Question

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.

Answer #1

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

So

1) Depression in freezing point = 5.12 X 0.663 = 3.39
^{0}C

the normal freezing point = 5.5^{0}C

So acutal freezing point = 5.5 - 3.3**9 =
2.11 ^{0}C**

2) elevation in boiling point = 2.65 X 0.663 = 1.756

Boiling point of benzene = 80.1^{0}C

So acutal boiling point = 80.1 + 1.756 =
**81.856 ^{0}C**

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