
Colligative
Properties of
Solutions
Solutions
part 2
Dr. Hisham E Abdellatef
Professor of pharmaceutical analytical chemistry
http://www.staff.zu.edu.eg/ezzat_hisham/browseMyFiles.asp?path=./use
rdownloads/physical%20chemistry%20for%20clinical%20pharmacy/
1
Colligative Properties of Solutions
There are four common types of colligative
properties:

1.
2.
3.
4.

Vapor pressure lowering
Freezing point depression
Boiling point elevation
Osmotic pressure
Vapor pressure lowering is the key to all
four of the colligative properties.
2
Lowering of Vapor Pressure and
Raoult’s Law

Addition of a nonvolatile solute to a solution
lowers the vapor pressure of the solution.



The effect is simply due to fewer solvent
molecules at the solution’s surface.
The solute molecules occupy some of the spaces
that would normally be occupied by solvent.
Raoult’s Law models this effect in ideal
solutions.
3
Lowering of Vapor Pressure and
Raoult’s Law

Derivation of Raoult’s Law.
0
Psolvent  X solvent Psolvent
where Psolvent  vapor pressure of solvent in solution
0
Psolvent
 vapor pressure of pure solvent
X solvent  mole fraction of solvent in solution
4
Lowering of Vapor Pressure and
Raoult’s Law

Lowering of vapor pressure, Psolvent, is defined as:
Psolvent  P
0
solvent
 Psolvent
0
0
 Psolvent
- ( X solvent)( Psolvent
)
 (1  X solvent)P
0
solvent
5
Lowering of Vapor Pressure and
Raoult’s Law


Remember that the sum of the mole fractions
must equal 1.
Thus Xsolvent + Xsolute = 1, which we can
substitute into our expression.
X solute  1 - X solvent
0
Psolvent  X solute Psolvent
which is Raoult' s Law
6
Lowering of Vapor Pressure and
Raoult’s Law

This graph shows how the solution’s vapor pressure
is changed by the mole fraction of the solute, which
is Raoult’s law.
7
Examples
The vapor pressure of water is 17.5 torr at 20°C.
Imagine holding the temperature constant while
adding glucose, C6H12O6, to the water so that the
resulting solution has XH2O = 0.80 and XGlu = 0.20.
What is , the vapor pressure of water over the solution
PA  X P
0
A A
PA  X P  0.80X 17.5torr
0
A A
= 14 torr
8
Glycerin, C3H8O3, is a nonvolatile nonelectrolyte with
a density of 1.26 g/mL at 25°C. Calculate the vapor
pressure at 25°C of a solution made by adding 50.0
mL of glycerin to 500.0 mL of water. The vapor
pressure of pure water at 25°C is 23.8 torr
9
The vapor pressure of pure water at 110°C is 1070
torr. A solution of ethylene glycol and water has a
vapor pressure of 1.00 atm at 110°C. Assuming that
Raoult's law is obeyed, what is the mole fraction of
ethylene glycol in the solution? Answer: 0.290
P°H2O =1070 torr
PH2O = 1 Atm = 760 torr
PH2O
torr
XH2O = --------- = 760
--------P°H2O 1070 torr =
XH2O + XEG = 1
0.7103 + XEG = 1
0.71028
1- 0.7103 = XEG
XEG =
0.28972
= 0.290
Many solutions do not obey Raoult's law exactly:
They are not ideal solutions.
If the intermolecular forces between solvent and
solute are weaker than those between solvent and
solvent and between solute and solute, then the solvent
vapor pressure tends to be greater than predicted by
Raoult's law.
Conversely, when the interactions between solute and
solvent are exceptionally strong, as might be the case
when hydrogen bonding exists, the solvent vapor
pressure is lower than Raoult's law predicts.
Although you should be aware that these departures
from ideal solution occur, we will ignore them for the
remainder of this chapter.
11
More Examples
Sucrose is a nonvolatile, nonionizing solute in
water. Determine the vapor pressure lowering, at
27°C, of a solution of 75.0 grams of sucrose,
C12H22O11, dissolved in 180. g of water. The
vapor pressure of pure water at 27°C is 26.7 torr.
Assume the solution is ideal.
1 m ol Suc
nSuc  75.0 gSuc
 0.219m ol
342.3g Suc
1 m ol Water
nWater  180gWater
 9.99m ol
18g Watyer
nwater
9.991
X Water 

