Colligative Properties

Consider three beakers:
 50.0 g of ice
 50.0 g of ice + 0.15 moles NaCl
 50.0 g of ice + 0.015 moles sugar (sucrose)

What will the temperature of each beaker be?
Tf = Kf  m
 Beaker 1: 0C
 Beaker 2: -5.58 C
 Beaker 3: -.558 C

Colligative Properties

The reduction of the freezing point of a
substance is an example of a colligative
property:
 A property of a solvent that depends on
the total number of solute particles
present

There are four colligative properties to
consider:
 Vapor pressure lowering (Raoult’s Law)
 Freezing point depression
 Boiling point elevation
 Osmotic pressure
Colligative Properties – Vapor Pressure

A solvent in a closed container
reaches a state of dynamic
equilibrium.

The pressure exerted by the vapor in
the headspace is referred to as the
vapor pressure of the solvent.

The addition of any nonvolatile solute
(one with no measurable vapor
pressure) to any solvent reduces the
vapor pressure of the solvent.
Colligative Properties – Vapor Pressure

Nonvolatile solutes reduce the ability of the
surface solvent molecules to escape the
liquid.
 Vapor pressure is lowered.

The extent of vapor pressure lowering
depends on the amount of solute.
 Raoult’s Law quantifies the amount of
vapor pressure lowering observed.
Colligative Properties – Vapor Pressure

Raoult’s Law:
PA = XAPOA
where PA = partial pressure of the solvent
vapor above the solution (ie with
the solute)
XA = mole fraction of the solvent
PoA = vapor pressure of the pure
solvent
Colligative Properties – Vapor Pressure
Example: The vapor pressure of water at 20oC
is 17.5 torr. Calculate the vapor pressure of an
aqueous solution prepared by adding 36.0 g of
glucose (C6H12O6) to 14.4 g of water.
Given: PoH2O= 17.5 torr
mass solute = 36.0 g of glucose=.2 mol
mass solvent = 14.4 g of water= .8 mol
Find: PH2O
Raoult’s Law: PA = XAPOA
Colligative Properties – Vapor Pressure
Solution:
Raoult’s Law: PA = XAPOA
PoH2O= 17.5 torr
mass solute = 36.0 g of glucose=.2 mol
mass solvent = 14.4 g of water= .8 mol
So, .8/(.2 +.8) x 17.5 torr =
Answer: 14.0 torr
Colligative Properties – Vapor Pressure
Example: The vapor pressure of pure water at
110oC is 1070 torr. A solution of ethylene
glycol and water has a vapor pressure of 1.10
atm at the same temperature. What is the mole
fraction of ethylene glycol in the solution?
Both ethylene glycol and water are liquids.
How do you know which one is the solvent and
which one is the solute?
Colligative Properties – Vapor Pressure
Given: PoH2O = 1070 torr
Psoln = 1.10 atm
Find: XEG
Raoult’s Law: PA = XAPOA
Solution:
Answer: XEG = 0.219
Colligative Properties – Vapor Pressure

Ideal solutions are those that obey Raoult’s
Law.

Real solutions show approximately ideal
behavior when:
 The solution concentration is low
 The solute and solvent have similarly sized
molecules
 The solute and solvent have similar types
of intermolecular forces.
Colligative Properties – Vapor Pressure

Raoult’s Law breaks down when solventsolvent and solute-solute intermolecular
forces of attraction are much stronger or
weaker than solute-solvent intermolecular
forces.
Raoult’s Law: Mixing Two Volatile Liquids
 Since
BOTH liquids are volatile and
contribute to the vapor, 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 = cAP°A + cBP°B
Also, cA + cB = 1
(since there are 2 components)
Boiling Point
Elevation
Colligative Properties – BP Elevation

The addition of a
nonvolatile solute
causes solutions to
have higher boiling
points than the pure
solvent.
 Vapor
pressure
decreases with
addition of nonvolatile solute.
 Higher
temperature is
needed in order for vapor
pressure to equal 1 atm.
Colligative Properties- BP Elevation

The change in boiling point is proportional to
the number of solute particles present and
can be related to the molality of the solution:
Tb = Kb.m
where Tb = boiling point elevation
Kb = molal boiling point elevation
constant
m = molality of solution
The value of Kb depends only on the identity
of the solvent (see Table 11.5).
Colligative Properties - BP Elevation
Example: Calculate the boiling point of an
aqueous solution that contains 20.0 mass %
ethylene glycol (C2H6O2, a nonvolatile liquid).
Solute = 20g
Solvent =80g
Kb (solvent) =.51 C/m
Tb = Kb  m
Colligative Properties - BP Elevation
Molality of solute:
Solute = 20g
Solvent =80g
Kb (solvent) =.51 C/m
Tb =
BP = 102.1oC
Tb = Kb  m
Colligative Properties - BP Elevation
Example: The boiling point of an aqueous
solution that is 1.0 m in NaCl is 101.02oC
whereas the boiling point of an aqueous
solution that is 1.0 m in glucose (C6H12O6) is
100.51oC. Explain why.
Answer: Van’t Hoff Factor
Van’t Hoff Factor (i)




