1. No category

SECTION 13.2
Colligative Properties
of Solutions
Teacher Notes and Answers
SECTION 2 Colligative Properties of
Solutions
1.dissolved salt
2.higher; lower
3.–1.93°C
4.freezing-point; freezing point; boiling-point;
boiling point
5.The arrow should point from right to left across
the membrane.
6.sucrose solution: 0.5°C; calcium chloride
solution: 1.5°C
7.the attraction between ions of opposite charge
Practice
A.–0.426°C
B.–118.1°C
C.0.175 m
D.102.7°C
E.2.0 m
Fa.0.75°C
Fb.1.5 m
G.–7.4°C
H.2.6°C
I.0.054 m
Review
1a.freezing-point depression
1b.freezing-point depression
2a.–3.9°C/m
2b.acetic acid
3.the more-concentrated solution
4a.–0.744°C
4b.No, at this concentration, clustering of ions will
cause the freezing point depression to be less
than expected.
5.solution B, solution A, solution C; For a given
solution, the freezing-point depression and the
boiling-point elevation depend only on the
solute concentration. Because the freezing point
of solution B is depressed the least, that solution
must have the lowest solute concentration, so
its boiling point would be elevated the least.
Similarly, the freezing point of solution C is
depressed the most, so solution C must have the
highest solute concentration and thus the highest
boiling-point elevation.
I o n s i n A q u e o u s S o l u t i o n s a n d C o l l i g at i v e P r o p e r t i e s
1
SECTION 13.2
Colligative Properties
of Solutions
The presence of solutes in a solution affects the properties
of the solution. Some of these properties do not depend on
the nature of the solutes. Colligative properties are properties
that depend on the concentration of solute particles but not
on their identity. Recall that molality is the amount of solute
per kilogram of solvent. This section covers the colligative
properties of solutions, and how they change with changes
in molality.
Key Terms
colligative properties
nonvolatile substance
freezing-point depression
molal freezing-point constant
boiling-point elevation
molal boiling-point constant
semipermeable membrane
osmosis
osmotic pressure
Most of the solutes discussed in this section are
nonvolatile substances. A nonvolatile substance has little
tendency to become a gas under the existing conditions.
Critical Thinking
1.
Apply In ocean water, which of the following is a
nonvolatile substance: dissolved carbon dioxide,
dissolved oxygen, or dissolved salt?
Lowering vapor pressure depends on nonelectrolyte
solute concentration.
Pure water
Aqueous solution of nonvolatile solute
Used to represent C12H22O11, sucrose
When a nonvolatile substance isUsed
dissolved
in a solvent, the
to represent H2O, water
boiling point of the solvent increases. The diagram at the right
helps to explain why this occurs. A volume of an aqueous
solution contains fewer water molecules than the same
volume of pure water. Therefore, fewer molecules must exist
in the gas phase for the processes of evaporation and
condensation to balance. The equilibrium vapor pressure of
the solution is less than the equilibrium vapor pressure of
pure water.
At a given temperature, the vapor
pressure of water over pure water is
greater than the vapor pressure of water
over an aqueous solution of sucrose.
2
C H A P TER 1 3
Pure water
Used to represent C12H22O11, sucrose
Used to represent H2O, water
Aque
Vapor Pressure vs. Temperature for a Pure Solvent
and a Solution with a Nonvolatile Solute
1.0
en
t
re
Pu
Solid
lv
so
∆Psolvent
at bp of
pure solvent
io
n
Vapor pressure of solvent (atm)
Liquid
l
So
ut
The graph shows that a solution has
a higher boiling point and a lower
freezing point than the pure solvent.
This is because the vapor pressure
of a solution is lower than the vapor
pressure of the solvent alone. Note
that Δ​tf​​is the change in freezing
point temperature, Δ​t​b​is the change
in boiling point temperature, and
Δ​Psolvent
​
​is the change in vapor
pressure of the solvent at the
boiling point.
Gas
tf
solution
∆t f
tb
tf
tb
pure
solution
pure
solvent
solvent ∆t b
Temperature (˚C)
Remember that a liquid boils when its equilibrium vapor
pressure is equal to the atmospheric vapor pressure. If
equilibrium vapor pressure is lowered for a solution versus a
pure substance, then the solution must be raised to a higher
temperature before it can reach the boiling point. This is
shown in the graph above. The lower curve on the graph
above (the blue curve representing the solution) reaches a