 0.978541
nWater  nUc 9.991 0.2191
0
PWater  PWater
X Water  26.7 torr X 0.97854 26.13
Vapor Pressure Lowered = 26.7-26.1= 0.6
12
solution is made by mixing 52.1 g of propyl
chloride, C3H8Cl, and 38.4 g of propyl bromide,
C3H8Br. What is the vapor pressure of propyl
chloride in the solution at 25°C? The vapor pressure
of pure propyl chloride is 347 torr at 25°C and that
of pure propyl bromide is 133 torr at 25°C. Assume
that the solution is an ideal solution.
1 m ol CP
nCP  52.1 g CP
 0.6633
78.54g CP
1 m ol CB
nCB  38.4 g CB
 0.312
122.99g CB
nPCr
0.6633
X PC 

 0.67996
nPC  nPB 0.6633 0.3122
0
PPC  PPC
X PC  347X 0.679964 235.95  236Torr
13
. At 25°C a solution consists of 0.450 mole of
pentane, C5H12, and 0.250 mole of cyclopentane,
C5H10. What is the mole fraction of cyclopentane in
the vapor that is in equilibrium with this solution?
The vapor pressure of the pure liquids at 25°C are
451 torr for pentane and 321 torr for cyclopentane.
Assume that the solution is an ideal solution.
0
PPen  PPen
X Pen  .450X 451 202.95
0
PCPen  PCPen
X CPen  0.250X 321 80.25
P V
P V
PV
n
; n Pen  Pen ; nCPen  CPen
RT
RT
RT
PCPenV
nCPen
PCPen
RT
X CPen 


nCPen  n Pen PCPenV PPenV PCPen  PPen

RT
RT
80.25
PCPen 
 0.283
80.25  202 .95
14
Fractional Distillation


Distillation is a technique used to separate solutions
that have two or more volatile components with
differing boiling points.
A simple distillation has a single distilling column.


Simple distillations give reasonable separations.
A fractional distillation gives increased separations
because of the increased surface area.

Commonly, glass beads or steel wool are inserted into the
distilling column.
15
Boiling Point Elevation

Addition of a nonvolatile solute to a solution
raises the boiling point of the solution above
that of the pure solvent.



This effect is because the solution’s vapor
pressure is lowered as described by Raoult’s law.
The solution’s temperature must be raised to
make the solution’s vapor pressure equal to the
atmospheric pressure.
The amount that the temperature is elevated
is determined by the number of moles of
solute dissolved in the solution.
16
Boiling Point Elevation

Boiling point elevation relationship is:
Tb  K b m
where : Tb  boiling point elevation
m  molal concentrat ion of solution
K b  molal boiling point elevation constant
for the solvent
17
Boiling Point Elevation

Example 14-4: What is the normal boiling
point of a 2.50 m glucose, C6H12O6, solution?
Tb  K b m
Tb  (0.512 0 C/m)( 2.50m)
Tb  1.280 C
Boiling Point of the solution = 100.00 C + 1.280 C = 101.280 C
18
Boiling-Point Elevation
The addition of a nonvolatile solute lowers the vapor
pressure of the solution.
At any given temperature,
the vapor
pressure
of the
solution
is lower
than that
of the
pure
liquid
19
The increase in boiling point relative to that of the
pure solvent, Tb, is directly proportional to the
number of solute particles per mole of solvent
molecules.
Molality expresses the number of moles of solute per
1000 g of solvent, which represents a fixed number
of moles of solvent
Tb  Kb m
B.Point (°C)
Solvent
Water, H2O
Benzen, C6H6
Ethanol, C2H6O
Carbon tetrachloride,
CCl4
Chloroform,
CHCl3
100.0
80.1
78.4
76.8
61.2
Kb
(°C/m)
0.52
2.53
1.22
5.02
3.63
Freezing P.
(°C)
Kf (°C/m)
0.00
5.5
-114.0
-22
-63.5
1.86
5.12
1.99
29.8
4.68
20
Automotive antifreeze consists of ethylene glycol,
C2H6O2, a nonvolatile nonelectrolyte. Calculate the
boiling point of a 25.0 mass percent solution of
ethylene glycol in water.
21
Calculate the boiling point of a solution of 2.0 molal
of NaCl. Kb, water= 0.52 °C /mola.
t = Kbm
NaCl(aq)  Na+ + Cl2.0 m
2.0 m
2.0 m
2.0 m + 2.0 m = 4.0m
t = (0.52 °C/molal)(4.0 molal) =2.08 °C
BP = NBP +t = 100.00°C +2.08 °C = 102.08° C
22
Freezing Point Depression


Addition of a nonvolatile solute to a solution
lowers the freezing point of the solution
relative to the pure solvent.
See table 14-2 for a compilation of boiling
point and freezing point elevation constants.
23
Freezing Point Depression

Relationship for freezing point depression is:
Tf  K f m
where: Tf  freezing point depression of solvent
m  molal concentration of soltuion
K f  freezing point depression constant for solvent
24
Freezing Point Depression

Notice the similarity of the two relationships
for freezing point depression and boiling point
elevation.
Tf  K f m vs.Tb  K b m

Fundamentally, freezing point depression and boiling
point elevation are the same phenomenon.


