Colligative properties depend only on the number of dissolved particles in solution and not on their identity. The key colligative properties are vapor pressure lowering, boiling point elevation, and freezing point depression. Vapor pressure lowering occurs because solute particles decrease the number of solvent particles that can evaporate from the surface. Boiling point elevation and freezing point depression occur because adding solute particles lowers the vapor pressure of the solvent, requiring more energy for evaporation or freezing. The degree of change in boiling point or freezing point depends on the molality of the solution.

2. Solutions have different properties than
either the solutes or the solvent used to
make the solution.
Those properties can be divided
into:
Colligative
properties
Non-
Colligative
properties
Non-colligative properties depend on the identity of the
dissolved species and the solvent.

3. Colligative properties depend only on the
number of dissolved particles
(molecules or ions, small or large) in
solution and not on their identity.
I. Osmotic pressure
II. Vapor pressure lowering
III. Boiling point elevation
IV. Freezing point depression
Colligative properties

4. Explanation
compare the properties of:
1.0 M aq. sugar solution to a 0.5 M aq. solution of NaCl.
Despite the conc. of NaCl = ½ the conc. of sucrose both
solutions have precisely the same number of dissolved
particles, why?
because each NaCl unit creates two particles upon
dissolution a Na+ and a Cl-.
Both solutions have the same freezing point, boiling
point, vapor pressure, and osmotic pressure because
those colligative properties of a solution only depend on
the number of dissolved particles.
Other non-colligative properties including e.g. viscosity,
surface tension, and solubility are different.

5. Diffusion in liquids:
 Substance tend to move or diffuse from regions of
higher concentration to region of lower concentration
so the differences in concentration disappear.
 By placing water (solvent) on conc. aqueous salt
solution salt moves into water layer &
water moves to salt solution
till the new solution becomes uniform.

6.  Osmosis:
on separating conc. solution from water by semi
permeable membrane (allow passage of solvent&
prevent solute) water will diffuse from solution of
low solute conc. to solution of higher solute conc.
 The diffusion of solvent through semipermeable
membrane is called osmosis

7.  The solvent moves into the tube and the height of the
solution will rise till the hydrostatic pressure exerted
by the column of solution equal to osmotic pressure
What is then the Osmotic pressure?
 This is the external pressure that must be applied to
the solution in order to prevent it being diluted by the
entry of solvent via osmosis.

8. Osmotic pressure of Non-electrolytes
(not ionized)
Osmotic pressure  concentration
twice concentration twice osmotic pressure
Osmotic pressure  number of molecules
Osmotic pressure of 2 solutions having the same
molal concentration are identical

9. Example:
Explain why?
Solution contains 34.2 g sucrose (mol wt 342) in 1000 g
water has the same osmotic pressure as dextrose
solution (mol wt 180) contains 18 g/1000 g water?
 No. of moles of sucrose = wt/mol wt
= 34.2/342 = 0.1 molal
 No. of moles of dextrose = wt/mol wt
= 18/180 = 0.1 molal
So that the 2 solutions are iso-osmotic

10. Van’t Hoff Equation
 Van’t Hoff stated that there is a proportionality
between osmotic pressure & concentration &
absolute temp. (not applied in conc. solution)
 He Suggested a relationship analogue to the
equation for an ideal gas :
osmotic pressure ( ) in a diluted solution was equal to the
pressure exerted by the solute if it was a gas occupying the
same volume
PV = nRT
  c
  T
 = R c T

11. V = nRT
 = n/V RT
 = c R T
 = m R T
 = Osmotic pressure
T = absolute temperature
n = number of moles
V = volume in liter
c = molarity (n/V)
m = molality (n/wt)
Morse equation
 = c R T
 = m R T
Van’t Hoff equation

12.  Example:
What is the osmotic pressure of 1 g sucrose
(mol wt = 342) dissolved in 100 ml water at
25oC ? R= 0.082
Moles of sucrose = wt/M wt = 1/342
= 0.0029/100 ml
0.0029 moles in 100 ml
X moles in 1000 ml = 0.029 moles
Molarity = 0.029/1000 ml
 = cRT
= 0.029 x 0.082 x (25+273)
= 0.7 atm.

