An aqueous solution containing 0.10 g KIO3 ( formula weight =214.0) was treated with an excess of KI solution. The solution was acidified with HCl . The liberated I2 consumed 45 ml of thiosulphate solution to decolorize the blue starch iodine complex. Calculate the molarity of Na2S2O3 ??

My solution is

Redox change
IO3- ---> I2
2IO3- + 12H+ + 10e- = I2 + 6H2O
n factor of KIO3 = 5
Eq wt of KIO3 = 214/5
Gm eq of KIO3 = 0.10/(214/5) = Gm eq of I2
Let the molarity of Na2S2O3 = x
N factor of Na2S2O3 = 1
(0.10 x 5)/(214) = (45/1000) x (x) x 1
or x = 0.052M
But the solution is 0.062M
What is wrong in my solution.

## Comments

Sort by:

TopNewesthave u tried it ?? – Rudraksh Singh · 2 weeks ago

Log in to reply

http://exxamm.com/QuestionBank/practice/1119367210 THIS is the link bro ,, but sorry it is not the full length paper – Rudraksh Singh · 2 weeks, 1 day ago

Log in to reply

Hey this is iit 1998 problem ... Thanks for responding bro !! – Rudraksh Singh · 2 weeks, 1 day ago

Log in to reply

– Md Zuhair · 2 weeks, 1 day ago

Can u send me the link please, for the paper?Log in to reply

Thanks for tagging me. Let me try, then i will see ur solution, Then if it is different than urs, i will post it here. – Md Zuhair · 2 weeks, 1 day ago

Log in to reply

@Tapas Mazumdar , @Md Zuhair , @Akshat Sharda – Rudraksh Singh · 2 weeks, 1 day ago

Log in to reply

@Prakhar Bindal , @Aniket Sanghi , @Shubham Dhull , @Harsh Shrivastava , @A E please guys help me out here ,,, – Rudraksh Singh · 2 weeks, 1 day ago

Log in to reply