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`.

Explanation

`2Kl +H_(2)SO_(4) +H_(2)O_(2) to K_(2)SO_(4) +2H_(2)O + I_(2)`
`2Na_(2)S_(2)O_(3) +I_(2) to Na_(2)S_(4)O_(6) +2NaI `
m.e of `H_(2)O_(2)` in 5 mL = m.e of `I_(2)` = m.e of `Na_(2)S_(2)O_(3)`
` :. ` m.e of `H_(2)O_(2)` in 50 mL = `0.1 xx20 = 2 ` ...(Eqn.1)
` :. ` m.e of `H_(2)O_(2)` in 1000 mL = `2/50 xx1000 = 40`
Equivalents per litre ` = 2/50 xx1000 = 40 `
Equivalents per litre ` = 40/1000` ..(Eqn . 3)
Grams per litre of `H_(2)O_(2) = 40/1000 xx 17` ...(Eqn .4)
`(" eq. wt . of "H_(2)O_(2) = 34/2 = 17 )`