This document discusses colligative properties, which are physical properties of solutions that depend on the concentration of solute particles rather than the identity of the solute. It defines key colligative properties like boiling point elevation, freezing point depression, and osmotic pressure. The document also explains how these properties obey mathematical relationships like Raoult's law. Several examples are provided to demonstrate how to use these relationships and colligative property equations to calculate values like vapor pressure, boiling point, freezing point, osmotic pressure, and molar mass of a solute.

1. 13.5 Colligative properties
Dissolving solute in pure liquid will change all
physical properties of liquid, Density, Vapor
Pressure, Boiling Point, Freezing Point, Osmotic
Pressure
Colligative Properties are properties of a liquid
that change when a solute is added.
The magnitude of the change depends on the
number of solute particles in the solution, not on
the identity of the solute particles.

2. Lowering the vapor pressure
The presence of a non-volatile solute means that fewer
solvent particles are at the solution’s surface, so less
solvent evaporates
A liquid in a closed container will establish equilibrium with its
vapor. When that equilibrium is reached, the pressure exerted
by the vapor is called the vapor pressure

3. Raoult’s Law
Describes vapor pressure lowering mathematically
.
The lowering of the vapour pressure when a
non-volatile solute is dissolved in a volatile
solvent (A) can be described by Raoult’s Law:
PA =XAPA
PA = vapour pressure of solvent A above the solution
XA = mole fraction of the solvent A in the solution
PA = vapour pressure of pure solvent A
only the solvent (A) contributes to
the vapour pressure of the solution

4. (
mol H2O
mol H2O + mol C12H22O11
(
Example: What is the vapor pressure of water above a sucrose
(MW=342.3 g/mol) solution prepared by dissolving 158.0 g of
sucrose in 641.6 g of water at 25 ºC? The vapor pressure of
pure water at 25 ºC is 23.76 mmHg.
Moles C12H22O11 =
(
(158 g C12H22O11)
1 mol C12H22O11
342.3 g C12H22O11
( =0.462mol
=
H2O =
X 35.6
35.6+ 0.462
=0.987
P =(0.987)(23.76 mmHg) = 23.5 mmHg
soln
P =X P
soln H2O H2O
Moles H2O =
(
(641.6 g H2O)
1 mol H2O
18 g H2O
( =35.6mol

5. Example: Glycerin (C3H8O3) is a nonvolatile nonelectrolyte with a
density of 1.26 g/mL at 25C of 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 23.5 torrr and its density is 1.0 g/mL. Calculate
the vapor pressure lowering
Moles C3H8O3 =
(
(50 mLC3H8O3)
1.26 g C3H8O3
1 mLC3H8O3
( =0.684mol
(
1 mol C3H8O3
92.1 g C3H8O3
(
Moles H2O =
(
(500 mLH2O)
1.0 g H2O
1 mLH2O
( =27.8mol
(
1 mol H2O
18 g H2O
(
=
(
mol H2O
mol H2O + mol C3H8O3
(
H2O =
X
(
27.8
27.8+ 0.684
( =0.976
P =(0.976)(23.8 torr) = 23.2 torr
H2O
P =X P
H2O H2O H2O

6. Example: 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?
=
760 torr
1070 torr
=0.710
P =X P
H2O H2O H2O H2O
P = 1070 torr
H2O
P = 1 atm = 760 torr
H2O
P
H2O =
X
H2O
P
XH2O XC2H6O2
+ =1
XC2H6O2
=10.71=0.290

7. Mixtures of Volatile Liquids
Both liquids evaporate & contribute to the vapor pressure

8. Raoult’s Law: Mixing Two Volatile Liquids
Since both liquids are volatile and contribute to the
vapour, the total vapor pressure can be represented
using Dalton’s Law:
PT = PA + PB
The vapor pressure from each component follows
Raoult’s Law:
PT = XAPA + XBPB
 
Also, XA + XB = 1 (since there are 2 components)

9. Benzene and Toluene
A two solvent (volatile) system
The vapor pressure from each component follows
Raoult's Law.
Benzene - Toluene mixture:
Recall that with only two components, XBz + XTol = 1
Benzene: when XBz = 1, PBz = PBz = 384 torr &
when XBz = 0 , PBz = 0
Toluene: when XTol = 1, PTol = PTol = 133 torr &
when XTol = 0, PBz = 0



10. Example: A mixture of benzene (C6H6) and toluene (C7H8)
containing 1.0 mol of benzene and 2.0 mol of toluene. What is
the total vapor pressure of the solution? [vapor pressures of
pure benzene and toluene are 75 torr and 22 torr, respectively]
C6H6 = =0.33
1
X
1+2
C7H8 = =0.67
2
X
1+2
PT = XAPA + XBPB
 
PT = [(0.33)(75 torr)] + [(0.67)((22 torr)]
= 24.75 torr + 14.75 torr
= 39.5 torr

11. Boiling-point elevation and Freezing-point depression
 In a solution of a nonvolatile solute, boiling and
freezing points differ from those of the pure solvent
 The boiling point of the solution is higher than
that of the pure liquid
Boiling point is elevated when solute inhibits solvent from escaping.
 The freezing point of the solution is lower than
that of the pure liquid
Freezing point is depressed when solute inhibits solvent from crystallizing.

