EFFECTS OF
CONCENTRATION ON
THE COLLIGATIVE
PROPERTIES OF
SOLUTIONS
Objectives:
1. Describe the effect of concentration on the colligative
properties of solutions (STEM_GC11PPIIId-f-115)
 Sub-tasks:
1. Define colligative and colligative properties.
2. Describe the effect of solute concentration on various
solution properties: vapor pressure, boiling point, freezing
point, and osmotic pressure.
3. Differentiate the colligative properties of nonelectrolyte
solutions and of electrolyte solutions. (STEM_GC11PP-IIIdf-116 ),
4. Calculate boiling point elevation and freezing point
depression from the concentration of a solute in a solution.
(STEM_GC11PP-IIId-f-117)

Colligative Properties
 Colligative came from a Latin word “colligatus”
which means “depending on the collection” or it
means, “grouped together.
 Changes in colligative properties depend only on the
number of solute particles present, not on the identity
of the solute particles.
 Among colligative properties are
 Vapor pressure lowering
 Boiling point elevation
 Melting point depression
 Osmotic pressure
 Let try the figure out this statement:
 ” It is a hot summer day and you have a
picnic at the park or beach front with your
classmates, friends or relatives with
watermelon and “dirty ice cream”.
Mmmmmm….. tastes good… refreshing….
The ice cream is an old-fashioned
homemade kind ice cream. The kind of
where the maker has a tub full of mix of
ingredients immersed in a bigger tub filled
with ice and salt. But wait a minute, why salt?
Why the ice cream vendor does add salt to
the ice?
Colligative Properties of Nonelectrolyte Solutions
Nonvolatile solutes lower the vapor pressure of a solvent
by an amount proportional to the solute mole fraction.
Vapor-Pressure Lowering
P1 = X1 P
0
1
Raoult’s law
P 10 = vapor pressure of pure solvent
X1 = mole fraction of the solvent
If the solution contains only one solute:
X1 = 1 – X2
P 10 - P1 = DP = X2 P 10
X2 = mole fraction of the solute
13.6
Vapor-Pressure
Lowering
Calculate the vapor pressure of a solution made by
dissolving 50.0 g glucose, C6H12O6, in 500 g of
water. The vapor pressure of pure water is 47.1 torr
at 37°C .
Ans. 46.63 torr
Boiling point is the temperature where
the vapor pressure of the liquid equals
the prevailing atmospheric pressure.
Water boils at 100 0C when its vapor
pressure equals the atmospheric
pressure. When a nonvolatile solute,
such as NaCl + H2O is added to a pure
water or solvent, boiling point becomes
higher than that of pure water.
Boiling-Point Elevation
When a nonvolatile solute, such as NaCl + H2O is added
to a pure water or solvent, boiling point becomes higher
than that of pure water.
SOLVENT
BOILING POINT
(0C)
100
Kb (0C
kg/mol)
0.51
Acetone
55.95
1.71
CCl4
76.8
5.02
Benzene
80.1
2.63
Chloroform
61.2
3.63
Water
Boiling point of solution and Kb of common solvents.
Boiling-Point Elevation
DTb = 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
DTb > 0
DTb = Kb m
m is the molality of the solution
Kb is the molal boiling-point
elevation constant (0C/m)
13.6
The colligative law states
that, “the boiling point
elevation is directly
proportional to the molal
concentration of the solution.”
A solution of 10.0 g of a nonvolatile, nondissociating
compound dissolved in 0.200 kg of benzene boils at
81.2 °C. Calculate the molecular weight of the
compound.
Ans. 119.5 g/mol
Freezing-Point Depression
The normal freezing point of a liquid is the
temperature at which a liquid becomes a solid at 1
atm.
When a solute is added to a pure solvent, the
solute particle interrupts and reduces the attractive
forces that will hold the solvent molecules together to
form into a solid state. Temperature must be lowered
to enable the solvent molecule to bind and come
closer together. The presence of a nonvolatile solute
lowers the freezing point of a pure solvent.
