http://www.ck12.org Chapter 16. Solutions
The boiling point of a solution can be calculated using the following expression:
∆Tb=kb×m×iwhere∆Tbis the increase in boiling point compared to the pure solvent,kbis the boiling point elevation constant for
the given solvent,mis the molality (not molarity) of the solution, andiis the van’t Hoff factor.
Example 16.14
Eugenol, the active ingredient in cloves, has the formula C 10 H 12 O 2. Eugenol is a molecular substance that does not
dissociate into ions when dissolved in a solvent. What is the boiling point of a solution in which 0.144 g of eugenol
is dissolved in 10.0 g benzene? Thekbvalue for benzene is 2.53 °C/m, and the boiling point of pure benzene is 80.1
°C.
Answer:
∆Tb=kb×m×i
The van’t Hoff factor (i) for a nondissociating substance is 1, and thekbvalue for benzene is given, so we only need
to calculate the molality of the solution. To do this, we need to determine the moles of the solute and the mass of the
solvent in kilograms.
molesC 10 H 12 O 2 = 0 .144 g C 10 H 12 O 2 × 164 1 mol C.22 g C^1010 HH^1212 OO^22 = 8. 77 × 10 −^4 mol C 10 H 12 O 2
mass of benzene= 10 .0 g benzene×1000 g1 kg = 0 .0100 kg benzene
molality=^8.^77 ×^10
− (^4) mol C 10 H 12 O 2
0 .0100 kg benzene =^0.^0877 m
∆Tb= 2. 53 ◦C/m× 0. 0877 m× 1 = 0. 222 ◦C
The boiling point is increased by 0.222 °C compared to the pure solvent. Because the boiling point of pure benzene
is 80.1 °C, the boiling point of the solution would be 80.3 °C.
Freezing Point Depression
The addition of solute to a pure solvent affects not only the boiling point of the solution, but the freezing point as
well. This phenomenon is calledfreezing point depression, and it can be calculated in essentially the same way as
boiling point elevation:
∆Tf=kf×m×i
where∆Tfis the decrease in freezing point compared to the pure solvent,kfis the freezing point depression constant
for the given solvent,mis the molality (not molarity) of the solution, andiis the van’t Hoff factor.
Example 16.15
Calculate the freezing point of a solution in which 5.00 g of biphenyl (C 12 H 10 ) and 7.50 g of naphthalene (C 10 H 8 )
are dissolved in 200.0 g of benzene. None of these substances dissociate into ions. The normal freezing point of
benzene is 5.5 °C, and its kfvalue is 5.12 °C/m.
Answer:
∆Tf=kf×m×i
We are given kf, andiis 1 for nondissociating substances. To determine the molality of the solution, we will need to
know the total moles of solute and the mass of the solvent in kilograms. Because colligative properties like freezing
point depression do not depend on the identity of the solute, it does not matter that we have two different types of
solute molecules.