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A solutioncontaining 0.730 g of camphor(molarmass = 152 )in 36.8g of acetone (b.p. 56.30^(@)C) boils at 56.55^(@)C. A solutionof 0.564 g of an unknown compound in the same weightof solventboils at 56.46^(@)C . Calculatethe molarmass of the unknown compound. |
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Answer» Solution :In this problem , THEVALUE of `k_(b)` is not given.The firstdata is used to calculate `k_(b)` WHICHIS used to calculatethe molar mass from the second data. (i) Calculationof `k_(b)` foracetone . `k_(b) = (Delta_(b) xx M_(B) xx w_(A))/(w_(B) xx 1000)` `Delta T_(b) = 56.55 - 56.30 = 0.25^(@)C, M_(B)= 152` `w_(B) = 0.730 g, w_(B) = 36.8 g` `therefore ""k_(b) = (0.25 xx 152 xx 36.8)/(0.736 xx 1000) = 1.92 K m^(-1)` (ii) Calculation of molarmass ofunknown COMPOUND. `M_(B) = (k_(b) xx 1000 xx w_(B))/(Delta T_(b) xx w_(A))` `k_(b) = 1.92 km^(-1) , Delta T_(b) = 56.46 = 56.30` ` = 0.16^(@)C` `w_(B) = 0.564 g, w_(A) = 36.8 g ` `M_(B) = (1.92 xx 1000 xx 0.564)/(0.16 xx 36.8) = 183.9 g mol^(-1)` |
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