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An queous solution containing 2.14 g KIO...

An queous solution containing 2.14 g `KIO_3` was treated with 100 ml of 0.4 M Kl solution, the weight of `I_2` produced is-

A

6.096 (g)

B

7.62 (g)

C

30.48 (g)

D

18.288 g

Text Solution

Verified by Experts

The correct Answer is:
A

`underset("mole")(KIO_3)+underset(0.01)(5KI)tounderset(L.R.)underset(0.04)(3K_2O+3I_2)`
`n_(i_2)=3/5xxn_(Kl)=3/5xx0.04=0.012/5 implies W_(l_2)=0.12/5xx254=6.096 g`
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