Principles of Chemistry: A Molecular Approach, 1st Ed.
Nivaldo Tro
Chapter 12
Solutions
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
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Solutions
• homogeneous mixtures
 Composition may vary from one sample to another.
 appears to be one substance, though really contains
multiple materials
• Most homogeneous materials we encounter
are actually solutions.
 e.g., air and seawater
• Nature has a tendency toward spontaneous
mixing.
 Generally, uniform mixing is more energetically
favorable.
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Solutions
• When table salt is mixed with water, it
seems to disappear, or become a
liquid—the mixture is homogeneous.
 The salt is still there, as you can tell from
the taste, or simply boiling away the water.
• Homogeneous mixtures are called
•
•
solutions.
The component of the solution that
changes state is called the solute.
The component that keeps its state is
called the solvent.
 If both components start in the same state,
the major component is the solvent.
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Seawater
• Drinking seawater will dehydrate you and
give you diarrhea.
• The cell wall acts as a barrier to solute
moving.
• The only way for the seawater and the cell
solution to have uniform mixing is for water
to flow out of the cells of your intestine and
into your digestive tract.
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Common Types of Solution
Solution Phase
gaseous solutions
liquid solutions
solid solutions
Solute
Phase
gas
gas
liquid
solid
solid
Solvent
Phase
gas
liquid
liquid
liquid
solid
Example
air (mostly N2 & O2)
soda (CO2 in H2O)
vodka (C2H5OH in H2O)
seawater (NaCl in H2O)
brass (Zn in Cu)
• Solutions that contain Hg and some other metal
•
are called amalgams.
Solutions that contain metal solutes and a metal
solvent are called alloys.
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Brass
Type
Color
% Cu
% Zn
Density
g/cm3
MP
°C
Tensile
Strength
psi
Uses
Gilding
reddish
95
5
8.86
1066
50K
pre-83 pennies,
munitions, plaques
Commercial
bronze
90
10
8.80
1043
61K
doorknobs,
grillwork
Jewelry
bronze
87.5
12.5
8.78
1035
66K
costume jewelry
Red
golden
85
15
8.75
1027
70K
electrical sockets,
fasteners & eyelets
Low
deep
yellow
80
20
8.67
999
74K
musical
instruments,
clock dials
Cartridge
yellow
70
30
8.47
954
76K
car radiator cores
Common
yellow
67
33
8.42
940
70K
lamp fixtures,
bead chain
Muntz metal
yellow
60
40
8.39
904
70K
nuts & bolts,
brazing rods
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Solubility
• When one substance (solute) dissolves in
•
another (solvent) it is said to be soluble.
Salt is soluble in water.
Bromine is soluble in methylene chloride.
When one substance does not dissolve in
another it is said to be insoluble.
Oil is insoluble in water.
• The solubility of one substance in another
depends on two factors—nature’s
tendency towards mixing, and the types
of intermolecular attractive forces.
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Mixing and the Solution Process
Entropy
• Most processes occur because the
end result has less potential energy.
• However, formation of a solution
does not necessarily lower the
potential energy of the system.
• When two ideal gases are put into
the same container, they
spontaneously mix.
 even though the difference in
attractive forces is negligible
• The gases mix because the energy
of the system is lowered through the
release of entropy.
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Mixing and the Solution Process
Entropy
• Entropy is the measure of
energy dispersal throughout the
system.
• Energy has a spontaneous
drive to spread out over as
large a volume as it is allowed.
• By each gas expanding to fill
the container, it spreads its
energy out and lowers its
entropy.
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Intermolecular Forces and the
Solution Process
• Energy changes in the formation of most solutions
•
also involve differences in attractive forces
between the particles.
In order for the solvent and solute to mix, you
must overcome
1. all of the solute–solute attractive forces.
2. some of the solvent–solvent attractive forces.
 Both processes are endothermic.
• At least some of the energy to do this comes from
making new solute–solvent attractions.
 which is exothermic
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Relative Interactions and Solution Formation
• When the solute-to-solvent attractions are weaker
than the sum of the solute-to-solute and solventto-solvent attractions, the solution will only form if
the energy difference is small enough to be
overcome by the entropy.
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Solubility
• There is usually a limit to the solubility of one
substance in another.
Gases are always soluble in each other.
Two liquids that are mutually soluble are said
to be miscible.
Alcohol and water are miscible.
Oil and water are immiscible.
• The maximum amount of solute that can be
•
dissolved in a given amount of solvent is called
the solubility.
