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 iodine consumed 45.0 mL of thiosulphate solution to decolourise the blue starch-iodine complex. Calculate the molarity of the sodium thiosulphate solution.

Explanation

`2KIO_(3)+10KI+12HCl to 12 KCl +6I_(2)+6H_(2)O`
[`2Na_(2)S_(2)O_(3)+I_(2) to 2NaI+Na_(2)S_(4)O_(6)]xx6`
`overlineul(underset(2"moles")(2KIO_(3))+10KI+underset(12"moles")(12Na_(2)S_(2)O_(3))+12HCl to 12KCl+12NaI+6Na_(2)S_(4)O_(6)+6H_(2)O)`
No. of moles of `KIO_(3)=(0.10)/(214)`
No. of moles of `Na_(2)S_(2)O_(3)` required for `(0.10)/(214)` moles of `KIO_(3)`
`=(12)/(2)xx(0.10)/(214)=(0.60)/(214)`
Molarity of `Na_(2)S_(2)O_(3)=(0.6)/(214)xx(1000)/(45)=0.0623 M`