vapor pressure of 1 atm at a higher temperature.
The graph also shows that a nonvolatile substance
decreases the freezing point of the solution versus the pure
solvent. The lower vapor pressure allows the solution to
remain in liquid form over a larger range of temperatures.
The presence of solute molecules interferes with the
formation of ordered structures within the solvent. The
solution must reach a lower temperature to get the average
kinetic energy of liquid molecules low enough to form the
more orderly structures of a solid.
The lowering of vapor pressure resulting from the presence
of solute depends on the concentration of solute particles, not
on their identity. Therefore, vapor-pressure lowering is a
colligative property.
T I P The vapor pressure of
water is lowered by
5.5 × ​10​–4​atm by a 1 m solution of
any nonelectrolyte solute.
! Remember
The freezing point of a substance
is the temperature at which the
solid and liquid forms of the
substance are at equilibrium at
atmospheric pressure.
READING CHECK
2. Salt water has a
boiling point and a
freezing point than pure water.
I o n s i n A q u e o u s S o l u t i o n s a n d C o l l i g at i v e P r o p e r t i e s
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The freezing point depression is directly proportional
to the molal concentration of a solution.
Adding solute to a solution lowers its freezing point. One
reason people sprinkle salt on sidewalks is to prevent ice from
forming by lowering the freezing point of water. The salt also
dissolves in ice as it melts, preventing the ice from refreezing.
Scientists have found that the freezing point of a
1 m solution of any nonelectrolyte in water is –1.86°C instead
of 0.00°C. In other words, 1 mol of solute dissolved in water
lowers the freezing point of water by 1.86°C. Through
experiments, scientists have found that 2 mol of solute lowers
the freezing point of water by 3.72°C. By studying the effect
of different concentrations of solute on the freezing point,
scientists have found that the concentration of solute and the
change in freezing point are directly proportional.
The freezing-point depression, Δ​tf​​, is the difference
between the freezing point of the pure solvent and that of a
solution of a nonelectrolyte in the solvent. Because this
quantity and concentration are directly proportional, they can
be related by a constant, as in the equation below.
Δ​t​f​= ​Kf​ ​ m
In the equation, m is molality, and Δ​tf​​is expressed in °C.
The constant, K
​ ​f​, is called the molal freezing-point constant.
The molal freezing-point constant is the freezing point of the
solvent in a 1-molal solution of a nonvolatile, nonelectrolyte
solute. As stated above, the molal freezing-point constant for
water is –1.86°C. The table shows the molal freezing-point
constants for other substances.
Molal Freezing-Point Constants
Solvent
Normal freezing
point (°C)
Molal freezing-point
constant, K
​ ​f​(°C/m)
16.6
-3.90
Acetic acid
Camphor
Ether
178.8
-39.7
-116.3
-1.79
Naphthalene
80.2
-6.94
Phenol
40.9
-7.40
Water
4
C H A P TER 1 3
0.00
-1.86
CONNECT
Even though the concepts of
freezing and melting are familiar,
scientists have only just begun to
understand these processes. Starting
several degrees below the melting
point of ice, a fluid surface layer
forms that is a few molecules thick.
This “quasiliquid layer” could help
explain many features of ice, such
as the unique shapes of snowflakes
or the physics of ice skating.
Critical Thinking
3. Calculate Assuming that
ocean water has 1.04 mol of
dissolved salt, what is ocean
water’s freezing point?
SAMPLE PROBLEM
What is the actual freezing point of water in a solution of
17.1 g of sucrose, C
​ 12
​ ​​H​22​​O​11​, in 200. g of water?
SOLUTION
1 ANALYZE
Determine the information that is given and unknown.
Given: s olute: 17.1 g ​C12
​ ​​H​22​​O​11​
solvent: 200. g H
​ 2​ ​O
Unknown: freezing point of the solution
2 PLAN
Describe how to use the given information to find the unknown.
First, determine the amount of solute in moles. Then, determine
the molality of the solution. Finally, use the equation for
freezing-point depression to determine the amount that the
freezing point decreases below 0°C.
3 SOLVE
Use the plan above to calculate the unknown.
molar mass of C
​ 12
​ ​​H​22​​O​11​
= 12(12.01 g/mol) + 22(1.0 g/mol) + 11(16.00 g/mol)
= 342.34 g/mol
1 mol ​C​12​​H​22​​O​11​
17.1 g ​C12
​ ​​H​22​​O​11​× _________________
​ 
  