The solubility of one substance in another varies
with temperature and pressure.
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Will It Dissolve?
• Chemist’s Rule of Thumb—
Like Dissolves Like
• A chemical will dissolve in a
•
solvent if it has a similar
structure to the solvent.
When the solvent and solute
structures are similar, the
solvent molecules will attract
the solute particles at least as
well as the solute particles to
each other.
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Classifying Solvents
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Example 12.1a Predict whether the following
vitamin is soluble in fat or water.
The 4 OH groups
make the molecule
highly polar and it will
also H-bond to
water.
Vitamin C is water
soluble.
Vitamin C
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Example 12.1b Predict whether the following
vitamin is soluble in fat or water.
The 2 C=O groups are
polar, but their geometric
symmetry suggests their
pulls will cancel and the
molecule will be
nonpolar.
Vitamin K3 is fat soluble.
Vitamin K3
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Practice—Decide if the following are more
soluble in hexane, C6H14,or water.
nonpolar molecule
more soluble in C6H14
naphthalene
formaldehyde
stearic acid
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polar molecule
more soluble in H2O
nonpolar part dominant
more soluble in C6H14
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Energetics of Solution Formation:
The Enthalpy of Solution
• In order to make a solution you must:
1. overcome all attractions between the solute
particles—endothermic.
2. overcome some attractions between solvent
molecules—endothermic.
3. form new attractions between solute particles and
solvent molecules—exothermic.
• The overall DH for making a solution depends on
the relative sizes of the DH for these 3 processes.
DHsol’n = DHsolute + DHsolvent + DHmix
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Solution Process
1. add energy in to overcome all solute–solute attractions
3. form new solute–solvent attractions, releasing energy
2. add energy in to overcome some solvent–solvent attractions
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Energetics of Solution Formation
If the total energy cost for
breaking attractions between
particles in the pure solute
and pure solvent is greater
less than
the energy
than
the energy
released
released
in
in
making the new attractions
between the solute and
solvent, the overall process
will be endothermic
exothermic. .
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Heats of Hydration
• For aqueous solutions of ionic compounds, the
energy added to overcome the attractions
between water molecules and the energy released
in forming attractions between the water
molecules and ions is combined into a term called
the heat of hydration.
 attractive forces between ions = lattice energy
 DHsolute = –DHlattice energy
 attractive forces in water = H-bonds
 attractive forces between ion and water = ion–dipole
 DHhydration = heat released when 1 mole of gaseous ions
dissolves in water = DHsolvent + DHmix
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Heats of Solution for
Ionic Compounds
• For an aqueous solution of an ionic
compound, the DHsolution is the difference
between the heat of hydration and the lattice
energy.
DHsolution = DHsolute+ DHsolvent + DHmix
DHsolution = −DHlattice energy+ DHsolvent + DHmix
DHsolution = DH
−DH
− DH+lattice
DHenergy
hydration
lattice energy
hydration
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Comparing Heat of Solution to
Heat of Hydration
• Since the lattice energy is always exothermic, the size and
•
•
sign on the DHsol’n tells us something about DHhydration.
If the heat of solution is large and endothermic, then the
amount of energy it costs to separate the ions is more than
the energy released from hydrating the ions.
 DHhydration < DHlattice when DHsol’n is (+)
If the heat of solution is large and exothermic, then the
amount of energy it costs to separate the ions is less than
the energy released from hydrating the ions.
 DHhydration > DHlattice when DHsol’n is (−)
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Ion–Dipole Interactions
• When ions dissolve in water, they become
hydrated.
• Each ion is surrounded by water molecules.
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Practice—What is the lattice energy of KI if
DHsol’n = +21.5 kJ/mol and the DHhydration = −583 kJ/mol?
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Practice—What is the lattice energy of KI if
DHsol’n = +21.5 kJ/mol and the DHhydration = −583 kJ/mol?
Given: DHsol’n = +21.5 kJ/mol, DHhydration = −583 kJ/mol
Find: DHlattice, kJ/mol
Conceptual
DHsol’n, DHhydration
DHlattice
Plan:
Relationships:
DHsol’n = DHhydration− DHlattice
Solution:
Check:
The unit is correct, and the lattice energy
being exothermic is correct.
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Solution Equilibrium
• The dissolution of a solute in a solvent is an
•
•
•
equilibrium process.
Initially, when there is no dissolved solute, the
only process possible is dissolution.
