(1)` be the normality of `I_(2)` solution
Since one equivalent of `H_(2)O` produces 1 equilvent of `I_(2)`
`therefore` 50mL of `N_(1) I_(2)` solution =50mL of `N_(1)H_(2)O_(2)` solution
`therefore N_(1)=(20xx0.1)/(50)=0.04` N or Streangth of `H_(2)O_(2) "solution" =0.04xx17=0.68 gL^(-1)`