The only differences are the size of the effect which is
reflected in the sizes of the constants, Kf & Kb.
This is easily seen on a phase diagram for a solution.
25
Freezing Point Depression
26
Freezing Point Depression

Example 14-5: Calculate the freezing point of
a 2.50 m aqueous glucose solution.
Tf  K f m
Tf  (1.86 0 C/m)( 2.50m)
Tf  4.650 C
Freezing Point of solution = 0.000 C - 4.650 C = - 4.650 C
27
Freezing Point Depression

Example 14-6: Calculate the freezing point of
a solution that contains 8.50 g of benzoic acid
(C6H5COOH, MW = 122) in 75.0 g of
benzene, C6H6.
You do it!
28
Freezing Point Depression
1. Calculate molality!
? mol C 6 H 5 COOH 8.50 g C 6 H 5 COOH


kg C 6 H 6
0.0750 kg C 6 H 6
1 mol C 6 H 5 COOH
 0.929m
122 g C 6 H 5 COOH
2. Calculate the depression for this solution.
Tf  K f m
Tf  (5.12 0 C/m)( 0.929m)  4.76 0 C
F.P. = 5.480 C - 4.76 0 C = 0.72 0 C
29
Determination of Molecular Weight by
Freezing Point Depression
The size of the freezing point depression
depends on two things:

1.
2.

The size of the Kf for a given solvent, which are
well known.
And the molal concentration of the solution which
depends on the number of moles of solute and
the kg of solvent.
If Kf and kg of solvent are known, as is
often the case in an experiment, then we
can determine # of moles of solute and use
it to determine the molecular weight.
30
Determination of Molecular Weight by
Freezing Point Depression

Example 14-7: A 37.0 g sample of a new
covalent compound, a nonelectrolyte, was
dissolved in 2.00 x 102 g of water. The
resulting solution froze at -5.58oC. What is
the molecular weight of the compound?
31
Determination of Molecular Weight by
Freezing Point Depression
Tf  K f m thus the
Tf 5.580 C
m

 3.00m
0
K f 1.86 C
In this problem there are
200 mL  0.200 kg of water.
? mol compound in 0.200 kg H 2 O = 3.00 m  0.200 kg
 0.600 mol compound
37 g
Thus the molar mass is
 61.7 g/mol
0.600 mol
32
Colligative Properties and Dissociation
of Electrolytes

Electrolytes have larger effects on boiling point
elevation and freezing point depression than
nonelectrolytes.



This is because the number of particles released in
solution is greater for electrolytes
One mole of sugar dissolves in water to produce
one mole of aqueous sugar molecules.
One mole of NaCl dissolves in water to produce
two moles of aqueous ions:

1 mole of Na+ and 1 mole of Cl- ions
33
Colligative Properties and Dissociation
of Electrolytes

Remember colligative properties depend on the
number of dissolved particles.


Since NaCl has twice the number of particles we can
expect twice the effect for NaCl than for sugar.
The table of observed freezing point
depressions in the lecture outline shows this
effect.
34
Colligative Properties and Dissociation
of Electrolytes

Ion pairing or association of ions prevents the
effect from being exactly equal to the number
of dissociated ions
35
Colligative Properties and Dissociation
of Electrolytes


The van’t Hoff factor, symbol i, is used to
introduce this effect into the calculations.
i is a measure of the extent of ionization or
dissociation of the electrolyte in the solution.
i
Tf actual
Tf if nonelectrolyte
36
Colligative Properties and Dissociation
of Electrolytes

i has an ideal value of 2 for 1:1 electrolytes like NaCl,
KI, LiBr, etc.
2O
Na + Cl- H

Na +aq   Cl-aq  2 ions

formula unit
i has an ideal value of 3 for 2:1 electrolytes like
K2SO4, CaCl2, SrI2, etc.
2O
Ca 2+ Cl-2 H

Ca 2aq+   2 Cl-aq  3 ions
formula unit
37
Colligative Properties and Dissociation
of Electrolytes



Example 14-8: The freezing point of 0.0100 m NaCl
solution is -0.0360oC. Calculate the van’t Hoff factor
and apparent percent dissociation of NaCl in this
aqueous solution.
meffective = total number of moles of solute
particles/kg solvent
First let’s calculate the i factor.
Tf  actual 
K f meffective meffective
i


Tf  if nonelectrolyte
K f mstated
mstated
38
Colligative Properties and Dissociation
of Electrolytes
Tf  actual 
K f meffective meffective
i


