Chapter 13
Properties of Solutions
Sections 5 - 6
Adapted from:
Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.;
and Bruce E. Bursten
John D. Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall, Inc.
1
Colligative Properties
Changes in colligative properties
depend only on the number of solute
particles present, not on the identity of
the solute particles.
2
Colligative Properties
Among colligative properties are
1. Vapor pressure lowering
2. Boiling point elevation
3. Melting point depression
4. Osmotic pressure
3
1
Vapor Pressure
Because of solutesolvent IMF, higher
concentrations of
nonvolatile solutes
make it harder for
solvent to escape to
the vapor phase.
4
Question
1. How should this
affect the boiling
point?
5
Boiling Point Elevation and
Freezing Point Depression
Adding a
nonvolatile solute
causes solutions
to have higher
boiling points and
lower freezing
points than the
pure solvent.
6
2
Raoult’s Law
PA = XAPA
where
XA is the mole fraction of the solvent 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.
7
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.
8
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.
9
3
Colligative Change
Tb = Kb  m
Tf = Kf  m
10
Question
2. What should be the change in boiling
and freezing points of water if 3.00 m of
glucose is added to it?
1.53˚C (101.53˚C)
5.6˚C (-5.6˚C)
11
Question
3. What should be the change in boiling
and freezing points of benzene if 2.00
m of naphthalene is added to it?
5.06˚C (85.2˚C)
10.2˚C (-4.7˚C)
12
4
Boiling Point Elevation and
Freezing Point Depression
Note that 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
13
Colligative Properties of
Electrolytes
Solutions of electrolytes (which dissociate in
solution) should show greater changes than
those of non-electrolytes.
14
Question
4. How much greater of a change should
1.0 m NaCl make to 100g of water
than 1.0 m of glucose?
15
5
Question
5. It doesn’t though… why not?
16
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.
17
The van’t Hoff Factor
6. Should the van’t Hoff Factor increase or
decrease with solute concentration?
18
6
The van’t Hoff Factor
van’t Hoff factor, i
Tf = Kf  m  i
i=
Tf (measured)
Tf (calculated for non-electrolyte)
19
Osmosis
The process of particles moving from high
to low concentration across a semipermeable membranes.
20
21
7
Osmotic Pressure
The pressure required to stop osmosis,
known as osmotic pressure, , is
=(
n
)RT = MRT
V
where M is the molarity of the solution
R = 0.0821 L-atm/mol-K
If the osmotic pressure is the same on both
sides of a membrane the solutions are
isotonic.
22
Osmosis in Blood Cells
Hypertonic
Water will flow out
of the cell, and
crenation results.
23
Osmosis in Cells
Hypotonic
Water will flow into
the cell, and
hemolysis results.
24
8
Molar Mass from
Colligative Properties
Ex: a solution of an unknown non-volatile
electrolyte was prepared by dissolving
0.250 g of the substance in 40.0 g of CCl4.
The boiling point of the solution was
0.357˚C higher than that of the pure
solvent.
What is the molar mass of the solute?
25
Molar Mass from
Colligative Properties
STEP 1 – determine the number
molality.
Tb = Kb  m → m = Tb
Kb
Kb = 5.02˚C/m
26
Molar Mass from
Colligative Properties
STEP 1 – determine the number
molality.
m = Tb
Kb
0.357˚C = 0.0711 m
5.02˚C/m
27
9
Molar Mass from
Colligative Properties
STEP 2 – determine the number
of moles of solute.
mol
m=
→
kg solvent
solute
0.0400 kg CCl4 * 0.0711 mol
kg CCl
4
0.00284 mol
28
Molar Mass from
Colligative Properties
STEP 3 – determine the molar
mass.
Molar mass =
0.250 g
0.00284 mol
g
mol
= 88.0 g/mol
29
Colloids:
Suspensions of particles larger than individual
ions or molecules, but too small to be settled
out by gravity (10 - 2000 Ǻ)
30
10
Tyndall Effect Colloidal
suspensions can
scatter rays of
light.
31
Colloids in Biological
Systems
Some molecules
have a polar,
hydrophilic
(water-loving) end
and a nonpolar,
hydrophobic
(water-hating) end.
32
Colloids in Biological
Systems
Sodium stearate
Shows the ability to emulsify
33
11