   ​= 0.0500 mol C
​ 12
​ ​​H​22​​O​11​
342.34 g ​C12
​ ​​H​22​​O​11​
The mass of the solvent is 200. g = 0.200 kg.
amount of solute (mol)
molality = ​ _____________________
  
  
 ​
mass of solvent (kg)
0.0500 mol C
​ ​12​​H​22​​O​11​
  
  
= ​ ____________________
 ​
0.200 kg H
​ 2​ ​O
= 0.250 m
The freezing-point depression constant for water is −1.86°C/m.
Δ​tf​​= ​Kf​​m = (−1.86°C/m)(0.250 m) = −0.465°C
Therefore, the freezing point of the solution is −0.465°C.
4 CHECK
YOUR
WORK
Check the final answer to see if it is reasonable.
The correct units were used in each step and the final answer
has the correct number of significant digits.
I o n s i n A q u e o u s S o l u t i o n s a n d C o l l i g at i v e P r o p e r t i e s
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PRACTICE
A.A
solution consists of 10.3 g of the nonelectrolyte glucose,
​C6​ ​​H​12​​O​6​, dissolved in 250. g of water. What is the freezingpoint depression of the solution?
What is the molar mass of ​C6​ ​​H​12​​O​6​?
1 mol ​C​6​​H​12​​O​6​
 ​
moles of solute = 10.3 g ​C6​ ​​H​12​​O​6​× ______________________
​ 
   
  
g ​C6​ ​​H​12​​O​6​
=
amount of solute (mol)
molality of solution = _____________________
​ 
  
