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Dry air was passed successively through solution of `5g` of a solute in `180g` of water and then through pure water. The loss in weight of solution was `2.50 g` and that of pure solvent `0.04g`. The molecualr weight of the solute is:A. 31.25B. 3.125C. 312.5D. None |
Answer» Correct Answer - A `P^(@)-P_(S) prop ` loss in weight of water chamber. and `P_(S) prop ` loss in weight of solution chamber. `(P^(@)-P_(S))/-P_(S) = n_(2)/n_(1) =(W_(2) xx Mw_(1))/(Mw_(2) xx W_(1))` or `0.04/2.50 = (5 xx 18)/(Mw_(2) xx 180)` `:. Mw_(2) =31.25` |
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