20mL of 0.1 `Na_(2)S_(2)O_(3)`solution is required to reduce `I_(2)` liberated when Ki in excess is added to 25mL of `H_(2)O_(2)`solution .Calculate persent concentration (x/v) of `H_(2)O_(2)`
`=(0.034)/(25)xx100=0.136%`

Explanation

The reaction of KI with `H_(2)O_(2)`:
`H_(2)O_(2)+2I^(-)+2H^(+)toI_(2)+2H_(2)O`
The reaction of `I_(2)`with `Na_(2)S_(2)O_(3)`:
`2Na_(2)S_(2)O_(3)+I_(2)toNa_(2)S_(4)O_(6)+2NaI`
`therefore`2 mol of `Na_(2)S_(2)O_(3)`-=1 mol of `I_(2)` -=`(1)/(2)`-=1 mol of `H_(2)O_(2)`
or , 1 mol of `Na_(2)S_(2)O_(3)-=(1)/(2)mol of I_(2) -=(1)/(2)` mol of `H_(2)O_(4)` in reaction (2), the equivalent mass of `Na_(2)S_(2)O_(3)`
`(2xx"molecular mass "of Na_(2)S_(2)O_(3))/(2)`
=Molecular mass of `Na_(2)S_(2)O_(3)`
`therefore` 1000mL of 1M `Na_(2)S_(2)O_(3)` solution
`(1)/(2)xx34g=17g of H_(2)O_(2)`[Molar mass of `H_(2)O_(2)=34g]`
or, 1000mL of 1N `Na_(2)S_(2)O_(3)` solution =17g of `H_(2)O_(2)` [`therefore`Normality of `Na_(2)S_(2)O_(3)` in the reaction = its molarity]
or ,1mL of 1N `Na_(2)S_(2)O_(3)` solution `=17xx10^(-3)g of H_(2)O_(2)`
or, 20mL, of 0.1 `Na_(2)S_(2)O_(3)` solution
`-=17xx10^(-3)xx20xx0.1-=0.034g of H_(2)O_(2)`