  
 ​=
mass of solvent (kg)
Δ​t​f​= ​Kf​​m =
B. If
0.500 mol of a nonelectrolyte solute are dissolved in
500.0 g of ether, what is the freezing point of the solution?
What is the molality of the solution?
What is the freezing-point depression of the solution?
What is the normal freezing point of ether?
What is the freezing point of the solution?
6
C H A P TER 1 3
SAMPLE PROBLEM
A water solution containing an unknown quantity of a
nonelectrolyte solute is found to have a freezing point of
–0.23°C. What is the molal concentration of the solution?
SOLUTION
1 ANALYZE
Determine the information that is given and unknown.
Given: freezing point of solution = –0.23°C
Unknown: molality of the solution
2 PLAN
Write an equation that can be used to find the unknown.
The equation for the freezing-point depression of a
solution is Δ​tf​​= ​Kf​​m. Therefore, the molality of the
solution can be written as follows.
Δ​tf​​
m = ___
​   ​ 
​K​f​
The value of Δ​t​f​can be derived from the given
information and the freezing point of pure water.
3 SOLVE
Substitute the given information into the equation.
The freezing-point depression of the solution is –0.23°C,
because the freezing point of the solution is 0.23°C below
0.00°C. The molal freezing-point constant, K
​ f​​, for water is
found in the table on page 425.
Δ​t​f​
-0.23°C 
m = ​ ___ ​ = ​ __________
 ​ 
= 0.12 m
​K​f​ -1.86°C/m
4 CHECK
YOUR
WORK
Check the final answer to see if it is reasonable.
The correct units were used in each step and the final
answer has the correct number of significant digits.
PRACTICE
C. In
a laboratory experiment, the freezing point of an
aqueous solution of glucose is found to be –0.325°C.
What is the molal concentration of this solution?
I o n s i n A q u e o u s S o l u t i o n s a n d C o l l i g at i v e P r o p e r t i e s
7
A solution with a nonelectrolyte solvent will have a
higher boiling point than pure solvent.
Adding solute to a solution raises its boiling point. The vapor
pressure of the solution is lower than it would be for the pure
solvent. More heat must be added to the solution before the
vapor pressure equals the atmospheric pressure. A liquid will
boil when these two quantities are equal.
Scientists have found that the boiling point of a
1 m solution of any nonelectrolyte in water is 100.51°C instead
of 100.00°C. Just as with freezing-point depression, scientists
have found that the concentration of solute and the change in
freezing point are directly proportional.
The boiling-point elevation, Δ​tb​ ​, is the difference between
the boiling point of the pure solvent and that of a solution of a
nonelectrolyte in the solvent. This quantity and the molality of
a solution are related by the equation below.
Δ​t​b​= ​Kb​ ​m
READING CHECK
4. The molal
constant is always a negative value
because the
Once again, m is molality, and Δ​t​b​is expressed in °C. The
constant, ​Kb​ ​, the molal boiling-point constant, is the boiling
point of the solvent in a 1-molal solution of a nonvolatile,
nonelectrolyte solute. The molal freezing-point constant for
water is 0.51°C. The molal boiling-point constants, and for
comparison the molal-freezing point constants, for several
substances are shown in the table.
of a solution is always lower.
Conversely, the molal
constant is
always a positive value because the
of a solution is always higher.
Molal Freezing-Point and Boiling-Point Constants
Solvent
Normal
freezing
point (°C)
Molal freezingpoint constant,
​ f​​ (°C/m)
K
Normal
boiling
point (°C)
Molal boilingpoint constant,
K
​ b​ ​ (°C/m)
16.6
-3.90
117.9
3.07
207.4
5.61
Acetic acid
Camphor
Ether
178.8
-39.7
-116.3
-1.79
34.6
2.02
Naphthalene
80.2
-6.94
217.7
5.80
Phenol
40.9
-7.40
181.8
3.60
-1.86
100.0
0.51
Water
8
0.00
C H A P TER 1 3
SAMPLE PROBLEM
What is the boiling point of a solution made from 20.1 g of a
nonelectrolyte solute and 400.0 g of water? The molar mass of
the solute is 62.0 g/mol.
SOLUTION
1 ANALYZE
Determine the information that is given and unknown.
Given: s olute mass = 17.1 g, molar mass of solute
= 62.0 g/mol
solvent: 400.0 g of water
Unknown: boiling-point elevation
2 PLAN
Describe how to find the unknown value.
First, determine the amount of solute in moles. Then,
determine the molality of the solution. Finally, use the
equation for boiling-point elevation to determine the
amount that the boiling point increases above 100°C.
3 SOLVE
Use the plan above to calculate the unknown value.
    ​= 0.324 mol solute
20.1 g solute × ___________
​ 1 mol solute
62.0 g solute
The mass of the solvent is 400.0 g = 0.4000 kg.
amount of solute (mol) _______________
  
  
  
