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

We know,
`KIO_(3)+5KIrarr3K_(2)O+3I_(2)`
i.e, `2I^(5+)+10erarrI_(2)^(0)`
`2I^(-)rarrI_(2)^(0)+2e`
Now the liberated `I_(2)` reacts with `Na_(2)S_(2)O_(3)` to give
`I_(2)+2erarr2I^(-)`
`2S_(2)O_(3)^(2-)rarrS_(4)O_(6)^(2-)+2e`
`because` Milli-mole ratio is `I_(2):S_(2)O_(3)::1:2`
Thus, milli`-`mole of `Na_(2)S_(2)O_(3) "used"xx(1)/(2)`
`=45xxMxx(1)/(2)` ( `M` is molarity of thiosulphate)
Also milli-mole ratio of `KIO_(3): I_(2)::1:3`
Thus, `((0.1)/(214))/((45M)/(2))=(1)/(3)`
`therefore M=(0.1xx1000xx3xx3)/(214xx45)=0.062`