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Chemistry 205
Colligative Properties
Freezing Point Depression
• To measure the change in temperature of
water as it freezes, and to determine its
freezing point.
• To measure the drop in the freezing
temperature (freezing-point depression)
caused by the dissolution of a solute in
• To study the effect of different solutes
(electrolytes and non-electrolytes) on the
freezing point of water.
Concentration Units (Review!)
The concentration of a solution is the amount of solute
present in a given quantity of solvent or solution.
Percent by Mass
mass of solute
x 100%
% by mass =
mass of solute + mass of solvent
mass of solute x 100%
mass of solution
Mole Fraction (X)
moles of A
XA =
sum of moles of all components
Ref. Chang, Sect. 12.3
Concentration Units Continued
Molarity (M)
M =
moles of solute
liters of solution
Molality (m)
m =
moles of solute
mass of solvent (kg)
Ref. Chang, Sect. 12.3
• A pure liquid such as water or ethanol has
characteristic physical properties: melting
point, boiling point, density, vapor pressure,
• Addition of a soluble solute to the liquid forms
a homogeneous mixture called a solution.
• The solvent of the solution assumes physical
properties that are no longer definite, but
dependent on the amount of solute added.
These properties are called colligative
Colligative Properties of Nonelectrolyte
Colligative properties are properties that depend only on the
number of solute particles (whether atoms, molecules or ions) in
solution and not on the nature of the solute particles.
Interesting Colligative properties:
Vapor-Pressure Lowering
Boiling-Point Elevation
Freezing-Point Depression
Vapor-Pressure Lowering
The vapor pressure of a solution (nonvolatile solute
and a solvent) is less than that of a pure solvent.
P1 = X1 P 10
Raoult’s law
P1 = partial pressure of a solvent over a
P 10 = vapor pressure of pure solvent
X1 = mole fraction of the solvent
If the solution contains only one solute, X1 = 1 – X2,
X2 = mole fraction of solute
P1 = (1 – X2) P 01
P 10 - P1 = ∆P = X2 P 10
Boiling-Point Elevation
Boiling point of a liquid: The temperature at which the vapor
pressure equals the external atmospheric pressure. (Normal
boiling point: external pressure = 1atm.)
∆Tb = Tb – T b0
T b0 is the boiling point of
the pure solvent
T b is the boiling point of
the solution
Tb > T b0
∆Tb > 0
∆Tb = Kb m
m is the molality of the solution
Kb is the molal boiling-point
elevation constant (0C/m)
Ref. Chang, Sect. 12.6
Freezing-Point Depression
Freezing point of a liquid: is the temperature at which solid
and liquid phases coexist in equilibrium. (Normal freezing
point: external pressure = 1atm.)
∆Tf = T 0f – Tf
is the freezing point of
the pure solvent
T f is the freezing point of
the solution
T f > Tf
∆Tf > 0
∆Tf = Kf m
m is the molality of the solution
Kf is the molal freezing-point
depression constant (0C/m)
Ref. Chang, Sect. 12.6
Ref. Chang, Table 12.2
Colligative Properties of Electrolyte Solutions
0.1 m NaCl solution
0.1 m Na+ ions & 0.1 m Cl- ions
0.1 m NaCl solution
0.2 m ions in solution
van’t Hoff factor (i) =
actual number of particles in soln after dissociation
number of formula units initially dissolved in soln
i should be
Boiling-Point Elevation
∆Tb = iKb m
Freezing-Point Depression
∆Tf = iKf m
Ref. Chang, Sect. 12.7
For electrolyte solutions: Measured colligative properties are
smaller than those calculated.
Explanation: Formation of ion pairs (cations and anions held
together by electrostatic forces of attraction). Presence of ion
pairs reduces the number of particles in solution, causing a
Ref. Chang, Table 12.3
reduction in the colligative properties.
What is the molality of a 5.86 M ethanol (C2H5OH)
solution whose density is 0.927 g/mL?
moles of solute
moles of solute
m =
M =
mass of solvent (kg)
liters of solution
Assume 1 L of solution:
5.86 moles ethanol = 270 g ethanol
927 g of solution (1000 mL x 0.927 g/mL)
mass of solvent = mass of solution – mass of solute
= 927 g – 270 g = 657 g = 0.657 kg
moles of solute
m =
mass of solvent (kg)
5.86 moles C2H5OH
= 8.92 m
0.657 kg solvent
Ref. Chang, Sect. 12.3
What is the freezing point of a solution containing 478 g
of ethylene glycol,C2H6O2, (antifreeze) in 3202 g of
water? The molar mass of ethylene glycol is 62.01 g.
∆Tf = Kf m
Kf water = 1.86 0C/m
moles of solute
m =
mass of solvent (kg)
478 g x
1 mol
62.01 g
= 2.41 m
3.202 kg solvent
∆Tf = Kf m = 1.86 0C/m x 2.41 m = 4.48 0C
∆Tf = T 0f – Tf
Tf = T 0f – ∆Tf = 0.00 0C – 4.48 0C = -4.48 0C
Ref. Chang, Sect. 12.6
Refer to the hand-out.
• In Part I, determine the freezing point of water.
• In Part II, calculate the number of moles of
solute and the molality of each solution.
• Knowing that the molal freezing-point
depression constant, Kf is equal to 1.86°C
kg/mol, determine the expected freezing point
of each solution and compare it with the
observed value.
∆Tf = i Kf x m
Practice Exercises
1. Which of the following aqueous solutions has the
highest boiling point? Kb for water is 0.52°C/m.
Which has the highest freezing point?
0.2 m KCl, 0.2 m Na2SO4, 0.3 m Ca(NO3)2
2. What is the molarity of a solution of 10% by mass
cadmium sulfate, CdSO4 (molar mass = 208.46
g/mol)? The density of the solution is 1.10 g/mL.
1. Highest boiling point: 0.3 m Ca(NO3)2
Highest freezing point: 0.2 m KCl
2. 0.528 mol/L
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