Chapter 16
Solutions
16.1 Properties of Solutions
16.2 Concentrations of Solutions
16.3 Colligative Properties
of Solutions
16.4 Calculations Involving
Colligative Properties
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CHEMISTRY
& YOU
How can you grow a tree made out of
crystals?
Remember, the
crystallization of a
solute from solution is
a physical change
that is different from
freezing.
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Solution Formation
Solution Formation
What factors affect how fast a
substance dissolves?
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Solution Formation
Granulated sugar dissolves faster than
sugar cubes, and both granulated sugar
and sugar cubes dissolve faster in hot tea
or when you stir the tea.
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Solution Formation
The compositions of the solvent and the
solute determine whether or not a
substance will dissolve.
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Solution Formation
The compositions of the solvent and the
solute determine whether or not a
substance will dissolve.
Factors that affect how fast a substance
dissolves include:
• Agitation
• Temperature
• Particle size of the solute
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Solution Formation
Agitation
If the contents of the glass are stirred, the
crystals dissolve more quickly.
• The dissolving process
occurs at the surface of the
sugar crystals.
• Stirring speeds up the
process because fresh
solvent (the water) is
continually brought in
contact with the surface of
the solute (sugar).
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Solution Formation
Agitation
Agitation (stirring or shaking) affects only
the rate at which a solid solute dissolves.
• It does not influence the
amount of solute that will
dissolve.
• An insoluble substance
remains undissolved
regardless of how
vigorously or for how long
the solvent/solute system
is agitated.
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Solution Formation
Temperature
Temperature also influences the rate at
which a solute dissolves.
• Sugar dissolves much more
rapidly in hot tea than in
iced tea.
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Solution Formation
Temperature
At higher temperatures, the kinetic energy
of water molecules is greater than at
lower temperatures, so the
molecules move faster.
• The more rapid motion of the
solvent molecules leads to an
increase in the frequency of the
force of the collisions between
water molecules and the
surfaces of the sugar crystals.
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Solution Formation
Particle Size of the Solute
The rate at which a solute dissolves also
depends upon the size of the solute
particles.
• The smaller particles in
granulated sugar expose a
much greater surface area
to the colliding water
molecules.
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Solution Formation
Particle Size of the Solute
The dissolving process is a surface
phenomenon.
• The more surface area of
the solute that is exposed,
the faster the rate of
dissolving.
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Which of the following will not speed
up the rate at which a solid solute
dissolves?
A.
B.
C.
D.
Increasing the temperature
Stirring the mixture
Crushing the solute
Decreasing the temperature
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Which of the following will not speed
up the rate at which a solid solute
dissolves?
A.
B.
C.
D.
Increasing the temperature
Stirring the mixture
Crushing the solute
Decreasing the temperature
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Solubility
Solubility
How can you describe the
equilibrium in a saturated solution?
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Solubility
What is
happening in
this figure?
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Solubility
What is
happening in
this figure?
• Particles move from
the solid into the
solution.
• Some dissolved particles move from the solution back
to the solid.
• Because these two processes occur at the same rate,
no net change occurs in the overall system.
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Solubility
Such a solution
is said to be
saturated.
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Solubility
Such a solution
is said to be
saturated.
• A saturated solution contains the maximum
amount of solute for a given quantity of
solvent at a constant temperature and
pressure.
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Solubility
In a saturated solution, a state of
dynamic equilibrium exists between
the solution and any undissolved
solute, provided that the temperature
remains constant.
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Solubility
The solubility of a substance is the
amount of solute that dissolves in a given
quantity of a solvent at a specified
temperature and pressure to produce a
saturated solution.
• Solubility is often expressed in grams of
solute per 100 g of solvent (g/100 g H2O).
• Sometimes the solubility of a gas is
expressed in grams per liter of solution (g/L).
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Solubility
A solution that contains less solute than a
saturated solution at a given temperature
and pressure is an unsaturated
solution.
• If additional solute is added to an
unsaturated solution, the solute will dissolve
until the solution is saturated.
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Solubility
Some liquids—for example, water and
ethanol—are infinitely soluble in each
other.
• Two liquids are miscible if they dissolve
in each other in all proportions.
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Solubility
Liquids that are insoluble in each other are
immiscible.
• Oil and water
are examples
of immiscible
liquids.
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The solubility of a substance is often
expressed as which of the following?
A.
B.
C.
D.
grams of solute per 100 liters of solvent
grams of solute per 1 cm3 of solvent
grams of solute per 100 grams of solvent
grams of solute per 100 grams of solution
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The solubility of a substance is often
expressed as which of the following?
A.
B.
C.
D.
grams of solute per 100 liters of solvent
grams of solute per 1 cm3 of solvent
grams of solute per 100 grams of solvent
grams of solute per 100 grams of solution
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Factors Affecting Solubility
Factors Affecting Solubility
What factors affect the solubility of
a substance?
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Factors Affecting Solubility
Temperature affects the solubility of
solid, liquid, and gaseous solutes in a
solvent; both temperature and
pressure affect the solubility of
gaseous solutes.
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Interpret Graphs
Temperature
Solubility (g/100g H2O)
The solubility of most solid substances
increases as the temperature of the solvent
increases.
Temperature (°C)
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Interpret Graphs
Temperature
• For a few
substances,
solubility
decreases with
temperature.
Solubility (g/100g H2O)
The solubility of most solid substances
increases as the temperature of the solvent
increases.
Temperature (°C)
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Interpret Data
Solubilities of Substances in Water at Various Temperatures
Solubility (g/100 g H2O)
Substance
Formula
0°C
Barium hydroxide
Ba(OH)2
1.67
Barium sulfate
BaSO4
0.00019
0.00025
Calcium hydroxide
Ca(OH)2
0.189
0.173
Potassium chlorate
KClO3
4.0
7.4
19.3
56.0
Potassium chloride
KCl
27.6
34.0
42.6
57.6
Sodium chloride
NaCl
35.7
36.0
37.0
39.2
Sodium nitrate
NaNO3
74
88.0
114.0
Aluminum chloride
AlCl3
30.84
31.03
Silver nitrate
AgNO3
122
222.0
455.0
733
Sucrose (table sugar)
C12H22O11 179
230.9
260.4
487
Hydrogen
H2
0.00019
0.00016
0.00013
0.0
Oxygen
O2
0.0070
0.0043
0.0026
0.0
Carbon dioxide
CO2
0.335
0.169
0.076
0.0
20°C
31.89
50°C
100°C
—
—
—
0.00034
—
31.60
0.07
182
33.32
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Factors Affecting Solubility
Temperature
A supersaturated solution contains
more solute than it can theoretically hold
at a given temperature.