13. Osmotic pressure of Electrolytes
 They dissociate into ions
 increased number of particles (ions) formed
 so increase osmotic pressure
 NaCl ionized into 2 ions each has the same effect on
osmotic pressure as molecules
so  will be twice the solution contains the same
molal concentration of non-ionized substance.
K2SO4 3 times
FeCl3 4 times

14. V = i n RT  = i m RT
(i = dissociation factor )
For strong electrolyte
i = number of ions produced by ionization of
strong electrolytes (in diuted solution)
For weak electrolyte
i = total no. of particles (ions or molec) divided by
initial no. before ionization
For non-electrolyte
i = 1

15.  When non-volatile solute is dissolved in solvent, the
vapor pressure of solvent is lowered
 Solvent molecules on the surface which can escape
into vapor is replaced by solute molecules have little
(if any) vapor pressure.
Why is vapor pressure lowering a colligative property?
It only depends on number of dissolved solute particles.

16.  For ideal solution of non-electrolytes, vapor
pressure of solution follows Raoult’s law:
the vapor pressure of a solution, P, equals the mole
fraction of the solvent, X, multiplied by the vapor
pressure of the pure solvent, Po .
PA = XAPo
A
BA
A
o
AA
o
XX
P
PP


)1(
PA= v.p of solution
Po
A= v.p of pure solvent
XA = mole fraction of solvent
XB= mole fraction of solute
i.e relative vapor pressure lowering is equal
to the mole fraction of the solute XB
A
o
AA
o
P
PP 

17.  The absolute lowering of vapor pressure
(Po
A –PA) or ΔP of solution is given by
Po
A –PA= XB Po
A
Lowering of vapor pressure is
proportional to mole fraction of
solute
P  XB

18.  Problem: calculate the lowering of vapor pressure & vapor
pressure of solution containing 50g dextrose (mol wt 180) in
1000 g water (mole wt 18). The vapor pressure of water =
17.535 mm/Hg
nA (no of moles of solvent)= 1000/18 = 55 moles
nB (no of moles of solute) =50/180 = 0.278 mole
Mole fraction of dextrose XB = nB / (nA+nB)
= 0.278 / (55.5+0.278)
= 0.00498
Vapour pressure lowering
Po
A – PA = XB Po
A
= 0.00498 x 17.535
= 0.0873 mm
Vapor pressure of solution
PA = 17.535 - 0.0873 = 17.448 mm

19.  The boiling point of a liquid is defined as the temperature at
which the vapor pressure of that liquid equals the
atmospheric pressure(760mm Hg).
For a solution:
the vapor pressure of the
solvent is lower at any
given temperature.
Therefore, a higher
temperature is required to
boil the solution than the
pure solvent.

20. If we represent the difference in boiling point between the pure
solvent and a solution as Tb,
we can calculate that change in boiling point from the following
formula:
m: molality (because molality is temperature independent).
Kb: boiling point elevation constant that depends on the
particular solvent being used. (Kb water = 0.51)
i: van't Hoff factor and represents the number of dissociated
moles of particles per mole of solute

21.  Normal freezing or melting
point: is the temp. at which
solid & liquid are in
equilibrium under 1 atm.
 Addition of solute will
decrease the vapor pressure
and so will decrease the
freezing point
liquid
vapor
solid
Temp.
pressure
760
0
4.58
 Tf
 Tb
In order for a liquid to freeze it must achieve a very ordered state that
results in the formation of a crystal.
If there are impurities in the liquid, i.e. solutes, the liquid is inherently
less ordered. Therefore, a solution is more difficult to freeze than the
pure solvent so a lower temperature is required to freeze the liquid

22. In analogy to the boiling point elevation, we can calculate the
amount of the freezing point depression using the following
formula:
the –ve sign because the freezing point of the solution is less
than that of the pure solvent.
m: molality (because molality is temperature independent).
Kf: molal depression constant that depends on the
particular solvent being used. (Kb water = 1.86)
i: van't Hoff factor and represents the number of dissociated
moles of particles per mole of solute
N.B freezing point depression and boiling point elevation are a
direct result of the lowering of vapor pressure