Shortly, solute particles can start to recombine to
reform solute molecules, but the rate of
dissolution >> rate of deposition and the solute
continues to dissolve.
Eventually, the rate of dissolution = the rate of
deposition—the solution is saturated with solute
and no more solute will dissolve.
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Solubility Limit
• A solution that has the maximum amount of solute
dissolved in it is said to be saturated.
 depends on the amount of solvent
 depends on the temperature
 and pressure of gases
• A solution that has less solute than saturation is
•
said to be unsaturated.
A solution that has more solute than saturation is
said to be supersaturated.
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How Can You Make a Solvent Hold
More Solute Than It Is Able To?
• Solutions can be made saturated at non-room
•
•
conditions, then allowed to come to room
conditions slowly.
For some solutes, instead of coming out of
solution when the conditions change, they get
stuck in-between the solvent molecules and the
solution becomes supersaturated.
Supersaturated solutions are unstable and lose all
the solute above saturation when disturbed.
 e.g., shaking a carbonated beverage
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Adding Solute to a Supersaturated
Solution of NaC2H3O2
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Temperature Dependence of
Solubility of Solids in Water
• Solubility is generally given in grams of solute that
•
will dissolve in 100 g of water.
For most solids, the solubility of the solid
increases as the temperature increases.
 when DHsolution is endothermic
• Solubility curves can be used to predict whether a
solution with a particular amount of solute
dissolved in water is saturated (on the line),
unsaturated (below the line), or supersaturated
(above the line).
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Solubility Curves
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Temperature Dependence of Solid
Solubility in Water (g/100 g H2O)
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Purification by Recrystallization
• One of the common operations performed
by a chemist is purifying a material.
• One method of purification involves
dissolving a solid in a hot solvent until the
solution is saturated.
• As the solution slowly cools, the solid
crystallizes out leaving impurities behind.
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Recrystallization of KNO3
• KNO3 can be purified by
•
dissolving a little less
than 106 g in 100 g of
water at 60 ºC, then
allowing it to cool slowly.
When it cools to 0 ºC,
only 13.9 g will remain in
solution; the rest will
precipitate out.
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Practice—Decide if each of the following solutions is
saturated, unsaturated, or supersaturated.
50 g KNO3 in 100 g H2O @ 34 ºC
saturated
50 g KNO3 in 100 g H2O @ 50 ºC
unsaturated
50 g KNO3 in 50 g H2O @ 50 ºC
supersaturated
100 g NH4Cl in 200 g H2O @ 70 ºC
unsaturated
100 g NH4Cl in 150 g H2O @ 50 ºC
supersaturated
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Temperature Dependence of
Solubility of Gases in Water
• Solubility is generally given in moles of
solute that will dissolve in 1 liter of solution.
• generally lower solubility than ionic or polar
covalent solids because most are nonpolar
molecules
• For all gases, the solubility of the gas
decreases as the temperature increases.
The DHsolution is exothermic because you do not
need to overcome solute–solute attractions.
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Temperature Dependence of Gas
Solubility in Water (g/100 g H2O)
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Pressure Dependence of Solubility
of Gases in Water
• The larger the partial pressure of a gas in
contact with a liquid, the more soluble the
gas is in the liquid.
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Henry’s Law
• The solubility of a gas
(Sgas) is directly
proportional to its partial
pressure, (Pgas).
Sgas = kHPgas
• kH is called Henry’s Law
Constant.
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Relationship between Partial
Pressure and Solubility of a Gas
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Pressure
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Example 12.2 What pressure of CO2 is required to
keep the [CO2] = 0.12 M in soda at 25 °C?
Given:
Find:
Conceptual
Plan:
S = [CO2] = 0.12 M
P of CO2, atm
[CO2]
P
Relationships: S = k P, k = 3.4 × 10–2 M/atm
H
H
Solution:
Check: The unit is correct, and the pressure higher than 1 atm
meets our expectation from general experience.
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Practice—How many grams of NH3 will dissolve in
0.10 L of solution when its partial pressure is 7.6 torr?
(kH = 58 M/atm)
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Practice—How many grams of NH3 will dissolve in
0.10 L of solution when its partial pressure is 7.6 torr?
Given: P of NH3 = 7.6 torr; 0.10 L
Find: mass of NH3, g
Conceptual
Plan:
Relationships:
S = kHP, kH = 58 M/atm, 1 atm = 760 torr, 1 mol = 17.04 g
Solution:
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Concentrations
• Solutions have variable composition.
