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A stream of dry air was passed through a bulb containing a solution of 7.5 g of a compound dissolved in 75 g of water and then through another bulb containing pure water. The loss of mass in the first bulb was 2.81 g and in the second bulb was 0.054 g. Calculate the molecular mass of the compound. |
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Answer» Solution :` M_(2)= (P^(0)(W_(2)M_(1)))/((P^(0) - P)W_(1))` OR `M_(2) = (W_(2) M_(1))/(W_(1)) XX ((P^(0))/(P^(0)-P))` `("Total LOSS in MASS of solution bulb and the solvent bulb")/("Loss in mass of solvent bulb")` = `(7.5 xx 18)/(75) xx (2.864)/(0.054)` `M_(2) = 95.47`. |
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