12.  The diagram below shows how a phase diagram is
affected by dissolving a solute in a solvent.
 The black curve represents the pure liquid and the blue
curve represents the solution.
 Notice the changes in the boiling & freezing points.
Phase diagrams for a pure solvent and for a solution of nonvolatile solute

13.  The increase of boiling point, Tb is directly
proportional to the concentration of the solution
expressed by its molality, m.
Where, Tb = BP. Elevation
Tb = BP of solvent in solution
Tb° = BP of pure solvent
m = molality , kb = BP Constant
Tb = (Tb –Tb ) = kbm

Boiling-point elevation
 The boiling-point elevation is proportional to the
concentration of solute particles, regardless of whether the
particles are molecules or ions
 A 1 m aqueous solution of NaCl is 1 m Na+ and 1 m Cl-,
making 2 m in total solute particles
 The boiling-point of elevation of a 1 m aqueous solution of
NaCl is (2m)(0.51 C/m) = 1C.

14.  The decrease of freezing point, Tf is directly
proportional to the concentration of the solution
expressed by its molality, m.
Where, Tf = FP depression
Tf = FP of solvent in solution
Tf°= FP of pure solvent
m = molality
kf = FP depression constant
Freezing-point depression
Tf = (Tf –Tf) = kfm


15.  The van 't Hoff factor, i : The van 't Hoff factor is the
ratio between the actual concentration of particles
produced when the substance is dissolved, and the
concentration of a substance as calculated from its
mass.
 For most non-electrolytes dissolved in water, the van' t
Hoff factor is essentially 1.
 For most ionic compounds dissolved in water, the van't
Hoff factor is equal to the number of discrete ions in a
formula unit of the substance.
Tf (calculated for nonelectrolyte)
i =
Tf (measured)
For NaCl, van’t Hoff factor is 2, because NaCl consists of
one Na+ and on Cl- per formula unit

16. Example: Automotive antifreeze consists of ethylene glycol,
(CH2(OH)CH2(OH), a nonvolatile noneletrolyte. Calculate the
boiling point and freezing point of a 25.0 mass % solution of
ethylene glycol in water.[kb=0.51 (C/m) and kf=1.86 (C/m).
Let us consider we have 1000 g of solution:
Mass of ethylene glycol = 250 g
Mass of water = 750 g
Molality=
=5.37m
(
1 mol C2H6O2
62.1 g C2H6O2
(
(
moles C2H6O2
Kg H2O
( (
1000 g H2O
1 kg H2O
(
(
250 g C2H6O2
750 g H2O
(
=
Tb = (Tb –Tb ) = kbm

Tf = (Tf –Tf ) = kfm

= (0.51 C/m )  (5.37 m)
= 2.7 C
= (1.86 C/m )  (5.37 m)
= 10.0 C
Tb=Tb +Tb = 2.7 C +100 C

= 102.7 C
Tf=Tf – Tf = 0 C – 10 C

= – 10.0 C

17. Example: Calculate the freezing point of a solution containing
0.600 kg of CHCl3 and 42.0 g of eucalyptol (C10H18O), a
fragrant substance found in the leaves of eucalyptus trees
[kf=4.68 (C/m) and Tf=  63.5 C for chloroform].
Molality=
=0.45m
(
1 mol C10H18O
154 g C10H18O
(
(
moles C10H18O
Kg CHCl3
( (
42 g C10H18O
0.60 kg CHCl3
(
=
Tf = (Tf –Tf ) = kfm

= (4.68 C/m )  (0.45 m)
= 2.1 C
Tf =Tf – Tf = – 63.5 C – 2.1 C

= – 65.6 C

18. Example: List the following aqueous solutions in order of their
expected freezing point :
0.050 m CaCl2, 0.15 m NaCl, 0.10 m HCl, 0.050 m
CH3COOH, 0.10 m C12H22O11
CaCl2, NaCl and HCl are stronger electrolytes
CH3COOH is week electrolyte
C12H22O11 is nonelectrolyte
0.050 m CaCl2  0.050 m in Ca2+ and 0.10 m in Cl-  0.15 m in particles
0.15 m NaCl  0.15 m in Na+ and 0.15 m in Cl-  0.30 m in particles
0.10 m HCl  0.10 m in H+ and 0.10 m in Cl-  0.20 m in particles
0.050 m CH3COOH  between 0.050 m and 0.10 m in particles
0.050 m C12H22O11  0.10 m in particles
0.15 m NaCl (lowest freezing-point ), 0.10 m HCl , 0.050 m CaCl2, 0.050 m
C12H22O11, 0.050 m CH3COOH (highest freezing-point )