NaCl is a strong electrolyte and should
dissolve into two distinct ions
Glucose is also dissolvable, but will only
form one molecule.
Therefore NaCl should lower the vapor
pressure by twice as much as glucose does.
i= moles of particles in solution
moles of solute dissolved
van’t Hoff Factor
However, One
mole of NaCl in
water does not
really give rise to
two moles of ions.
van’t Hoff Factor
Some Na+ and Cl−
reassociate for a
short time, so the
true concentration
of particles is
somewhat less
than two times the
concentration of
NaCl.
The van’t Hoff Factor


Reassociation is
more likely at higher
concentration.
Therefore, the
number of particles
present is
concentration
dependent, and (i)
can only be
determined
experimentally.
The van’t Hoff Factor
We modify the
previous equations
by multiplying by
the van’t Hoff
factor, i
Tf = i  Kf  m
Colligative Properties - BP Elevation
Example: A solution containing 4.5 g of glycerol,
C3H5(OH)3 a nonvolatile nonelectrolyte, in 100.0 g
of ethanol, C2H6O, has a boiling point of 79.0oC. If
the normal BP of ethanol is 78.4oC, calculate the
molar mass of glycerol.
Given: Tb = 79.0oC - 78.4oC = 0.6oC
mass solute = 4.5 g
mass solvent = 100.0 g = 0.100 kg
Kb = 1.22oC/m (Table 11.5)
Find:
molar mass (g/mol)
Tb = i  Kb  m
Freezing Point
Depression
Colligative Properties - Freezing Pt Depression

The addition of a
nonvolatile solute
causes solutions to
have lower freezing
points than the pure
solvent.

Solid-liquid equilibrium
line rises ~ vertically
from the triple point,
which is lower than that
of pure solvent.
 Freezing
point of
the solution is
lower than that of
the pure solvent.
Colligative Properties - Freezing Pt Depression

The magnitude of the freezing point depression is
proportional to the number of solute particles and
can be related to the molality of the solution.
Tf =i  Kf  m
where Tf = freezing point depression
Kb = molal freezing point depression
constant
m = molality of solution
The value of Kf depends only on the identity of
the solvent (see Table 11.5).
Colligative Properties - Freezing Pt Depression
Example: Calculate the freezing point
depression of a solution that contains 5.15 g of
benzene (C6H6) dissolved in 50.0 g of CCl4.
Given: mass solute =
mass solvent =
Kf solvent =
Find:
Tf =39.3
Tf = i  Kf  m
Colligative Properties - Freezing Pt Depression
Example: Which of the following will give the
lowest freezing point when added to 1 kg of
water: 1 mol of Co(C2H3O2)2, 2 mol KCl, or 3
mol of ethylene glycol (C2H6O2)? Explain why.
KCl don’t forget the van’t Hoff Factor applies
here as well we end up with 4 mol of ions the
most out of this group.
Osmosis
Colligative Properties - Osmosis

Some substances form semipermeable
membranes, allowing some smaller particles
to pass through, but blocking other larger
particles.

In biological systems, most semipermeable
membranes allow water to pass through, but
solutes are not free to do so.
If two solutions with identical concentration
(isotonic solutions) are separated by a
semipermeable membrane, no net movement
of solvent occurs.

Colligative Properties - Osmosis

Osmosis: the net movement of a
solvent through a semipermeable
membrane toward the solution
with greater solute concentration.

In osmosis, there is net movement
of solvent from the area of lower
solute concentration to the area of
higher solute concentration.
 Movement of solvent from high
solvent concentration to low
solvent concentration
Colligative Properties - Osmosis

Osmosis plays an important
role in living systems:
 Membranes of red blood
cells are semipermeable.

Placing a red blood cell in a
hypertonic solution (solute
concentration outside the cell
is greater than inside the cell)
causes water to flow out of the
cell in a process called
crenation.
Colligative Properties

Placing a red blood cell in a hypotonic
solution (solute concentration outside the
cell is less than that inside the cell) causes
water to flow into the cell.
 The cell ruptures in a process called
hemolysis.
Colligative Properties - Osmosis

Other everyday examples of osmosis:
A
cucumber placed in brine solution loses
water and becomes a pickle.
A
limp carrot placed in water becomes firm
because water enters by osmosis.
 Eating
large quantities of salty food
causes retention of water and swelling of
tissues (edema).
Osmotic pressure


Osmosis is the spontaneous movement of water across a semipermeable 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
P = iMRT
where P – osmotic pressure (atm)
i = van’t Hoff factor
M = molarity
R = 0.08206 L∙atm/mol∙K
T = Kelvin temperature
Colloids:
Suspensions of particles larger than
individual ions or molecules, but too small
to be settled out by gravity.
Tyndall Effect


Colloidal suspensions
can scatter rays of
light.
This phenomenon is
known as the Tyndall
effect.
Colloids in Biological Systems
Some molecules
have a polar,
hydrophilic (waterloving) end and a
nonpolar,
hydrophobic (waterhating) end.
Colloids in Biological Systems
Sodium
stearate is one
example of
such a
molecule.
Colloids in Biological Systems
These molecules
can aid in the
emulsification of
fats and oils in
aqueous solutions.
Surfactants

Change the surface properties so that two
things that would not normally mix do
 Emulsifying agent
 Soap
 Detergent
Hydrophobic – water-fearing (nonpolar)
Hydrophilic – water-loving (polar)
Action of soap on oil