• The crystallization of a supersaturated
solution can be initiated if a very small
crystal, called a seed crystal, of the solute
is added.
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Factors Affecting Solubility
The rate at which excess solute deposits upon
the surface of a seed crystal can be very rapid.
The solution
is clear before
a seed crystal
is added.
Crystals begin to
form immediately
after the addition
of a seed crystal.
Excess solute
crystallizes rapidly.
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CHEMISTRY
& YOU
How do you think crystal-growing kits
work? Use what you know about
solubility and supersaturated solutions
to explain your answer.
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CHEMISTRY
& YOU
How do you think crystal-growing kits
work? Use what you know about
solubility and supersaturated solutions
to explain your answer.
Crystal-growing kits usually begin
with a supersaturated solution.
When a seed crystal is added to
the solution, crystals rapidly begin
to grow because the
supersaturated solution contains
more solute than is theoretically
possible.
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Factors Affecting Solubility
Temperature
The effect of temperature on the
solubility of gases in liquid solvents is
opposite that of solids.
• The solubilities of most gases are greater
in cold water than in hot.
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Factors Affecting Solubility
Pressure
Changes in pressure have little effect on
the solubility of solids and liquids, but
pressure strongly influences the solubility
of gases.
• Gas solubility increases as the partial
pressure of the gas above the solution
increases.
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Factors Affecting Solubility
Pressure
Carbonated beverages are a good example.
• These drinks
contain large
amounts of carbon
dioxide (CO2)
dissolved in water.
• Dissolved CO2
makes the liquid fizz
and your mouth
tingle.
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Factors Affecting Solubility
Pressure
Carbonated beverages are a good example.
• The drinks are
bottled under a high
pressure of CO2
gas, which forces
larger amounts of
the gas into solution.
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Factors Affecting Solubility
Pressure
Carbonated beverages are a good example.
• When the container
is opened, the
partial pressure of
CO2 above the liquid
decreases.
• Immediately,
bubbles of CO2 form
in the liquid and
escape from the
open bottle.
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Factors Affecting Solubility
Pressure
How is the partial pressure of carbon
dioxide gas related to the solubility of
CO2 in a carbonated beverage?
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Factors Affecting Solubility
Pressure
How is the partial pressure of carbon
dioxide gas related to the solubility of
CO2 in a carbonated beverage?
• The relationship is described by Henry’s
law, which states that at a given
temperature, the solubility (S) of a gas in a
liquid is directly proportional to the pressure
(P) of the gas above the liquid.
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Factors Affecting Solubility
Pressure
• As the pressure of the gas above the
liquid increases, the solubility of the
gas increases.
• As the pressure of the gas decreases,
the solubility of the gas decreases.
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Factors Affecting Solubility
Pressure
You can write the relationship in the form
of an equation.
S1 S2
=
P1 P2
• S1 is the solubility of a gas at one pressure, P1.
• S2 is the solubility at another pressure, P2.
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Sample Problem 16.1
Using Henry’s Law
If the solubility of a gas in
water is 0.77 g/L at 3.5 atm of
pressure, what is its solubility
(in g/L) at 1.0 atm of
pressure? (The temperature is
held constant at 25°C.)
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Sample Problem 16.1
1 Analyze List the knowns and the unknown.
Use Henry’s law to solve for the
unknown solubility.
KNOWNS
UNKNOWN
P1 = 3.5 atm
S2 = ? g/L
S1 = 0.77 g/L
P2 = 1.0 atm
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Sample Problem 16.1
2 Calculate Solve for the unknowns.
• State the equation for Henry’s law.
S1 S2
=
P1 P2
Isolate S2 by multiplying
both sides by P2:
S
S
P2  1 = 2  P2
P1
P2
• Solve Henry’s law for S2. Substitute the
known values and calculate.
S1  P2
0.77 g/L  1.0 atm
S2 =
=
= 0.22 g/L
P1
3.5 atm
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Sample Problem 16.1
3 Evaluate Does the result make sense?
• The new pressure is approximately one-third
of the original pressure.
• So, the new solubility should be approximately
one-third of the original.
• The answer is correctly expressed to two
significant figures.
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Explain why an opened container of a
carbonated beverage is more likely to
go flat sitting on the counter than in
the refrigerator.
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Explain why an opened container of a
carbonated beverage is more likely to
go flat sitting on the counter than in
the refrigerator.
The solubility of a gas in a liquid
increases with decreasing temperature.
More carbon dioxide will remain in
solution at the colder temperature found
in the refrigerator.
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Key Concepts
Factors that determine how fast a
substance dissolves are stirring,
temperature, and surface area.
In a saturated solution, a state of dynamic
equilibrium exists between the solution
and any undissolved solute, provided that
the temperature remains constant.
Temperature affects the solubility of solid,
liquid, and gaseous solutes in a solvent;
both temperature and pressure affect the
solubility of gaseous solutes.
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Key Equation
S1 S2
Henry’s law:
=
P1 P2
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Glossary Terms
• saturated solution: a solution containing the
maximum amount of solute for a given amount
of solvent at a constant temperature and
pressure; an equilibrium exists between
undissolved solute and ions in solution
• solubility: the amount of a substance that
dissolves in a given quantity of solvent at
specified conditions of temperature and
pressure to produce a saturated solution
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Glossary Terms
• unsaturated solution: a solution that contains
less solute than a saturated solution at a given
temperature and pressure
• miscible: describes liquids that dissolve in each
other in all proportions
• immiscible: describes liquids that are insoluble
in each other; oil and water are immiscible
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Glossary Terms
• supersaturated solution: a solution that
contains more solute than it can theoretically
hold at a given temperature; excess solute
precipitates if a seed crystal is added
• Henry’s law: at a given temperature, the
solubility of a gas in a liquid is directly
proportional to the pressure of the gas above
the liquid
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Chapter 16
Solutions
16.1 Properties of Solutions
16.2 Concentrations of Solutions
16.3 Colligative Properties
of Solutions
16.4 Calculations Involving
Colligative Properties
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CHEMISTRY
& YOU
How can you describe the concentration
of a solution?