23. Colligative Properties
 Changes in Vapor Pressure: Raoult’s Law
 The vapor pressure over the solution is lower than the vapor
pressure of pure solvent.
 The vaopr pressure of the solvent Psolv is propotional to the
relative number of solvent molecules in the solution, the solvent
vapor pressure is proportional to the solvent mole fraction.
 Colligative Properties Definitions

24. Colligative Properties

25. Colligative Properties
 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

26. COLLIGATIVE
PROPERTIES

27. AFTER STUDYING THIS SECTION, STUDENTS
SHOULD KNOW:
 How to explain the affects that a
nonvolatile substance has on vapor
pressure, boiling point and freezing point
 How to determine which solution will have
a higher boiling point, lower freezing point
or lower vapor pressure based on the
number of nonvolatile substances

28. COLLIGATIVE PROPERTIES
 Vapor pressure lowering
 Boiling point elevation
 Freezing point depression

29.  Decrease in Vapor Pressure
 The pressure exerted by a vapor that is in equilibrium with its
liquid in a closed system
 A solution with a nonvolatile substance (one that is not easily
vaporized) contains a lower vapor pressure than a pure solvent
 When you dissolve salt or sugar into water, the molecules / ions
are surrounded by water molecules which “trap” them into the
solvent
 Therefore, fewer molecules will contain enough kinetic energy to
escape the vapor of the liquid

30. Boiling Point Elevation
 The difference in temperature at which the vapor
pressure of the liquid phase equals the atmospheric
pressure
 When a non-volatile substance is added to the
solvent, the vapor pressure decreases, so more kinetic
energy has to be added to raise the vapor pressure of
the liquid
 Depends on the concentration of particles, not the
type of particles

31. Freezing Point Depression
 The difference in temperature between the freezing
point of a solution and that of the pure solvent
 Amount of change in freezing point depends on the
number of particles dissolved and not the type

32. AFTER READING SECTION 18.3, YOU
SHOULD KNOW:
 How to explain the affects that a nonvolatile
substance has on vapor pressure, boiling point
and freezing point
 How to determine which solution will have a
higher boiling point, lower freezing point or lower
vapor pressure based on the number of
nonvolatile substances

33. How Vapor Pressure Depression Occurs
• Solute particles take up space in a solution.
• Solute particles on surface decrease the
number of solvent particles on the surface.
• Less solvent particles can evaporate which
lowers the vapor pressure of a liquid.

34. Vapor Pressure
As solute molecules are added
to a solution, the solvent
become less volatile
(=decreased vapor
pressure).
Therefore, the vapor pressure
of a solution is lower than
that of the pure solvent.

35. Vapor Pressures of Pure Water and a Water Solution
The vapor pressure of water over pure water is greater than the
vapor pressure of water over an aqueous solution containing a
nonvolatile solute.
Solute particles take up
surface area and lower
the vapor pressure

36. Raoult’s Law
 Vapor pressure of a solution varies directly as
the mole fraction of solvent:
PA = XAPA
where
• XA is the mole fraction of compound A
• PA is the normal vapor pressure of A at that temperature
NOTE: This is one of those times when you want to make
sure you have the vapor pressure of the solvent (NOT the
solute!!!)

37. Raoult’s Law (continued)
• A solution of water and glucose with a mole
fraction of 0.5 will have a vapor pressure
that is 0.5 times the vapor pressure of water
alone.

38. SAMPLE EXERCISE– Calculation of
Vapor-Pressure Lowering
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 (1 torr = 1 mmHg) (you
can look these up!).

39. Solution
Analyze: Our goal is to calculate the vapor pressure of a solution, given the
volumes of solute and solvent and the density of the solute.
Plan: We can use Raoult’s law (Equation 13.10) to calculate the vapor
pressure of a solution. The mole fraction of the solvent in the solution, XA, is
the ratio of the number of moles of solvent (H2O) to total solution (moles
C3H8O3 + moles H2O).
Solve: To calculate the mole fraction of water in the solution, we must determine the number of
moles of C3H8O3 and H2O:

40. SAMPLE EXERCISE 13.8 continued
The vapor pressure of the solution has been lowered by 0.6 torr relative to that of
pure water.
We now use Raoult’s law to calculate the vapor pressure of water for the solution:

41.  PRACTICE EXERCISE
 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

42. Ideal Solution
• All intermolecular attractions are the same.
• This is like an Ideal Gas – do you think Ideal
Solutions are the same as Real Solutions?
• Solute-Solute
• Solvent-Solvent
• Solute-Solvent
Attractions are the
same

43. Boiling Point Elevation and
Freezing Point Depression
Solute-solvent
interactions also
cause solutions to
have higher boiling
points and lower
freezing points than
the pure solvent.

