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.

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## Comments

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TopNewest@Prakhar Bindal , @Aniket Sanghi , @Shubham Dhull , @Harsh Shrivastava , @A E please guys help me out here ,,,

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@Tapas Mazumdar , @Md Zuhair , @Akshat Sharda

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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.

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http://exxamm.com/QuestionBank/practice/1119367210 THIS is the link bro ,, but sorry it is not the full length paper

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have u tried it ??

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Hey this is iit 1998 problem ... Thanks for responding bro !!

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Can u send me the link please, for the paper?

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