0.1 g KIO_(3) and excess KI when treated with HCl, the iodine is liberated. The liberated iodine required 45 mL solution thiosulphate for titration . The molarity of sodium thisoulphate will be equivalent to :

0.1 g KIO_(3) and excess KI when treated with HCl, the iodine is liber

1.	0.1 g KIO_(3) and excess KI when treated with HCl, the iodine is liberated. The liberated iodine required 45 mL solution thiosulphate for titration . The molarity of sodium thisoulphate will be equivalent to :
Answer» 0.252 M 0.126 M 0.0313 M 0.0623 M Solution :The reaction involved are : `IO_(3)^(-)+5I^(-)+6H^(+) to 3I_(2)+H_(2)O` `2Na_(2)S_(2)O_(3)+I_(2) to 2 NaI+Na_(2)S_(4)O_(6)` Number of moles of `I_(2)=3xx " Number of moles of " KIO_(3)` `=(3xx0.1)/(214)` Number of moles of `Na_(2)S_(2)O_(3)=2xx " Number of moles of " I_(2)` `(Mxx45)/(1000)=(2xx3xx0.1)/(241)` M=0.623