   ​
molality = _____________________
​ 
 ​= ​ 0.324 mol solute
0.4000 kg H
​ 2​ ​O
mass of solvent (kg)
= 0.810 m
The boiling point elevation constant for water is 0.51°C/m.
Δ​t​b​= ​Kb​ ​m = (0.51°C/m)(0.810 m) = 0.41°C
The boiling point of the solution is the normal boiling point
of water plus the boiling point elevation.
boiling point of solution = b.p. of water + Δ​tb​ ​
4 CHECK
YOUR
WORK
= 100.00°C + 0.41°C
= 100.41°C
Check the final answer to see if it is reasonable.
The correct units were used in each step and the final
answer has the correct number of significant digits.
I o n s i n A q u e o u s S o l u t i o n s a n d C o l l i g at i v e P r o p e r t i e s
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PRACTICE
D.A
solution contains 450.0 g of sucrose, C
​ 12
​ ​​H​22​​O​11​, a
nonelectrolyte, dissolved in 250.0 g of water. What is the
boiling point of the solution?
What is the molar mass of C
​ 12
​ ​​H​22​​O​11​?
1 mol ​C12
​ ​​H​22​​O​11​
moles of solute = 450.0 g ​C12
​ ​​H​22​​O​11​× _______________
​ 
  
  
 ​
​g ​C12
​ ​​H​22​​O​11​
=
amount of solute (mol)
  
