Colligative Properties

Consider three beakers:
 50.0 g of ice
 50.0 g of ice + 0.15 moles NaCl
 50.0 g of ice + 0.15 moles sugar (sucrose)

What will
 Beaker
 Beaker
 Beaker
the temperature of each beaker be?
1:
2:
3:
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 reduced.

The extent of vapor pressure lowering depends
on the amount of solute.
 Raoult’s Law quantifies the amount of vapor
pressure lowering that is observed.
Colligative Properties – Vapor Pressure

Raoult’s Law:
PA = XAPOA
where PA = partial pressure of the solvent (A)
vapor above the solution (ie with
the solute)
XA = mole fraction of the solvent (A)
PoA = vapor pressure of the pure
solvent (A)
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.
Colligative Properties – Vapor Pressure
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
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 solvent-solvent
and solute-solute intermolecular forces of
attraction are much stronger or weaker than
solute-solvent intermolecular forces.
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:
DTb = Kb.m
where DTb = 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 13.4).
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 =
Solvent =
Kb (solvent) =
DTb = Kb  m
Colligative Properties - BP Elevation
Molality of solute:
DTb =
BP = 102.1oC
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.
Colligative Properties - BP Elevation
Example: A solution containing 4.5 g of glycerol,
a nonvolatile nonelectrolyte, in 100.0 g of
ethanol has a boiling point of 79.0oC. If the
normal BP of ethanol is 78.4oC, calculate the
molar mass of glycerol.
Given: DTb =
mass solute =
mass solvent =
Kb = 1.22oC/m (Table 13.4)
Find:
molar mass (g/mol)
DTb = Kb  m
Colligative Properties - BP Elevation
Step 1: Calculate molality of solution
Step 2: Calculate moles of solute present
Step 3: Calculate molar mass
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.
DTf = Kf  m
where DTf = freezing point depression
Kf = molal freezing point depression
constant
m = molality of solution
The value of Kf depends only on the identity of
the solvent (see Table 13.4).
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:
DTf
DTf = Kf  m
Colligative Properties - Freezing Pt Depression
Molality of solution:
DTf =
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.
Colligative Properties - Freezing Pt Depression

Reminder: You should be able to do the
following as well:

Calculate the freezing point of any solution
given enough information to calculate the
molality of the solution and the value of Kf

Calculate the molar mass of a solution given
the value of Kf and the freezing point
depression (or the freezing points of the
solution and the pure solvent).
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).