Question

An queous solution containing 2.14 g `KIO_3` was treated with 100 ml of 0.4 M Kl solution, the weight of `l_2` produced is-
A. 6.096 (g)
B. 7.62 (g)
C. 30.48 (g)
D. 18.288 g

Answer 1

Correct Answer - 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`