Freezing-Point Depression
DTf = T 0f – Tf
T
0
Tf
f
is the freezing point of
the pure solvent
is the freezing point of
the solution
T 0f > Tf
DTf > 0
DTf = Kf m
m is the molality of the solution
Kf is the molal freezing-point
depression constant (0C/m)
13.6
SOLVENT
FREEZING POINT
(0C)
0
Kf (0C kg/mol)
Benzene
5.5
5.12
Acetic Acid
16.6
3.90
CCl4
22.8
29.8
Camphor
179
39.7
Naphthalene
80.2
6.80
Water
Figure 1. Freezing point of Solution
Source: General Chemistry by: Tabajura, Jr. G.D.
1.86
How many grams of pyrazine (C4H4N2) would have
to be dissolved in 1.50 kg of carbon tetrachloride to
lower the freezing point by 4.4 °C? The freezing point
constant for carbon tetrachloride is 30. °C/m.
Ans. 17.6 g
13.6
Osmotic Pressure (p)
Osmosis is the selective passage of solvent molecules through a porous
membrane from a dilute solution to a more concentrated one.
A semipermeable membrane allows the passage of solvent molecules but
blocks the passage of solute molecules.
Osmotic pressure (p) is the pressure required to stop osmosis.
dilute
more
concentrated
13.6
Osmotic Pressure (p)
High
P
Low
P
p = MRT
M is the molarity of the solution
R is the gas constant
T is the temperature (in K)
13.6
A cell in an:
isotonic
solution
hypotonic
solution
hypertonic
solution
13.6
1. Hypertonic solution. If a cell is placed in a hypertonic solution,
there will be a net flow of water out of the cell, and the cell will lose
volume. A solution will be hypertonic to a cell if its solute
concentration is higher than that inside the cell, and the solutes
cannot cross the membrane.
2. Hypotonic solution. If the cell is placed in a hypotonic solution,
there will be a net flow of water into the cell, and the cell will gain
volume. If the solute concentration outside the cell is lower than
inside the cell, and the solutes cannot cross the membrane, then that
solution is hypotonic to the cell.
3. Isotonic solution. If the solute concentration outside the cell is the
same as inside the cell, and the solutes cannot cross the membrane,
then that solution is isotonic to the cell. The cell would be normal in
shape.
Sample Problem:
1. A very dilute solution, 0.0010 M sugar in water, is separated
from pure water by an osmotic membrane. What osmotic
pressure developed at 25oC? The gas constant R = 0.0821 L
atm / mol K.
2. A solution is made by dissolving 13.0 g of
sucrose, C12H22O11, in 117 g of water, producing
a solution with a volume of 125 mL at 20.°C.
What is the expected osmotic pressure at
20.°C?
Colligative Properties of Nonelectrolyte Solutions
Colligative properties are properties that depend only on the
number of solute particles in solution and not on the nature of
the solute particles.
Vapor-Pressure Lowering
P1 = X1 P 10
Boiling-Point Elevation
DTb = Kb m
Freezing-Point Depression
DTf = Kf m
Osmotic Pressure (p)
p = MRT
13.6
Colligative Properties of Electrolyte Solutions
0.1 m Na+ ions & 0.1 m Cl- ions
0.1 m NaCl solution
Colligative properties are properties that depend only on the
number of solute particles in solution and not on the nature of
the solute particles.
0.1 m NaCl solution
van’t Hoff factor (i) =
0.2 m ions in solution
actual number of particles in soln after dissociation
number of formula units initially dissolved in soln
i should be
nonelectrolytes
NaCl
CaCl2
1
2
3
13.7
Chemists compare the degree of
dissociation of electrolytes at
different dilutions by a quantity
called the van't Hoff factor,
van’t Hoff factor (i) as the number
of particles each solute formula
unit breaks apart into when it
dissolves
Colligative Properties of Electrolyte Solutions
Boiling-Point Elevation
DTb = i Kb m
Freezing-Point Depression
DTf = i Kf m
Osmotic Pressure (p)
p = iMRT
13.7
Determine the freezing point of
1.77 m solution of NaCl in H2O.
Solution
For NaCl, we need to remember to include the
van’t Hoff factor, which is 2. Otherwise, the
calculation of the freezing point is
straightforward:
ΔTf = (2)(1.77 m)(1.86°C/m) = 6.58°C
This represents the change in the freezing
point, which is decreasing. So we have to
subtract this change from the normal freezing
point of water, 0.00°C:
0.00 − 6.58 = −6.58°C