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

Moles of `KlO_(3) = (0.1)/(214) = 0.00047`
` :. ` Moles of `l_(2)` lliberated from `KlO_(3) = (0.00047)/2 = 0.000235`
Moles of Kl reacting ` = 0.00047 xx 5 = 0.00235 (. :. " n-factor for " KlO_(3) " and " Kl " are " 5 ad 1 "respectively " )`
Moles of `l_(2)` produced from `Kl = (0.00235)/2 = 0.001175`
Total moles of `l_(2)` produced and reacted `= 0.000235 + 0.001175 = 0.00141` equivalents of `l_(2)` reacted = `0.00141 xx 2 = 0.00282 `= equivalents of thiosulphate Solution :
Molarity = `(0.00282 xx 1000)/45 = 0.063` M (For thiosulphate n-factor = 1)