  
 ​=
molality of solution = _____________________
​ 
mass of solvent (kg)
Δ​t​b​= ​Kb​ ​m =
boiling point of solution =
+
E. If
the boiling-point elevation of an aqueous solution
containing a nonvolatile electrolyte is 1.02°C, what is the
molality of the solution?
F. The
boiling point of an aqueous solution containing a
nonvolatile nonelectrolyte is 100.75°C?
10
a. What
is the boiling point elevation of the solution?
b. What
is the molality of the solution?
C H A P TER 1 3
=
Osmotic pressure is determined by the concentration
of dissolved solute particles.
Another colligative property is related to the pressure
that fluids of different concentrations exert on each other.
This property is important to the function of the cells in
living things.
A cell is surrounded by a semipermeable membrane. A
semipermeable membrane allows the passage of some
particles while blocking the passage of others. For example, a
cell membrane might allow the passage of water molecules
but block the passage of large solute molecules.
Suppose the cell membrane separates two fluids that have
different concentrations of solute. Particles from both fluids
will strike the membrane at the same rate, but more of the
particles from the dilute fluid will be small enough to pass
through the membrane. The result is a net flow of water
from one side to the other. The movement of solvent
through a semipermeable membrane from the side of
lower concentration to the side of higher concentration is
called osmosis.
Pressure would have to be exerted on the membrane from
the side with the more concentrated solution in order to
prevent osmosis. Osmotic pressure is the external pressure
that is necessary to stop osmosis. Because this pressure is
dependent on the difference in concentrations, not on the
identity of the solutes, it is a colligative property.
(a)
(b)
Pure water
Semipermeable
membrane
Used to represent C12H22O11, sucrose
Sucrose solution
CONNECT
Regulation of osmosis is vital to the
life of a cell. Cells lose water and
shrink when placed in a solution of
higher concentration than their
interior fluid. Cells gain water and
swell when placed in a solution of
lower concentration. The cells of
vertebrates are protected from
swelling and shrinking by the blood
and lymph that surround the cells.
These fluids are kept at a
concentration equal to the
concentration inside cells.
LOOKING CLOSER
5. Draw an arrow on diagram
(b) below to represent the osmotic
pressure that is preventing
osmosis through the
semipermeable membrane.
(a) Pure water and an aqueous
sucrose solution are separated by a
semipermeable membrane. The net
movement of water molecules through
the membrane is from the pure water
side into the sucrose solution. (b) The
height of the solution rises until
the pressures on both sides of the
membrane balance. The difference in
the heights of the fluids exerts
pressure on the membrane toward the
pure water side. The difference in
concentrations exerts pressure on the
membrane toward the solution side.
Used to represent H2O, water
I o n s i n A q u e o u s S o l u t i o n s a n d C o l l i g at i v e P r o p e r t i e s
11
The total molality of all dissolved particles determines
changes in colligative properties.
The examples and problems discussed in this section so far
have involved nonelectrolytes. The same principles also apply
to electrolytes, so long as the dissociation and ionization of the
solutes are taken into account.
For example, consider two solutions, a 1 m solution of
sucrose and a 1 m solution of sodium chloride. Sucrose is a
nonelectrolyte. One mole of sucrose dissolves to form one
mole of solute particles. However, sodium chloride is a strong
electrolyte. Each mole of sodium chloride dissolves to form
two moles of solute particles, one mole of sodium ions and
one mole of chloride ions. Therefore, a sodium chloride
solution will have twice as many solute particles as a sugar
solution at the same molality. Since colligative properties are
dependent on particle concentration only, not the type of
particles, you would expect sodium chloride to have twice the
effect as sugar on the properties of solutions.
Calculated Values for Electrolyte Solutions
The changes in colligative properties caused by electrolytes is
proportional to the total molality of all dissolved particles, not
to formula units. The sample problem on the next page shows
how to apply a conversion factor that accounts for the number
of moles of ions that form from a mole of electrolyte.
Reverse osmosis can purify water.
Exerting more pressure on the
semipermeable membrane than is
coming from the osmotic pressure
can reverse the direction of osmosis.
So, if the membrane separates pure
water and salt water, the water
molecules will move from the salt
water to the pure water. Over time,
more and more pure water can
accumulate. The largest desalination plant in Florida uses reverse
osmosis to produce 12 million
gallons of fresh water a day.
Critical Thinking
6. Apply Assume that all three
solutions below have the same
molality. If the sodium chloride
solution has a boiling-point
elevation of 1.0°C, what are the
boiling-point elevations of the
other two solutions?
Na+
H2O
Cl–
C12H22O11
Sucrose solution
C12H22O11
H2O
C12H22O11(aq)
Ca2+
Sodium chloride solution
NaCl
H2O
Na+(aq) + Cl–(aq)
From left to right, these three solutions include 1 mol, 2 mol, and 3 mol of solute
particles per formula unit. A 1:2:3 ratio reflects the effect of equal concentrations
of these solutions on colligative properties.
12
CONNECT
C H A P TER 1 3
Calcium chloride solution
CaCl2
H2O
Ca2+(aq) + 2Cl–(aq)
SAMPLE PROBLEM
What is the expected change in the freezing point of water in a solution
of 62.5 g of barium nitrate, Ba​(N​O3​ ​)2​ ​, in 1.00 kg of water?
SOLUTION
1 ANALYZE
Determine the information that is given and unknown.
​ 2​ ​O
Given: solute: 62.5 g Ba​(N​O3​ ​)2​ ​, solvent: 1.00 kg H
Unknown: expected freezing-point depression
2 PLAN
Describe how to find the unknown value.
First, determine the amount of solute in moles and determine
the molality of the solution. Then convert the concentration
from molality of formula units to molality of ions. This
adjusted molality can then be used to calculate the freezingpoint depression of the solution.
3 SOLVE
Use the plan above to calculate the unknown.
molar mass of Ba​(N​O3​ ​)2​ ​
= 137.33 g/mol + 2(14.01 g/mol) + 6(16.00 g/mol)
= 261.35 g/mol
1 mol Ba​(N​O3​ ​)2​ ​
62.5 g Ba​(N​O​3​)2​ ​× _________________
​ 
  