44. Boiling Point Elevation
The change in boiling point is proportional to the molality of the
solution:
Tb = Kb  m
where Kb is the molal boiling point elevation constant, a property of the
solvent.
Tb is added to the normal boiling point of the solvent.

45. Freezing Point Depression
 The change in freezing point can be found similarly:
Tf = Kf  m
 Here Kf is the molal freezing point depression constant of the
solvent.
Tf is subtracted from the normal freezing point of the solvent.

46. Boiling Point Elevation and
Freezing Point Depression
In both equations, T
does not depend on what
the solute is, but only on
how many particles are
dissolved.
Tb = Kb  m
Tf = Kf  m

47. Colligative Properties of
Electrolytes
Because these properties depend on the number of particles
dissolved, solutions of electrolytes (which dissociate in solution)
show greater changes than those of nonelectrolytes.
e.g. NaCl dissociates to form 2 ion particles; It lowers freezing
points almost twice as much as methanol, a nonelectrolyte.
CaCl2 dissociates to form 3 ion particles; It lowers freezing
points almost three times as much as methanol.

48. Colligative Properties of
Electrolytes
However, a 1 M solution of NaCl does not show twice the change in
freezing point that a 1 M solution of methanol does.
It doesn’t act like there are really 2 particles.
The charged ions are still attracted to each other, and sometimes act
like one particle (instead of two.)

49. Molar Mass from
Colligative Properties
We can use the effects of a colligative property such
as osmotic pressure to determine the molar mass of a
compound.
We’ll do examples from the packet.

50. CALCULATIONS INVOLVING
COLLIGATIVE PROPERTIES
Section 18.4

51. AFTER READING SECTION
18.4, YOU SHOULD KNOW:
 How to calculate molality (m)
 How to calculate the molar mass of a
molecular compound from the freezing
point depression or boiling point elevation

52. MOLALITY
 Molality (m) – the number of moles of
solvent dissolved per kilogram of solvent
 m = mol solute / kg solvent
 Mole fraction
nA = moles of solute A nB moles
of solvent B

53. BOILING POINT ELEVATION
 Change in boiling point temperature
 ΔTb = Kb * m
 ΔTb = change in boiling pt temp
 Kb = molal boiling-point elevation constant
 Depends on the solvent
 Units (oC / m)
 m = molality (mole / kg)

54. FREEZING POINT
DEPRESSION
 Change in freezing point temperature
 ΔTf = Kf * m
 ΔTf = change in freezing pt temp
 Kf = molal freezing-point depression constant
 Depends on the solvent
 Units (oC / m)
 m = molality (mole / kg)

56. Concentration

57. CALCULATING CONCENTRATION
OF SOLUTIONS
1. Mass Percent = (mass of solute/mass of solution)100
2. Parts per million = (mass of solute/mass of solution)106
3. Mass/volume percent = (mass of solute/mL
solution)100
4. Volume percent = (mL solute / mL solution)100
5. Molarity = n/V = moles solute / L solution
6. Molality = moles of solute / kg solvent

58. CALCULATING CONCENTRATION
OF SOLUTIONS
1. What is the mass percent of a solution made by
adding 25.0 g of KCl to 100.0 mL of water?
2. What is the concentration of a solution made
by mixing 25.0 g of KCl in (a) 100.0 mL and
(b) 100.0 L of water in ppm?