Tf  if nonelectrolyte 
K f mstated
mstated
Tf  actual 
0.0360 0 C
Tf  actual   K f meffective  meffective 

Kf
1.86 0 C m
meffective
meffective 0.0194 m
 0.0194 m  i 

 194
.
mstated
0.0100 m
39
Colligative Properties and Dissociation
of Electrolytes


Next, we will calculate the apparent percent
dissociation.
Let x = mNaCl that is apparently dissociated.
40
Colligative Properties and Dissociation
of Electrolytes
NaCl 
 Na + Cl
H 2O
(0.0100  x)m
+
xm
-
xm
41
Colligative Properties and Dissociation
of Electrolytes
NaCl 
 Na + Cl
H 2O
(0.0100  x)m
+
xm
meffective  0.0100  x  x  x  m
-
xm
 0.0100  x  m  0.0194 m
x  0.0094 m
42
Colligative Properties and Dissociation
of Electrolytes
apparent % dissociati on =
mapp diss
mstated
100%
0.0094m

100%
0.0100m
 94%
43
Colligative Properties and Dissociation
of Electrolytes

Example 14-9: A 0.0500 m acetic acid
solution freezes at -0.0948oC. Calculate the
percent ionization of CH3COOH in this
solution.
You do it!
44
Colligative Properties and Dissociation
of Electrolytes
 H + + CH COOCH 3COOH 
3
0.0500  x m x m
xm
meff  0.0500  x   x  x  m  0.0500  x  m
Tf  K f  meff
meff
Tf
0.09480 C


 0.0510 m
0
Kf
1.86 C m
meff  0.0500  x  m  0.0510 m
x  0.0010 m
% ionized =
mionized
0.0010 m
 100% 
 100%
moriginal
0.0500 m
 2.0% ionized and 98.0% unionized
45
Osmotic Pressure
Osmosis is the net flow of a solvent
between two solutions separated by a
semipermeable membrane.


The solvent passes from the lower concentration
solution into the higher concentration solution.
Examples of semipermeable membranes
include:

1.
2.
3.
cellophane and saran wrap
skin
cell membranes
46
Osmotic Pressure
semipermeable membrane
sugar dissolved
H2O
in water
H2O
H2O
H2O
H2O
H2O
H2O
net2solvent
flow
O
47
Osmotic Pressure
48
Osmotic Pressure

Osmosis is a rate controlled phenomenon.


The solvent is passing from the dilute solution into
the concentrated solution at a faster rate than in
opposite direction, i.e. establishing an equilibrium.
The osmotic pressure is the pressure exerted
by a column of the solvent in an osmosis
experiment.
  MRT
where:  = osmotic pressure in atm
M = molar concentration of solution
L atm
mol K
T = absolute temperature
R = 0.0821
49
Osmotic Pressure
 For very dilute aqueous solutions, molarity
and molality are nearly equal.
 Mm
  mRT
for dilute aqueous solutions only
50
Osmotic Pressure
Osmotic pressures can be very large.

For example, a 1 M sugar solution has an osmotic
pressure of 22.4 atm or 330 p.s.i.

Since this is a large effect, the osmotic
pressure measurements can be used to
determine the molar masses of very large
molecules such as:

Polymers
Biomolecules like
1.
2.


proteins
ribonucleotides
51
Osmotic Pressure

Example 14-18: A 1.00 g sample of a biological
material was dissolved in enough water to give
1.00 x 102 mL of solution. The osmotic pressure
of the solution was 2.80 torr at 25oC. Calculate
the molarity and approximate molecular weight
of the material.
You do it!
52
Osmotic Pressure
  MRT  M 

RT
1 atm
? atm = 2.80 torr 
 0.00368 atm = 
760 torr
0.00368 atm
4
M =

150
.

10
M
L atm
0.0821 mol K 298 K
53
Osmotic Pressure
  MRT  M 

RT
1 atm
? atm = 2.80 torr 
 0.00368 atm = 
760 torr
0.00368 atm
4
M =

150
.

10
M
L atm
298 K
0.0821 mol
K 
?g
1.00 g
1L
4g



6
.
67

10
mol
4
mol 0.100 L 150
.  10 M
typical of small proteins
54
Osmotic Pressure

Water Purification by Reverse Osmosis
If we apply enough external pressure to an
osmotic system to overcome the osmotic
pressure, the semipermeable membrane
becomes an efficient filter for salt and other
dissolved solutes.



Ft. Myers, FL gets it drinking water from the Gulf
of Mexico using reverse osmosis.
US Navy submarines do as well.
Dialysis is another example of this phenomenon.
55
End of Chapter 2

Human Beings are solution chemistry in
action!
56