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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 |
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Answer» 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` |
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