59.  The following slides are
extra worked out examples.

60. CALCULATING CONCENTRATION OF
SOLUTIONS
 How many grams of a solution that is 32.7%
by mass NaCl would contain 45.0 g of NaCl?
Mass % = (mass of solute /mass of solution) 100
32.7 % = (45.0 g / x )100
x = 45.0 g / 0.327
x = mass of solution = 138 g

61. CALCULATING CONCENTRATION OF
SOLUTIONS
 How much solute is present in 756.1 mL of a
14.7% (mass/volume) HCl solution?
Mass/volume % = (masssolute/volumesolution) 100
14.7 % = (x / 756.1 mL )100
x = 0.147 (756.1 mL)
x = mass of solute = 111g

62. CALCULATING CONCENTRATION OF
SOLUTIONS
 How many grams of LiF are in 500 mL of a
solution that is 18.4% by mass and has a
density of 1.197 g/mL?
 Masssolution = densitysolution (Volumesolution)
= 1.197 g/mL (500 mL)
= 598.5 g solution
Mass% = (masssolute/masssolution) 100
18.4 % = (x / 598.5 g )100
x = 0.184 (598.5 g)
x = mass of solute = 110 g

63. CALCULATING CONCENTRATION OF
SOLUTIONS
 A 350 mL sample of drinking water was analyzed
and found to contain 0.0046 g of sulfate salts.
Calculate the concentration of sulfate salts in this
water sample?
 Masswater = densitywater (Volumewater)
= 1.00 g/mL (350 mL)
= 350 g solution
ppm = (masssolute/masssolution) 106
ppm = (0.0046g / 350 g ) 106
ppm = 13 ppm
Use ppm for trace amounts of solute.

64. CALCULATING CONCENTRATION OF
SOLUTIONS
 Calculate the molality of a solution composed of
53.0 g of KOH in 500 mL of water.
First calculate the moles of solute:
n = 53.0 g (1 mol / 56 g) = 0.946 mol solute
Next, m = moles of solute / kg solvent
m = (0.946 mol / 0.500 kg )
m = 1.89 molal solution

65. PRACTICE PROBLEMS #36
calculating the CONCENTRATION OF SOLUTIONS
___1. How many grams of a 45.0 % Mg(OH)2 solution can be made from
7.00 g of solid Mg(OH)2?
___2. How many kilograms of a 6.8% KC2H3O2 solution will contain 5.3
moles of KC2H3O2?
___3. How many grams of KOH are needed to make 250.0 mL of a solution
that is to contain 6.70 mg/mL of potassium ion?
___4. If 134.5 g of a 25.0 % sucrose solution was prepared, how many
grams of a 5.00% solution of sucrose would contain the same amount of
sugar?
___5. How many grams of solution, 10.0% KOH by mass, are required to
neutralize 25.0 mL of a 2.00 M HCl solution if the moles of HCl is equal
to the moles of KOH?
15.6 g
7.6 kg
2.41 g
673 g
28.0 g

66. Group study problem #36
calculating the CONCENTRATION OF SOLUTIONS
___1 How many grams of a solution, that is 76.3% by mass KBr, would
contain 38.95 g of KBr?
___2. What is the ppm and molality of a solution containing 75.0 g of
ethylene glycol, C2H6O2, in 200.0 g of water?
___3. How many grams of CuSO4 are needed to make 350.0 mL of a
solution that contains 10.5 mg/mL of Cu2+ ion? Calculate the molarity of
the resulting solution.
___4. Automobile battery acid is 38% H2SO4 and has a density of 1.29
g/mL. Calculate the molality of this solution.
___5. If 27.0 g of LiBr are dissolved in 50.0 g of water, the percent of
LiBr by mass is
___6. If 15.0 g of KNO3 is added to 75g of water, what is the mass
percent of KNO3 in the solution?

67. Raoult’s Law
 Psolv = Xsolv Po
solv
 Raoult’s Law applies to ideal solutions
Roult’s Law Diagram

68. Colligative Properties

69. Colligative Properties
 Raoult’s Law Adding a non volatile solute to a
solvent lowers the vapor pressure of the solvent
so the change in vapor pressure of the solvent
can be calculated as a function of the mole
fraction.

70. Colligative Properties

71. Colligative Properties

72. Colligative Properties

73. Colligative Properties
 The boiling point of a solution is related to
the solute concentration.
 The boiling point elevation  Tbp is directly
proportional to the molality of the solute:
  T bp = Kbp msolute
 molal boiling point elevation constant has
the units of degrees/molal (oC / m).