The federal government
and state governments
set standards limiting
the amount of
contaminants allowed in
drinking water.
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Molarity
Molarity
How do you calculate the molarity of
a solution?
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Molarity
The concentration of a solution is a
measure of the amount of solute that is
dissolved in a given quantity of solvent.
• A solution that contains a relatively small
amount of solute is a dilute solution.
• A concentrated solution contains a large
amount of solute.
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Molarity
In chemistry, the most important unit of
concentration is molarity.
• Molarity (M) is the number of moles of
solute dissolved in one liter of solution.
• Molarity is also known as molar
concentration.
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Molarity
The figure below illustrates the procedure
for making a 0.5M, or 0.5-molar, solution.
Add 0.5 mol of solute
to a 1-L volumetric
flask half filled with
distilled water.
Swirl the flask
carefully to
dissolve the
solute.
Fill the flask with
water exactly to
the 1-L mark.
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Molarity
To calculate the molarity of a solution,
divide the number of moles of solute
by the volume of the solution in liters.
moles of solute
Molarity (M) =
liters of solution
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Sample Problem 16.2
Calculating Molarity
Intravenous (IV) saline solutions are often
administered to patients in the hospital.
One saline solution contains 0.90 g NaCl in
exactly 100 mL of solution. What is the
molarity of the solution?
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Sample Problem 16.2
1 Analyze List the knowns and the unknown.
Convert the concentration from g/100 mL
to mol/L. The sequence is:
g/100 mL → mol/100 mL → mol/L
KNOWNS
solution concentration = 0.90 g NaCl/100 mL
molar mass NaCl = 58.5 g/mol
UNKNOWN
solution concentration = ?M
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Sample Problem 16.2
2 Calculate Solve for the unknown.
Use the molar mass to convert
g NaCl/100 mL to mol NaCl/100 mL.
Then convert the volume units so
that your answer is expressed in
mol/L.
The relationship
1 L = 1000 mL
gives you the
conversion factor
1000 mL/1 L.
Solution
0.90 g NaCl
1 mol NaCl
1000 mL
 58.5 g NaCl 
concentration = 100 mL
1L
= 0.15 mol/L
= 0.15M
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Sample Problem 16.2
3 Evaluate Does the result make sense?
• The answer should be less than 1M because
a concentration of 0.90 g/100 mL is the same
as 9.0 g/1000 mL (9.0 g/1 L), and 9.0 g is less
than 1 mol NaCl.
• The answer is correctly expressed to two
significant figures.
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Sample Problem 16.3
Calculating the Moles of Solute
in a Solution
Household laundry bleach is a dilute
aqueous solution of sodium
hypochlorite (NaClO). How many
moles of solute are present in 1.5 L
of 0.70M NaClO?
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Sample Problem 16.3
1 Analyze List the knowns and the unknown.
The conversion is: volume of solution → moles
of solute. Molarity has the units mol/L, so you
can use it as a conversion factor between
moles of solute and volume of solution.
KNOWNS
volume of solution = 1.5 L
solution concentration = 0.70M NaClO
UNKNOWN
moles solute = ? mol
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Sample Problem 16.3
2 Calculate Solve for the unknown.
Multiply the given
volume by the
molarity expressed
in mol/L.
Make sure that your volume units
cancel when you do these
problems. If they don’t, then you’re
probably missing a conversion
factor in your calculations.
0.70 mol NaCl
1.5 L 
= 1.1 mol NaClO
1L
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Sample Problem 16.3
3 Evaluate Does the result make sense?
• The answer should be greater than 1 mol but
less than 1.5 mol, because the solution
concentration is less than 0.75 mol/L and the
volume is less than 2 L.
• The answer is correctly expressed to two
significant figures.
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How much water is required to make
a 1.00M aqueous solution of NaCl, if
58.4 g of NaCl are dissolved?
A. 1.00 liter of water
B. enough water to make 1.00 liter of
solution
C. 1.00 kg of water
D. 100 mL of water
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How much water is required to make
a 1.00M aqueous solution of NaCl, if
58.4 g of NaCl are dissolved?
A. 1.00 liter of water
B. enough water to make 1.00 liter of
solution
C. 1.00 kg of water
D. 100 mL of water
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Making Dilutions
Making Dilutions
What effect does dilution have on
the amount of solute?
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Making Dilutions
Both of these solutions contain the same
amount of solute.
• You can tell by the color of solution (a) that it is
more concentrated than solution (b).
• Solution (a) has
the greater
molarity.
• The more dilute
solution (b) was
made from
solution (a) by
adding more solvent.
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Making Dilutions
Diluting a solution reduces the number
of moles of solute per unit volume, but
the total number of moles of solute in
solution does not change.
Moles of solute
Moles of solute
=
before dilution
after dilution
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Making Dilutions
Moles of solute
Moles of solute
=
before dilution
after dilution
The definition of molarity can be rearranged
to solve for moles of solute.
moles of solute
Molarity (M) = liters of solution (V)
Moles of solute = molarity (M)  liters of solution (V)
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Making Dilutions
The total number of moles of solute remains
unchanged upon dilution.
Moles of solute = M1  V1 = M2  V2
• M1 and V1 are the molarity and the volume of the initial
solution.
• M2 and V2 are the molarity and volume of the diluted
solution.
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Making Dilutions
The student is preparing 100 mL of 0.40M MgSO4
from a stock solution of 2.0M MgSO4.
She measures 20 mL She transfers the
of the stock solution
20 mL to a 100-mL
with a 20-mL pipet.
volumetric flask.
She carefully adds
water to the mark
to make 100 mL of
solution.
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Sample Problem 16.4
Preparing a Dilute Solution
How many milliliters of
aqueous 2.00M MgSO4
solution must be diluted with
water to prepare 100.0 mL of
aqueous 0.400M MgSO4?
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Sample Problem 16.4
1 Analyze List the knowns and the unknown.
Use the equation M1  V1 = M2  V2 to solve for
the unknown initial volume of solution (V1) that
is diluted with water.
KNOWNS
M1 = 2.00M MgSO4
M2 = 0.400M MgSO4
V2 = 100.0 mL of 0.400M MgSO4
UNKNOWN
V1 = ? mL of 2.00M MgSO4
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Sample Problem 16.4
2 Calculate Solve for the unknown.
Solve for V1 and substitute the known
values into the equation.
