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» <html><body><p>0.252 M <br/>0.126 M<br/>0.0313 M<br/>0.0623 M</p>Solution :The reaction involved are : <br/> `IO_(3)^(-)+<a href="https://interviewquestions.tuteehub.com/tag/5i-326324" style="font-weight:bold;" target="_blank" title="Click to know more about 5I">5I</a>^(-)+6H^(+) to 3I_(2)+H_(2)O` <br/> `2Na_(2)S_(2)O_(3)+I_(2) to 2 NaI+Na_(2)S_(4)O_(<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>)` <br/> Number of moles of `I_(2)=3xx " Number of moles of " KIO_(3)` <br/> `=(3xx0.1)/(214)` <br/> Number of moles of `Na_(2)S_(2)O_(3)=2xx " Number of moles of " I_(2)` <br/> `(Mxx45)/(1000)=(2xx3xx0.1)/(241)` <br/> M=0.623</body></html>