74. Colligative Properties
 What quantity of elthylene glycol,
HOCH2CH2OH must be added to 125 g of
water to raise the boiling point of 1.0oC?
 Kbp of water is + 0.5121oC/m

75. Colligative Properties
 The freezing point of a solution is related to the
solute concentration.
 The freezing point depression  Tfp is directly
proportional to the molality of the solute:
 Tfp = Kfp msolute
 molal freezing point depression constant has the
units of degrees/molal (oC / m).

76. Colligative Properties
 In the northern United States, summer cottages are
usually closed up for the winter. When doing so the
owners “winterize” the plumbing by putting
antifreeze, HOCH2CH2OH, in the toilet tanks. Will
adding 525 g of anitfreeze to 3 kg of water ensure
that the water will not freeze at -25oC.
 Kfp of water is -1.86 oC/m

77. Colligative Properties
 Colligative Properties and Molar Mass
Determination

78. Colligative Properties
 Van’t Hoff Factor i
 When an ionic compound dissolves in a solvent, the
number of ions determines the behavior of the
solvent.
 For example, 1 m NaCl will lower the f.p. of water
twice as much as 1 m sugar because NaCl breaks
into 2 ions

79. Colligative Properties
 Van’t Hoff Factor I Diagrams & Photo
 Actually only dilute solutions behave in this manner.
 Officially, i =  Tfp measured
 Tfp calculated
 So often the value of i approaches a whole
number.

80. Colligative Properties
Osmosis is the movement of solvent molecules through a
semipermeable membrane from a region of lower to a region
of higher solute concentration.

81. Colligative Properties
Osmotic Pressure the pressure created by the column above
the solution measured by the difference in height between
the solution in the tube and the level of water in the beaker.
Osmotic Terms

82. Colligative Properties
Osmotic Pressure is related to concentration.
Recall the Ideal Gas Law PV = nRT
Rearrange the equation so the concentration is (n/V)
P = (n/V) R T
Change the symbols P to P for osmotic pressure
and (n/V) to c
The resulting equation for Osmotic Pressure is
P = c R T R = .082 L x atm
mol x K

83. Colligative Properties
Osmosis
isotonic similar concentrations of solutes
hypotonic solution with lower solute concentrations
hypotonic solution with higher solute concentrations
reverse osmosis using pressure to purify water.

84. Colligative Properties
Colloids

85. Colligative Properties
Colloids are classified according to the state of dispersed phase
and the dispersing medium.
Hydrophobic ‘water fearing’ weak attractive forces
between water and surface of colloidal particles.
Hydrophilic ‘water loving’ are strongly attracted to the
water molecules
Emulsions are colloidal dispersions of one liquid in another
by using an emulsifying agent such as protein or soap.

86. Colligative Properties
Surfactants are emulsifying agents
Hydrocarbon end soluble in oil
Polar end soluble in water

87. Any Questions

88. Study Questions
 Define the following terms:
[Colligative properties, Osmotic pressure, vapour pressure, ideal solution, real solution, molar mass, electrolyte,
weight mass, gravity, molality, molarity, etc]
 Respond to the following questions:
 Explain the processes of the main colligative properties of the pharmaceutical materials and how the
properties vary with named factors
 What is gravity and how does it affect the movement of material substance
 Group work discussional questions:
 Give a detailed account of the variables that can be considered in the quantification of solutes in solution
system
 How many grams of a solution, that is 76.3% by mass KBr, would contain 38.95 g of KBr?
 What is the ppm and molality of a solution containing 75.0 g of ethylene glycol, C2H6O2, in 200.0 g of water?
 How many grams of CuSO4 are needed to make 350.0 mL of a solution that contains 10.5 mg/mL of Cu2+ ion?
Calculate the molarity of the resulting solution.
 Automobile battery acid is 38% H2SO4 and has a density of 1.29 g/mL. Calculate the molality of this
solution.
 If 27.0 g of LiBr are dissolved in 50.0 g of water, the percent of LiBr by mass is
 If 15.0 g of KNO3 is added to 75g of water, what is the mass percent of KNO3 in the solution?