M2  V2
0.400M  100.0 mL
V1 =
=
= 20.0 mL
M1
2.00M
Thus, 20.0 mL of the initial solution must be
diluted by adding enough water to increase the
volume to 100.0 mL.
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Sample Problem 16.4
3 Evaluate Does the result make sense?
• The initial concentration is five times larger
than the dilute concentration.
• Because the number of moles of solute does
not change, the initial volume of solution
should be one-fifth the final volume of the
diluted solution.
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100.0 mL of a 0.300M CuSO4·5H2O
solution is diluted to 500.0 mL.
What is the concentration of the
diluted solution?
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100.0 mL of a 0.300M CuSO4·5H2O
solution is diluted to 500.0 mL.
What is the concentration of the
diluted solution?
M1  V1 = M2  V2
M1  V1 0.300M  100.0 mL
M2 =
=
V2
500.0 mL
M2 = 0.0600M
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Percent Solutions
Percent Solutions
How do percent by volume and
percent by mass differ?
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Percent Solutions
Percent by volume of a solution is
the ratio of the volume of solute to
the volume of solution.
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Percent Solutions
Percent by volume of a solution is
the ratio of the volume of solute to
the volume of solution.
• Isopropyl alcohol (2-propanol) is sold
as a 91-percent solution by volume.
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Percent Solutions
Percent by volume of a solution is
the ratio of the volume of solute to
the volume of solution.
• Isopropyl alcohol (2-propanol) is sold
as a 91-percent solution by volume.
• You could prepare such a solution by
diluting 91 mL of pure isopropyl
alcohol with enough water to make
100 mL of solution.
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Percent Solutions
Percent by volume of a solution is
the ratio of the volume of solute to
the volume of solution.
• Isopropyl alcohol (2-propanol) is sold
as a 91-percent solution by volume.
• The concentration is written as 91
percent by volume, 91 percent
(volume/volume), or 91% (v/v).
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Percent Solutions
Percent by volume of a solution is
the ratio of the volume of solute to
the volume of solution.
volume of solute
Percent by volume (%(v/v)) =
 100%
volume of solution
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Sample Problem 16.5
Calculating Percent by Volume
What is the percent by volume of ethanol
(C2H6O, or ethyl alcohol) in the final
solution when 85 mL of ethanol is diluted
to a volume of 250 mL with water?
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Sample Problem 16.5
1 Analyze List the knowns and the unknown.
Use the known values for the volume of solute
and volume of solution to calculate percent by
volume.
KNOWNS
volume of solute = 85 mL ethanol
volume of solution = 250 mL
UNKNOWN
Percent by volume = ?% ethanol (v/v)
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Sample Problem 16.5
2 Calculate Solve for the unknown.
State the equation for percent by
volume.
Percent by volume (%(v/v)) =
volume of solute
volume of solution
 100%
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Sample Problem 16.5
2 Calculate Solve for the unknown.
Substitute the known values into the
equation and solve.
85 mL ethanol
Percent by volume (%(v/v)) =
 100%
250 mL
= 34% ethanol (v/v)
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Sample Problem 16.5
3 Evaluate Does the result make sense?
• The volume of the solute is about one-third
the volume of the solution, so the answer is
reasonable.
• The answer is correctly expressed to two
significant figures.
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Percent Solutions
Another way to express the concentration
of a solution is as a percent by mass, or
percent (mass/mass).
Percent by mass of a solution is the ratio
of the mass of the solute to the mass of
the solution.
Percent by mass (%(m/m)) =
mass of solute
mass of solution
 100%
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Percent Solutions
mass of solute
Percent by mass (%(m/m)) =
 100%
mass of solution
• Percent by mass is sometimes a convenient
measure of concentration when the solute is a
solid.
• You have probably seen information on food
labels expressed as a percent composition.
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CHEMISTRY
& YOU
What are three ways to calculate the
concentration of a solution?
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CHEMISTRY
& YOU
What are three ways to calculate the
concentration of a solution?
The concentration
of a solution can be
calculated in moles
solute per liter of
solvent, or molarity
(M), percent by
volume (%(v/v)), or
percent by mass
(%(m/m)).
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Sample Problem 16.6
Using Percent by Mass as a
Conversion Factor
How many grams of
glucose (C6H12O6) are
needed to make 2000 g of
a 2.8% glucose (m/m)
solution?
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Sample Problem 16.6
1 Analyze List the knowns and the unknown.
The conversion is mass of solution → mass of
solute. In a 2.8% C6H12O6 (m/m) solution, each
100 g of solution contains 2.8 g of glucose. Used
as a conversion factor, the concentration allows
you to convert g of solution to g of C6H12O6.
KNOWNS
mass of solution = 2000 g
percent by mass = 2.8% C6H12O6(m/m)
UNKNOWN
mass of solute = ? g C6H12O6
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Sample Problem 16.6
2 Calculate Solve for the unknown.
Write percent by mass as a conversion
factor with g C6H12O6 in the numerator.
2.8 g C6H12O6
100 g solution
You can solve this
problem by using
either dimensional
analysis or algebra.
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Sample Problem 16.6
2 Calculate Solve for the unknown.
Multiply the mass of the solution by the
conversion factor.
2000 g solution 
2.8 g C6H12O6
100 g solution
= 56 g C6H12O6
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Sample Problem 16.6
3 Evaluate Does the result make sense?
• The prepared mass of the solution is 20  100 g.
• Since a 100-g sample of 2.8% (m/m) solution
contains 2.8 g of solute, you need 20  2.8 g = 56 g
of solute.
• To make the solution, mix 56 g of C6H12O6 with
1944 g of solvent.
• 56 g of solute + 1944 g solvent = 2000 g of solution
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What is the mass of water in a 2000 g
glucose (C6H12O6) solution that is
labeled 5.0% (m/m)?
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What is the mass of water in a 2000 g
glucose (C6H12O6) solution that is
labeled 5.0% (m/m)?
% (m/m) =
mass of glucose
 100%
mass of solution
(% (m/m))  mass of solution
mass of glucose =
100%
mass of glucose = 2000 g  0.050 = 100 g C6H12O6
mass of water = 2000 g – 100 g = 1900 g H2O
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Key Concepts
To calculate the molarity of a solution,
divide the moles of solute by the volume
of the solution in liters.
Diluting a solution reduces the number of
moles of solute per unit volume, but the
total number of moles of solute in solution
does not change.
