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

Upon reaction between `KIO_(3)` and KI, `I_(2)` is liberated
`KIO_(3) + 5KI rarr 3K_(2)O + 3I_(2)`
This liberated `I_(2)` reacts with `Na_(3)S_(3)O_(3)`
One mole `KIO_(3)` liberated three moles of `I_(2)`
Moles of `KIO_(3) = 0.10 g // molwt. (214) = 0.000467 `
Liberated moles of `I_(2) = 3 xx 0.000467`
One mole of `I_(2)` requires two moles of `Na_(2)S_(2)O_(3)` for complete decolourisation of `I_(2)`
Moles of `Na_(2)S_(2)O_(3) = 2 (3 xx 0.000467)`
Molarity of `Na_(2)S_(2)O_(3) =` mol es/volume in litre `= 2xx(3xx0.000467)/(45.0 xx 10^(-3)) = 0.0623 M`