Q9 of 30 Page 1

18 g of glucose, C6H12O6 (Molar mass = 180 g mol-1) is dissolved in 1 kg of water in a sauce pan. At what temperature will this solution boil? (Kb for water =0.52 K kg mol-1, boiling point of pure water =373.15K)

Weight of solvent H2O (W1) = 1kg = 1000gm


Molar mass of solvent H2O = 18gm


Weight of solute (C6H12O6 ) (W2) = 18gm


Molar mass of solute (M2) = 180gm


Moles of solute = = 0.1moles


Kb for water =0.52 K kg mol-1


Boiling point of pure water =373.15K


Elevation in boiling point ΔTb =


=


=0.052K


ΔTb = Tb –b [Tb = Boiling point of solution]


[T°b =Boiling point of pure water]


0.052 = Tb – 373.15


Tb = 373.202K


Hence, boiling point of the solution is 373.202K


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