Percent by volume is the ratio of the
volume of solute to the volume of solution.
Percent by mass is the ratio of the mass of
the solute to the mass of the solution.
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Key Equations
moles of solute
Molarity (M) =
liters of solution
M1  V1 = M2  V2
volume of solute
Percent by volume =
 100%
volume of solution
mass of solute
Percent by mass =
 100%
mass of solution
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Glossary Terms
• concentration: a measurement of the amount
of solute that is dissolved in a given quantity of
solvent; usually expressed as mol/L
• dilute solution: a solution that contains a
small amount of solute
• concentrated solution: a solution containing
a large amount of solute
• molarity (M): the concentration of solute in a
solution expressed as the number of moles of
solute dissolved in 1 liter of solution
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Chapter 16
Solutions
16.1 Properties of Solutions
16.2 Concentrations of Solutions
16.3 Colligative Properties
of Solutions
16.4 Calculations Involving
Colligative Properties
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CHEMISTRY
& YOU
Why do you need salt
to make ice cream?
Ice-cream makers know
that if you add rock salt
to ice, the mixture
freezes at a few
degrees below 0°C.
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Describing Colligative
Properties
Describing Colligative Properties
What are three colligative
properties of solutions?
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Describing Colligative
Properties
Tea is not the same as pure water.
• Some of these differences in properties have
little to do with the specific identity of the
solute.
• Instead, they depend upon the mere
presence of solute particles in the solution.
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Describing Colligative
Properties
A colligative property is a property of
solutions that depends only upon the
number of solute particles, not upon their
identity.
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Describing Colligative
Properties
Three important colligative properties
of solutions are
• vapor-pressure lowering
• freezing-point depression
• boiling-point elevation
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Describing Colligative
Properties
Vapor-Pressure Lowering
Vapor pressure is the pressure exerted
by a vapor that is in dynamic equilibrium
with its liquid in a closed system.
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Describing Colligative
Properties
Vapor-Pressure Lowering
Vapor pressure is the pressure exerted
by a vapor that is in dynamic equilibrium
with its liquid in a closed system.
• A solution that contains a solute that is
nonvolatile (not easily vaporized) always
has a lower vapor pressure than the pure
solvent.
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Describing Colligative
Properties
Vapor-Pressure Lowering
Higher vapor
pressure
Pure solvent
Solvent particle
Equilibrium is
established between
the liquid and vapor
in a pure solvent.
Lower vapor
pressure
Solution containing
nonvolatile solute
Solute particle
In a solution, solute particles
reduce the number of solvent
particles able to escape the liquid.
Equilibrium is established at a
lower vapor pressure.
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Describing Colligative
Properties
Vapor-Pressure Lowering
Ionic solutes that dissociate have greater
effects on vapor pressure than does a
nondissociating solute.
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Describing Colligative
Properties
Vapor-Pressure Lowering
Ionic solutes that dissociate have greater
effects on vapor pressure than does a
nondissociating solute.
• Three moles of sodium
chloride dissolved in
water produce 6 mol of
particles because each
formula unit of NaCl
dissociates into two
ions.
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Describing Colligative
Properties
Vapor-Pressure Lowering
Ionic solutes that dissociate have greater
effects on vapor pressure than does a
nondissociating solute.
• Three moles of calcium
chloride dissolved in
water produce 9 mol of
particles because each
formula unit of CaCl2
dissociates into three
ions.
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Describing Colligative
Properties
Vapor-Pressure Lowering
Ionic solutes that dissociate have greater
effects on vapor pressure than does a
nondissociating solute.
• Three moles of glucose
dissolved in water
produce 3 mol of
particles because
glucose does not
dissociate.
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Describing Colligative
Properties
Vapor-Pressure Lowering
The decrease in a solution’s vapor pressure
is proportional to the number of particles the
solute makes in solution.
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Describing Colligative
Properties
Vapor-Pressure Lowering
Which solution has the lowest vapor pressure?
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Describing Colligative
Properties
Vapor-Pressure Lowering
Which solution has the lowest vapor pressure?
• The vapor-pressure lowering caused by 0.1 mol of NaCl
in 1000 g of water is twice that caused by 0.1 mol of
glucose in the same quantity of water.
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Describing Colligative
Properties
Vapor-Pressure Lowering
Which solution has the lowest vapor pressure?
• In the same way, 0.1 mol of CaCl2 in 1000 g of water
produces three times the vapor-pressure lowering as 0.1
mol of glucose in the same quantity of water.
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Describing Colligative
Properties
Freezing-Point Depression
When a substance freezes, the particles
of the solid take on an orderly pattern.
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Describing Colligative
Properties
Freezing-Point Depression
When a substance freezes, the particles
of the solid take on an orderly pattern.
• The presence of a solute in water disrupts the
formation of this pattern.
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Describing Colligative
Properties
Freezing-Point Depression
When a substance freezes, the particles
of the solid take on an orderly pattern.
• The presence of a solute in water disrupts the
formation of this pattern.
• As a result, more kinetic energy must be
withdrawn from a solution than from the pure
solvent to cause the solution to solidify.
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Describing Colligative
Properties
Freezing-Point Depression
The freezing point of a solution is lower
than the freezing point of the pure solvent.
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Describing Colligative
Properties
Freezing-Point Depression
The freezing point of a solution is lower
than the freezing point of the pure solvent.
• The difference in temperature between the
freezing point of a solution and the freezing
point of the pure solvent is called the
freezing-point depression.
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Describing Colligative
Properties
Freezing-Point Depression
Freezing-point depression is another
colligative property.
• The magnitude of the freezing-point
depression is proportional to the number of
solute particles dissolved in the solvent and
does not depend upon their identity.
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Describing Colligative
Properties
Freezing-Point Depression
The freezing-point depression
of aqueous solutions plays an
important role in helping keep
travelers safe in cold, icy
weather.
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Describing Colligative
Properties
Freezing-Point Depression
The freezing-point depression
of aqueous solutions plays an
important role in helping keep
travelers safe in cold, icy
weather.
• The truck spreads a layer of salt on
the icy road to make the ice melt.
• The melted ice forms a solution with
a lower freezing point than that of
pure water.
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Describing Colligative
Properties
Boiling-Point Elevation
The boiling point of a substance is the
temperature at which the vapor pressure of the
liquid phase equals atmospheric pressure.
