An aqueous solution containing 0.10 g `KIO_(3)` (formula weight`=214.0`) was treated with an excess of KI solution the solution was acidified with HCl. The liberated `I_(2)` consumed 45.0 " mL of " thiosulphate solution to decolourise the blue starch-iodine complex. Calculate the molarity of the sodium thosulphate solution.

Explanation

The reaction between `KIO_(3)` and `KI` is
`KIO_(3)+5KI+6"HCl"to6KCl+3H_(2)O+3I_(2)`
The reaction of `I_(2)` with thiosulphate ion is
`2S_(2)O_(3)^(2-)+I_(2)toS_(4)O_(6)^(2-)+2I^(ɵ)`
From these reactions, we can write
`1 mol KIO_(3)-=3 mol I_(2)`
`1 mol I_(2)-=2 mol S_(2)O_(3)^(2-)`
From the given mass of `KIO_(3)`, we get
Amount of `KIO_(3)=(0.10g)/(214.0g mol^(-1))=4.6729xx10^(-4)mol`
Amount of `I_(2)` generated `=3xx4.6729xx10^(-4)mol`
Amount of `S_(2)O_(3)^(2-)` consumed`=2(3xx4.6729xx10^(-4)mol)` ...(i)
Let m be the molality of thiosulphate solution.
From the given volume of thiosulphate solution, we get amount of `S_(2)O_(3)^(2-)` consumed`=MV=M(45.0xx10^(-3)L)` ..(ii)
Equating equation (i) and (ii) we get
`M(45.0xx10^(-3)L)=2(3xx4.6729xx10^(-4)mol)`
`M=(2(3xx4.6729xx10^(-4)mol))/(45.0xx10^(-3)L)=0.0623 mol L^(-1)`