A solution is made by dissolving 0.539 mol of nonelectrolyte solute in 813 g of benzene. Calculate the freezing point, Tf, and boiling point, Tb, of the solution. Constants may be found here.
the change in freezing point of depression in freezing point and elevation in boiling poin are colligative property and will depend upon the molality of solution as
Depression in freezing point = Kf X molality
Kf = constant (for benzene = 5.12
Elevation in boiling point = Kb X molallity
Kb= constant = 2.65 for benzene
Molality of solution = Moles of solute / weight of solvent in kg = 0.539 / 0.813 = 0.663 m
1) Depression in freezing point = 5.12 X 0.663 = 3.39 0C
the normal freezing point = 5.50C
So acutal freezing point = 5.5 - 3.39 = 2.110C
2) elevation in boiling point = 2.65 X 0.663 = 1.756
Boiling point of benzene = 80.10C
So acutal boiling point = 80.1 + 1.756 = 81.8560C
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