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Describing Colligative
Properties
Boiling-Point Elevation
The boiling point of a substance is the
temperature at which the vapor pressure of the
liquid phase equals atmospheric pressure.
• Adding a nonvolatile solute to a liquid solvent
decreases the vapor pressure of the solvent.
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Describing Colligative
Properties
Boiling-Point Elevation
The boiling point of a substance is the
temperature at which the vapor pressure of the
liquid phase equals the atmospheric pressure.
• Adding a nonvolatile solute to a liquid solvent
decreases the vapor pressure of the solvent.
• Because of the decrease in vapor pressure,
additional kinetic energy must be added to raise
the vapor pressure of the liquid phase of the
solution to atmospheric pressure and initiate
boiling.
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Describing Colligative
Properties
Boiling-Point Elevation
The boiling point of a solution is higher
than the boiling point of the pure solvent.
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Describing Colligative
Properties
Boiling-Point Elevation
The boiling point of a solution is higher
than the boiling point of the pure solvent.
• The difference in temperature between the
boiling point of a solution and the boiling point
of the pure solvent is the boiling-point
elevation.
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Describing Colligative
Properties
Boiling-Point Elevation
The fluid circulating through a car’s cooling
system is a solution of water and ethylene
glycol, or antifreeze.
• The antifreeze doesn’t just
lower the freezing point of
the water in the cooling
system.
• It also elevates the boiling
point, which helps protect
the engine from
overheating in the
summer.
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Describing Colligative
Properties
Boiling-Point Elevation
Boiling-point elevation is a colligative
property; it depends on the concentration of
particles, not on their identity.
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Describing Colligative
Properties
Boiling-Point Elevation
Boiling-point elevation is a colligative
property; it depends on the concentration of
particles, not on their identity.
• The magnitude of the boiling-point elevation is
proportional to the number of solute particles
dissolved in the solvent.
– The boiling point of water increases by 0.512°C
for every mole of particles that the solute forms
when dissolved in 1000 g of water.
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CHEMISTRY
& YOU
Solutes other than NaCl could be
used to produce the same freezingpoint depression in an ice-cream
machine. What factors do you think
make NaCl a good choice?
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CHEMISTRY
& YOU
Solutes other than NaCl could be
used to produce the same freezingpoint depression in an ice-cream
machine. What factors do you think
make NaCl a good choice?
NaCl, or rock salt, is readily
available, inexpensive, and nontoxic. It is an ionic compound. It
produces twice the freezing-point
depression of a molecular solid
such as sucrose, or table sugar.
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You have 500 mL of 1M solutions of
NaCl, Na2SO4, Na3PO4, and Al2(SO4)3.
Which solution will have the highest
boiling point?
A. NaCl(aq)
B. Na2SO4(aq)
C. Na3PO4(aq)
D. Al2(SO4)3(aq)
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You have 500 mL of 1M solutions of
NaCl, Na2SO4, Na3PO4, and Al2(SO4)3.
Which solution will have the highest
boiling point?
A. NaCl(aq)
B. Na2SO4(aq)
C. Na3PO4(aq)
D. Al2(SO4)3(aq)
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Key Concepts
Colligative properties of solutions
include vapor-pressure lowering,
freezing-point depression, and
boiling-point elevation.
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Glossary Terms
• colligative property: a property of a solution
that depends only upon the number of solute
particles, and not upon their identities; boilingpoint elevation, freezing-point depression, and
vapor-pressure lowering are colligative
properties
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Glossary Terms
• freezing-point depression: the difference in
temperature between the freezing point of a
solution and the freezing point of the pure
solvent
• boiling-point elevation: the difference in
temperature between the boiling point of a
solution and the boiling point of the pure
solvent
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BIG IDEA
The Mole and Quantifying Matter
Solubility, miscibility, concentration, and
colligative properties are used to describe
and characterize solutions.
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Chapter 16
Solutions
16.1 Properties of Solutions
16.2 Concentrations of Solutions
16.3 Colligative Properties
of Solutions
16.4 Calculations Involving
Colligative Properties
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CHEMISTRY
& YOU
How hot is a pot of boiling pasta?
Recall that
dissolved salt
elevates the
boiling point of
water.
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Molality and Mole
Fraction
Molality and Mole Fraction
What are two ways of expressing the
ratio of solute to solvent in a
solution?
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Molality and Mole
Fraction
Recall that colligative properties of
solutions depend only on the number of
solute particles dissolved in a given
amount of solvent.
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Molality and Mole
Fraction
Chemists use two ways to express
the ratio of solute particles to solvent
particles: in molality and in mole
fractions.
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Molality and Mole
Fraction
Molality (m) is the number of moles of
solute dissolved in 1 kilogram (1000 grams)
of solvent.
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Molality and Mole
Fraction
Molality (m) is the number of moles of
solute dissolved in 1 kilogram (1000 grams)
of solvent.
• Molality is also known as molal concentration.
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Molality and Mole
Fraction
Molality (m) is the number of moles of
solute dissolved in 1 kilogram (1000 grams)
of solvent.
• Molality is also known as molal concentration.
Molality (m) =
moles of solute
kilogram of solvent
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Molality and Mole
Fraction
Molality is not the same as molarity.
• Molality refers to moles of solute per
kilogram of solvent rather than moles of
solute per liter of solution.
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Sample Problem 16.7
Using Molality
How many grams of potassium
iodide must be dissolved in
500.0 g of water to produce a
0.060 molal KI solution?
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Sample Problem 16.7
1
Analyze List the knowns and the unknown.
• According to the definition of molality, the
final solution must contain 0.060 mol KI per
1000 g H2O.
• Use the molality as a conversion factor to
convert from mass of the solvent (H2O) to
moles of the solute (KI).
• Then use the molar mass of KI to convert
from mol KI to g KI.
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Sample Problem 16.7
1
Analyze List the knowns and the unknown.
The steps are as follows:
mass of H2O → mol KI → g KI.
KNOWNS
mass of water = 500.0 g = 0.5000 kg
solution concentration = 0.060m
molar mass KI = 166.0 g/mol
UNKNOWN
mass of solute = ? g KI
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Sample Problem 16.7
2
Calculate Solve for the unknown.
Identify the conversion factor based on
0.060m that allows you to convert from
g H2O to mol KI.
0.060 mol KI
1.000 kg H2O
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Sample Problem 16.7
2
Calculate Solve for the unknown.
