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A sugar syrup of weight 214.2 g contains 34.2gofsugar(C_(12)H_(22)O_(11)).Calculate(i)molal concentration (ii) mole fraction of sugar in the syrup. |
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Answer» Solution :Weight of sugar syrup = 214.2 g Weight of sugar present = 34.2 g MOLECULAR mass of sugar `(C_(12)H_(22)O_(11)) = 342` (i) CALCULATION of molality : Number of moles of sugar = `34.2/342 = 0.1` Weight of water `=214.2 - 34.2 = 180 g = 0.18 g` `therefore` Molecular of the solution `=("No. of moles of sugar")/("Weight of water in kg")` `=0.1/0.18 = 0.56 MOL kg^(-1)` Hence, the MOLAL concentration of the given solution is 0.56 m. (ii) Calculation of mole fraction : Number of moles of sugar `(n_(1))=0.1` Number of moles of water `(n_(2)) = 180/18 = 10` Total number of moles in solution `=n_(1) + n_(2) = 0.1 + 10 = 10.1` `therefore` The mole fraction of sugar `=n_(1)/(n_(1)+ n_(2)) = 0.1/10.1 = 9.9 xx 10^(-3)` |
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