   ​
261.35 g Ba​(N​O3​ ​)2​ ​
= 0.239 mol Ba​(N​O3​ ​)2​ ​
The mass of the solvent is 1.00 kg, so the concentration of the
formula unit Ba​(N​O​3​)2​ ​is 0.239 m. The dissociation equation
for barium nitrate is as follows.
​H​ ​O
+
  ​  
​  ​ 2⟶B​
   a​ ​ + 2NO​Ba​(N​O​3​)2​ ______
3
Each formula unit of Ba​(N​O​3​)2​ ​yields 3 ions in solution.
Therefore, the concentration of ions is 3 × 0.239 m = 0.717 m.
​Kf​​m
Δ​t​f​(expected) =
= (–1.86°C/m)(0.717 m)
= –1.33°C
4 CHECK
YOUR
WORK
Check the final answer to see if it is reasonable.
The correct units were used in each step and the final answer
has the correct number of significant digits.
I o n s i n A q u e o u s S o l u t i o n s a n d C o l l i g at i v e P r o p e r t i e s
13
PRACTICE
G.What
is the expected freezing-point depression for a
solution that contains 2.0 mol of magnesium sulfate
dissolved in 1.0 kg of water?
What is the molality of MgS​O​4​formula units in the solution?
What is the dissociation equation for magnesium sulfate?
What is the molality of ions in the solution?
Δ​tf​​= ​Kf​​m =
H.What
is the expected boiling-point elevation of water for a
solution that contains 150 g of sodium chloride dissolved
in 1.0 kg of water?
What is the molar mass of sodium chloride?
How many moles of sodium chloride are in the solution?
What is the molality of the NaCl formula units in the solution?
What is the dissociation equation for sodium chloride?
What is the molality of ions in the solution?
Δ​tb​ ​= ​Kb​ ​m =
I. The
freezing point of an aqueous sodium chloride solution
is –0.20°C. What is the molality of the solution?
( ) __________________
( )
m = ____
​ 
 
   ​=
m ions in solution
 ​= ​ 
( ) ( )
The molality of the solution is
14
C H A P TER 1 3
m NaCl.
Actual Values for Electrolyte Solutions
The method shown on the previous two pages calculates the
expected value of a colligative property for an electrolytic
solution. The actual values for strong electrolytes are almost
the same, but not exactly the same, as these expected values.
The table below shows some actual values.
Molal Freezing-Point Depressions for Aqueous Solutions of Ionic Solutes
Δ​t f​​, observed
_______________________
   
  
 ​
​ 
Concentration
(m)
Δ​tf​​, observed
(°C)
Δ​tf​​, nonelectrolyte
solution (°C)
KCl
0.1
0.01
0.001
-0.345
-0.0361
-0.00366
-0.186
-0.0186
-0.00186
1.85
1.94
1.97
MgS​O4​ ​
0.1
0.01
0.001
-0.225
-0.0285
-0.00338
-0.186
-0.0186
-0.00186
1.21
1.53
1.82
BaC​l2​ ​
0.1
0.01
0.001
-0.470
-0.0503
-0.00530
-0.186
-0.0186
-0.00186
2.53
2.70
2.84
Solute
Δ​tf​​, nonelectrolyte solution
The table shows that a dilute concentration of an
electrolyte that forms two ions in solution, such as KCl or
MgS​O4​ ​, has a freezing-point depression that is almost twice
the nonelectrolytic value. An electrolyte that forms three ions,
such as Ba​Cl​2​, has a value almost three times as large.
The table also shows that as concentration decreases, the
actual value is closer to the expected value. This suggests
a reason for the actual values not being the same as the
expected values. Ions in solution feel some attraction to
ions of opposite charge as they move through a fluid. This
attraction partially hinders the movement of the charges and
lowers the solute’s effective concentration. In more dilute
solutions a solute particle encounters oppositely charged ions
less frequently, so the attraction has less effect.
Another trend the table shows is that ions of greater charge
cause values that vary farther from expected values. Both
KCl and MgS​O4​ ​form two moles of ions in solution, but the
+
 