Identify the conversion factor based
on the molar mass of KI that allows
you to convert from mol KI to g KI.
166.0 g KI
1 mol KI
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Sample Problem 16.7
2
Calculate Solve for the unknown.
Multiply the known
solvent volume by the
conversion factors.
To make the 0.060-molal KI
solution, you would dissolve
5.0 g of KI in 500.0 g of water.
0.060 mol KI
166.0 g KI
0.5000 kg H2O  1.000 kg H O  1 mol KI = 5.0 g KI
2
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Sample Problem 16.7
3
Evaluate Does the result make sense?
• A 1-molal KI solution is one molar mass of KI
(166.0 g) dissolved in 1000 g of water.
• The desired molal concentration (0.060m) is
about 1/20 of that value, so the mass of KI
should be much less than the molar mass.
• The answer is correctly expressed to two
significant figures.
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Molality and Mole
Fraction
The concentration of a solution can also
be expressed as a mole fraction.
• The mole fraction of a solute in a solution is
the ratio of the moles of that solute to the total
number of moles of solvent and solute.
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Molality and Mole
Fraction
In a solution containing nA mol of solute A
and nB mol of solvent B, the mole fraction
of solute A (XA) and the mole fraction of
solvent B (XB) can be expressed as
follows:
XA =
nA
nA + nB
XB =
nB
nA + nB
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Molality and Mole
Fraction
Note that mole fraction is a dimensionless
quantity.
• The sum of the mole fractions of all the
components in a solution equals unity, or
one.
XA =
nA
nA + nB
XB =
nB
nA + nB
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Molality and Mole
Fraction
Ethylene glycol (EG) is added to water as
antifreeze.
• The figure below illustrates how to calculate
the mole fractions of the solute and solvent
for a solution of EG in water.
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Sample Problem 16.8
Calculating Mole Fractions
Ethylene glycol (EG, or C2H6O2)
is added to automobile cooling
systems to protect against cold
weather. What is the mole
fraction of each component in a
solution containing 1.25 mol of
ethylene glycol and 4.00 mol of
water?
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Sample Problem 16.8
1
Analyze List the knowns and the unknowns.
• The given quantities of solute (EG) and
solvent (water) are expressed in moles.
• Use the equations for mole fraction of a
solute and mole fraction of a solvent to solve
this problem.
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Sample Problem 16.8
1
Analyze List the knowns and the unknowns.
KNOWNS
moles of ethylene glycol (nEG) = 1.25 mol EG
moles of water (nH2O) = 4.00 mol H2O
UNKNOWNS
mole fraction EG (XEG) = ?
mole fraction H2O (XH2O) = ?
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Sample Problem 16.8
2
Calculate Solve for the unknowns.
Write the equation for
the mole fraction of
ethylene glycol (XEG)
in the solution.
XEG =
Note that the denominator for
each mole fraction is the
same: the total number of
moles of solvent and solute in
the solution.
nEG
nEG + nH2O
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Sample Problem 16.8
2
Calculate Solve for the unknowns.
Write the equation for
the mole fraction of
water (XH2O) in the
solution.
XH2O =
Note that the denominator for
each mole fraction is the
same: the total number of
moles of solvent and solute in
the solution.
nH2O
nEG + nH2O
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Sample Problem 16.8
2
Calculate Solve for the unknowns.
Substitute the known values into each
equation.
XEG =
XH2O =
nEG
=
nEG + nH2O
nH2O
nEG + n=H2O
1.25 mol
= 0.238
1.25 mol + 4.00 mol
4.00 mol
0.762mol
1.25 mol += 4.00
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Sample Problem 16.8
3
Evaluate Does the result make sense?
• The sum of the mole fractions of all the
components in the solution equals 1
(XEG + XH2O = 1.000).
• Each answer is correctly expressed to
three significant figures.
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What is the mole fraction of He in a
gaseous solution containing 4.0 g of He,
6.5 g of Ar, and 10.0 g of Ne?
A. 0.60
B. 1.5
C. 0.20
D. 0.11
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What is the mole fraction of He in a
gaseous solution containing 4.0 g of He,
6.5 g of Ar, and 10.0 g of Ne?
A. 0.60
B. 1.5
C. 0.20
D. 0.11
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Freezing-Point Depression
and Boiling-Point Elevation
Freezing-Point Depression and
Boiling-Point Elevation
How are freezing-point depression
and boiling-point elevation related to
molality?
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Freezing-Point Depression
and Boiling-Point Elevation
Depressions of freezing points and
elevations of boiling points are usually
quite small.
• To measure colligative properties accurately,
you would need a thermometer that can
measure temperatures to the nearest
0.001°C.
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Freezing-Point Depression
and Boiling-Point Elevation
Depressions of freezing points and
elevations of boiling points are usually
quite small.
• Another way to determine the magnitudes of
colligative properties is by calculating them.
• You can do this if you know the molality of
the solution and some reference data about
the solvent.
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Freezing-Point Depression
and Boiling-Point Elevation
The magnitudes of the freezing-point
depression (ΔTf) and the boiling-point
elevation (ΔTb) of a solution are directly
proportional to the molal concentration (m),
assuming the solute is molecular, not ionic.
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Freezing-Point Depression
and Boiling-Point Elevation
The magnitudes of the freezing-point
depression (ΔTf) and the boiling-point
elevation (ΔTb) of a solution are directly
proportional to the molal concentration (m),
assuming the solute is molecular, not ionic.
ΔTf
ΔTb
m
m
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Interpret Graphs
101
Vapor pressure (kPa)
ΔTf is the
difference
between the
freezing point
of the solution
and the
freezing point
of the pure
solvent.
Temperature (°C)
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Interpret Graphs
101
Vapor pressure (kPa)
ΔTb is the
difference
between the
boiling point of
the solution
and the boiling
point of the
pure solvent.
Temperature (°C)
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Freezing-Point Depression
and Boiling-Point Elevation
With the addition of a constant, the
proportionality between the freezing-point
depression and the molality m can be
expressed as an equation.
ΔTf = Kf  m
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Freezing-Point Depression
and Boiling-Point Elevation
With the addition of a constant, the
proportionality between the freezing-point
depression and the molality m can be
expressed as an equation.
ΔTf = Kf  m
• The constant, Kf, is the molal freezing-point
depression constant, which is equal to the
change in freezing point for a 1-molal solution
of a nonvolatile molecular solute.