19. Osmosis
 Osmosis is the spontaneous movement of water across
a semi-permeable membrane from an area of low solute
concentration to an area of high solute concentration
 Osmotic Pressure - The pressure that must be applied
to stop osmosis

20. The osmotic pressure obeys a law similar in form to the
ideal gas law
 V= nRT
 =(n/V)RT = MRT
, osmotic pressure of soln
V, volume soln
n, number of moles of solute
R, ideal-gas constant
M, molarity of soln
 Two solutions having identical osmotic pressure are isotonic
 Solution of lower osmotic pressure is hypotonic with respect to
more concentrated soln
 Solution of more concentrated solution is hypertonic with
respect to the dilute soln

21. Example: The average osmotic pressure of blood is 7.7 atm at
25 C. What molarity of glucose (C6H12O6) will be isotonic
with blood?
 = M RT
M =

RT
T = 273 + 25 = 298 K
R = 0.0821 L.atm/mol.K
M =
(0.0821 L.atm/mol.K)(298 K)
7.7 atm
 = 7.7 atm
= 0.31 atm
M = ?

22.  = (0.0020 (mol/L)) (0.0821 L.atm/mol.K)(293 K)
Example: What is the osmotic pressure at 20 C of a 0.0020 M
sucrose (C12H22O11) solution?
 = M RT
T = 273 + 20 = 293 K
R = 0.0821 L.atm/mol.K
= 0.048 atm
M = 0.0020 M = 0.002 (mol/L)
 = ?

23. Example: A solution of an unknown nonvolatile nonelectrolyte was
prepared by dissolving 0.25 g of the substance in 40.0 g of CCl4. The
boiling point of the resultant solution was 0.357 C higher than that of
the pure solvent. Calculate the molar mass of the solute.
Number of mole of solute in the solution =
=0.00284molsolute
Tb = kbm
= 0.0711 m
 The colligative properties of solutions provides a
useful means experimentally determining molar mass.
Molality =
Tb
Kb
=
0.357 C
5.02 C/m
(
0.0711mol solute
Kg CCl4
(
(0.040 kg CCl4)
Molar mass =
=88g/mol
(
0.25 g
0.00284mol
(

24. Example: Camphor (C10H16O) melts at 179.8 C, and it has a
particularly large freezing-point-depression constant, kf =40.0 C/m.
When 0.186 g of an organic substance of unknown molar mass is
dissolved In 22.01 g of liquid camphor, the freezing point of the
mixture is found to be 176.7 C. What is the molar mass of the
solute?
Number of mole of solute in the solution =
=0.0017molsolute
Tf = kfm
= 0.0775 m
=
Molality =
Tf
Kf
3.1 C
40.0 C/m
(
0.0775 mol solute
Kg C10H16O
(
(0.02201 kg C10H16O)
=110g/mol
Molar mass =
(
0.186 g
0.0.0017mol
(
Tf = 179.8 – 176.7 = 3.1 C

25. Example: The osmotic pressure of an aqueous solution of a certain
protein was measured to determine the protein’s molar mass. The
solution contained 3.50 mg of protein dissolved in sufficient water to
form 5.00 mLof solution. The osmotic pressure of the solution at 25C
was found to be 1.54 torr. Treating the protein as a nonelectrolyte,
calculate its molar mass.
Mole of solute in the solution =
= 8.28  10-5 mol /L
Molarity =
0.002026 atm
(0.0821 L.atm/mol.K )(298 K)
=8.45  103 g/mol
Molar mass =
(
0.0035 g
4.14  10-7 mol
(
T = 273 + 25 = 298 C
R = 0.0821 L.atm/mol.K
 = 1.54 torr =1.54/760
= 0.002026 atm
 = M RT
(5.00  10-3 L)(8.28  10-5 mol /L)
= 4.14  10-7 mol

26. Example: A sample of 2.05 g of polystyrene of uniform polymer chain
length was dissolved in enough toluene to form 0.100 L of solution.
The osmotic pressure of this solution was found to be 1.21 kPa at
25C. Calculate the molar mass of the polystyrene.
= 4.88  10-4 mol /L
Molarity =
0.01194 atm
(0.0821 L.atm/mol.K )(298 K)
=4.20  104 g/mol
Molar mass =
(
2.05 g
4.88  10-5 mol
(
T = 273 + 25 = 298 C
R = 8.314 kg.m2/S2.mol.K
 = 1.21 kPa=1210/101325 =
0.01194 atm
 = M RT
Mole of solute in the solution = (0.10 L) (4.88  10-4 mol /L)
= 4.88  10-5 mol