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A solution containing 0.73 g of camphor ( molar mass 152 g "mol"^(-1) ) in 36.8 g of acetone ( boiling point56.3^(@)C) boils at 56* 55^(@)C . A solution of 0.564 g of unknown compound in the same weight of acetone boils at 56.46^(@)C . Calculatethe molar massof the unknown compound . |
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Answer» Solution :Given : MASS of camphor = `W_(2) = 0*73` g Molar mass of camphor = `M_(2) = 152 g "mol"^(-1)` Mass of acetone = `W_(1) = 36*8 ` g Mass of unknown compound = ` W'_(2) = 0*564` g Molar mass of compound = `M'_(2)= ?` For solution of camphor in acetone , ` DeltaT_(B) = T_(b)- T^(@)._(b)` ` = 329 * 55 - 329*30 = 0*25 ` K `:. K_(b) = (Delta T_(b)xx W_(1)xxM_(2))/(W_(2) xx 1000)` ` = (0*25 xx 36*8 xx 152)/(0*73 xx 1000)` ` = 1*916 "Kkg mol"^(-1)` For solution of unknown compound in acetone, ` DeltaT_(b) = T_(b) - T^(@)._(b)` ` = 329*46 - 329*30 = 0*16 ` K ` :. K_(b) = (DeltaT_(b)xxW_(1)xxM'_(2))/(E'_(2)xx1000)` ` M'_(2) = (K_(b)xxW'_(2)xx1000)/(DeltaT_(b)xxW_(2))` ` = (1*916 xx1000 xx0.564)/(0*16 xx36*8)` ` M'_(2) = 185*5` g/mol Hence , the molar mass of the unknown compound is ` 183*5 "g mol"^(-1)` |
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