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Freezing-Point Depression
and Boiling-Point Elevation
The boiling-point elevation of a solution
can also be expressed as an equation.
ΔTb = Kb  m
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Freezing-Point Depression
and Boiling-Point Elevation
The boiling-point elevation of a solution
can also be expressed as an equation.
• The proportionality constant is Kb, the molal
boiling-point elevation constant, which is
equal to the change in boiling point for a 1molal solution of a nonvolatile molecular
solute.
ΔTb = Kb  m
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Interpret Data
The values of Kf and Kb depend upon the
solvent.
Kf and Kb Values for
Some Common Solvents
Kf
(°C/m)
Kb
(°C/m)
Acetic Acid
3.90
3.07
Benzene
5.12
2.53
Solvent
Camphor
37.7
5.95
Cyclohexane
20.2
2.79
Ethanol
1.99
1.19
Nitrobenzene
7.00
5.24
Phenol
7.40
3.56
Water
1.86
0.512
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Sample Problem 16.9
Calculating the Freezing-Point
Depression of a Solution
Antifreeze protects a car from freezing. It
also protects it from overheating.
Calculate the freezing-point depression of
a solution containing exactly 100 g of
ethylene glycol (C2H6O2) antifreeze in
0.500 kg of water.
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Sample Problem 16.9
1
Analyze List the knowns and the unknown.
• Calculate the number of moles of C2H6O2
and the molality of the solution.
• Then calculate the freezing-point depression
using ΔTf = Kf  m.
KNOWNS
UNKNOWN
mass of C2H6O2 = 100 g
ΔTf = ?°C
mass of water = 0.500 kg
Kf for H2O = 1.86°C/m
molar mass C2H6O2 = 62.0 g/mol
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Sample Problem 16.9
2
Calculate Solve for the unknown.
• Use the molar mass of C2H6O2 to
convert the mass of solute to moles.
100 g C2H6O2 
1 mol C2H6O2
= 1.61 mol C2H6O2
62.0 g C2H6O2
• Calculate the molality of the solution.
m=
mol solute
kg solvent =
1.61 mol
= 3.22m
0.500 kg
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Sample Problem 16.9
2
Calculate Solve for the unknown.
Calculate the freezing-point depression.
ΔTf = Kf  m = 1.86°C/m  3.22m = 5.99°C
The freezing point of the solution is
0.00oC – 5.99°C = – 5.99°C.
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Sample Problem 16.9
3
Evaluate Does the result make sense?
• A 1-molal solution reduces the freezing
temperature by 1.86°C.
• So, a decrease of 5.99°C for an
approximately 3-molal solution is
reasonable.
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Sample Problem 16.10
Calculating the Boiling Point of a
Solution
What is the boiling point of a 1.50m NaCl
solution?
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Sample Problem 16.10
1
Analyze List the knowns and the unknown.
• Each formula unit of NaCl dissociates into two particles,
according to the equation NaCl(s) → Na+(aq) + Cl–(aq).
• Based on the total number of dissociated particles, the
effective molality is 2  1.50m = 3.00m.
• Calculate the boiling-point elevation (using the equation
ΔTb = Kb  m) and then add it to 100°C.
KNOWNS
UNKNOWN
Solution concentration = 1.50m NaCl
boiling point =
?°C
Kb for H2O = 0.512°C/m
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Sample Problem 16.10
2
Calculate Solve for the unknown.
• Calculate the boiling-point elevation,
making sure to use the molality of total
dissociated particles in solution.
ΔTb = Kb  m = 0.512°C/m  3.00m = 1.54°C
• Calculate the boiling point of the solution.
Tb = 100°C + 1.54°C = 101.54°C
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Sample Problem 16.10
3
Evaluate Does the result make sense?
• The boiling point increases about
0.5°C for each mole of solute
particles.
• The total change is reasonable.
• Because the boiling point of water is
defined as exactly 100°C, this value
does not limit the number of significant
figures in the solution of the problem.
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CHEMISTRY
& YOU
Does pasta cook at 100°C? After you
read Sample Problem 16.10, calculate the
boiling-point elevation for 2 L of water
containing a teaspoon of salt (about 20 g).
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CHEMISTRY
& YOU
Does pasta cook at 100°C? After you
read Sample Problem 16.10, calculate the
boiling-point elevation for 2 L of water
containing a teaspoon of salt (about 20 g).
1 mol NaCl
2 mol solute
20 g NaCl  58.5 g NaCl  1 mol NaCl = 0.7 mol solute
mol solute
0.7 mol
m = kg solvent =
2 kg
= 0.4m
ΔTb = Kb  m = 0.512°C/m  0.4m = 0.2°C
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How does the freezing-point depression
by CaCl2 for a given solvent compare to
the freezing-point depression caused by
the same molal concentration of
ethylene glycol (C2H6O2,)?
A. It is exactly the same.
B. It is twice as large.
C. It is three times as large.
D. It is four times as large.
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How does the freezing-point depression
by CaCl2 for a given solvent compare to
the freezing-point depression caused by
the same molal concentration of
ethylene glycol (C2H6O2,)?
A. It is exactly the same.
B. It is twice as large.
C. It is three times as large.
D. It is four times as large.
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Key Concepts
Chemists use two ways to express the
ratio of solute to solvent: in molality and
in mole fractions.
The magnitudes of freezing-point
depression and boiling-point elevation are
proportional to molality.
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Key Equations
Molality (m) =
moles of solute
kilograms of solvent
Mole fractions: XA =
nA
XB =
nA + nB
nB
nA + nB
ΔTf = Kf  m
ΔTb = Kb  m
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Glossary Terms
• molality (m): the concentration of solute in a solution
expressed as the number of moles of solute dissolved
in 1 kilogram (1000 g) of solvent
• mole fraction: the ratio of the moles of solute in
solution to the total number of moles of both solvent
and solute
• molal freezing-point depression constant (Kf): the
change in freezing point for a 1-molal solution of a
nonvolatile molecular solute
• molal boiling-point elevation constant (Kb): the
change in boiling point for a 1-molal solution of a
nonvolatile molecular solute
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BIG IDEA
The Mole and Quantifying Matter
Solution concentration can be quantified
in terms of molarity (moles of solute per
liter of solution), molality (moles of solute
per kilogram of solvent), percent by
volume, and percent by mass.
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