­charges on the M​g2+
​ ​and SO​24   ​ ions are twice that of ​K​ ​ and ​
Cl​–​ions. The variance between expected values and actual
values is larger because the attractive forces between ions in
MgS​O4​ ​solutions are greater.
READING CHECK
7. What attractive force accounts
for electrolytic solutions having
lower values of colligative
properties than expected?
I o n s i n A q u e o u s S o l u t i o n s a n d C o l l i g at i v e P r o p e r t i e s
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SECTION 13.2 REVIEW
VOCABULARY
1. What colligative property is displayed by each of the following situations?
a.Antifreeze is added to a car’s cooling system to prevent freezing when
the temperature is below 0°C.
b. Ice melts on sidewalks after salt has been spread on them.
REVIEW
2. Two moles of a nonelectrolytic solute are dissolved in 1 kg of an unknown
solvent. The solution freezes at 7.8°C below its normal freezing point.
a. What is the molal freezing-point constant of the unknown solvent?
b. Suggest a possible identity of the solvent.
3. If two solutions of equal amounts in a U-tube are separated by a
semipermeable membrane, will the level of the more-concentrated
solution or the less-concentrated solution rise?
4. a.Calculate the expected freezing-point depression of a
0.200 m KN​O​3​solution.
b.Will the value you calculated match the actual freezing-point depression
for this solution? Why or why not?
Critical Thinking
5. INFERRING RELATIONSHIPS The freezing-point depressions of aqueous
solutions A, B, and C are –2.3°C, –1.2°C, and –4.1°C, respectively. Predict the
order of the boiling-point elevations of these solutions, from lowest to
highest. Explain your ranking.
16
C H A P TER 1 3
Math Tutor Boiling and Freezing Points
of Solutions
The presence of a nonvolatile solute causes the freezing point of a solution to be lower
and the boiling point to be higher than those of the pure solvent. The freezing-point
depression, Δ​tf​​, is the amount that the freezing point is lowered. It is calculated by using
​ f​​m. The boiling-point elevation, Δ​tb​ ​, is the amount that the boiling
the formula Δ​tf​​= K
​ b​ ​m. To determine Δ​tf​​or Δ​tb​ ​,
point is elevated. It is found by using the equation Δ​tb​ ​= K
you need to know the molal concentration of the solution, m (moles of solute per
kilogram of solvent). You also need to know the molal freezing-point constant, K
​ f​​, or the
molal boiling-point constant, ​Kb​ ​.
Problem-Solving TIPS
• Make sure that you find the molal concentration, not the molar concentration.
• For electrolytes, calculate the total number of moles of ions in solution.
SAMPLE
What is the theoretical boiling point of a solution of 247 g of
potassium chloride, KCl, dissolved in 2.90 kg of water? Potassium
chloride is a strong electrolyte.
First, determine the molality of the KCl that dissolved. The amount
of KCl must be converted from grams to moles, using the molar mass
of KCl (74.55 g/mol).
1 mol KCl 
 ​ 
= 3.31 mol KCl
247 g KCl = 247 g KCl × ​ ___________
74.55 g KCl
solute particles (mol) ____________
KCl ​= 1.14 m
  
   ​= ​ 3.31 mol   
molality = ​ ___________________
2.90 kg ​H2​ ​O
mass of solvent (kg)
Because KCl is an electrolyte, the total moles of ions in solution must
be determined. The dissociation equation is as follows.
+
–
2
  ​K​ ​(aq) + C​l​ ​(aq)
⟶
KCl(s) ​ ______
​H​ ​O
So, 2 mol of ions form for every 1 mol of solute. Therefore,
a 1.14 m solution will form a solution with an ion concentration of
2 × 1.14 m = 2.28 m.
Finally, calculate the boiling-point elevation for the solution. The
molal boiling point constant for water is 0.51°C/m.
Δ​tb​ ​= ​Kb​ ​m = (0.51°C/m)(2.28 m) = 1.16°C
The boiling point is the sum of the solvent’s boiling point and Δ​tb​ ​.
boiling point of solution = 100.00°C + 1.16°C = 101.16°C
Practice Problems: Chapter Review practice problems 16–18
I o n s i n A q u e o u s S o l u t i o n s a n d C o l l i g at i v e P r o p e r t i e s
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