• To describe a solution, you need to describe the
•
•
components and their relative amounts.
The terms dilute and concentrated can be used
as qualitative descriptions of the amount of solute
in solution.
concentration = amount of solute in a given
amount of solution
 occasionally amount of solvent
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Solution Concentration
Molarity
• moles of solute per 1 liter of solution
• used because it describes how many
molecules of solute in each liter of solution
• if a sugar solution concentration is 2.0 M,
1 liter of solution contains 2.0 moles of
sugar, 2 liters = 4.0 moles sugar, 0.5 liters
= 1.0 mole sugar
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Molarity and Dissociation
• The molarity of the ionic compound allows you
•
•
•
•
to determine the molarity of the dissolved ions.
CaCl2(aq) = Ca2+(aq) + 2 Cl−(aq)
A 1.0 M CaCl2(aq) solution contains 1.0 mole of
CaCl2 in each liter of solution.
1 L = 1.0 mole CaCl2, 2 L = 2.0 moles CaCl2
Because each CaCl2 dissociates to give one
Ca2+, a 1.0 M CaCl2 solution is 1.0 M Ca2+.
1 L = 1.0 mole Ca+2, 2 L = 2.0 moles Ca2+
Because each CaCl2 dissociates to give 2 Cl−,
a 1.0 M CaCl2 solution is 2.0 M Cl−.
1 L = 2.0 moles Cl−, 2 L = 4.0 moles Cl−
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Solution Concentration
Molality, m
• moles of solute per 1 kilogram of solvent
defined in terms of amount of solvent, not
solution
like the others
• does not vary with temperature
because based on masses, not volumes
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Parts Solute in Parts Solution
• Parts can be measured by mass or volume.
• Parts are generally measured in same units.
 by mass in grams, kilograms, lbs. etc.
 by volume in mL, L, gallons, etc.
 mass and volume combined in grams and mL
• percentage = parts of solute in every 100 parts
solution
 If a solution is 0.9% by mass, then there are 0.9 grams
of solute in every 100 grams of solution.
or 0.9 kg solute in every 100 kg solution
• parts per million = parts of solute in every 1 million
parts solution
 If a solution is 36 ppm by volume, then there are 36 mL
of solute in 1 million mL of solution.
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Percent Concentration
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Parts per Million Concentration
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PPM
• grams of solute per 1,000,000 g of solution
• mg of solute per 1 kg of solution
• 1 liter of water = 1 kg of water
For aqueous solutions we often approximate the
kg of the solution as the kg or L of water.
For dilute solutions, the difference in density between
the solution and pure water is usually negligible.
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Parts per Billion Concentration
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Using Concentrations as
Conversion Factors
• Concentrations show the relationship between
the amount of solute and the amount of solvent.
 12%(m/m) sugar(aq) means 12 g sugar  100 g solution
 or 12 kg sugar  100 kg solution; or 12 lbs.  100 lbs. solution
 5.5%(m/v) Ag in Hg means 5.5 g Ag  100 mL solution
 22%(v/v) alcohol(aq) means 22 mL EtOH  100 mL solution
• The concentration can then be used to convert the
amount of solute into the amount of solution, or vice
versa.
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Example 12.3 What volume of 10.5% by
mass soda contains 78.5 g of sugar?
Given: 78.5 g sugar
Find: volume, mL
g sol’n
Conceptual g solute
Plan:
mL sol’n
Relationships: 100 g sol’n = 10.5 g sugar, 1 mL sol’n = 1.04 g
Solution:
Check:
The unit is correct; the magnitude seems reasonable
as the mass of sugar  10% the volume of solution.
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Preparing a Solution
• need to know amount of solution and
concentration of solution
• Calculate the mass of solute needed.
Start with amount of solution.
Use concentration as a conversion factor.
5% by mass 5 g solute  100 g solution
“Dissolve the grams of solute in enough
solvent to total the total amount of solution.”
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Practice—How would you prepare 250.0 mL of
19.5% by mass CaCl2? (d = 1.18 g/mL)
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Practice—How would you prepare 250.0 mL of
19.5% by mass CaCl2? (d = 1.18 g/mL)
Given: 250.0 mL solution
Find: mass CaCl2, g
Conceptual mL sol’n
Plan:
g sol’n
g solute
Relationships: 100 g sol’n = 19.5 g CaCl2, 1 mL sol’n = 1.18 g
Solution:
Answer:
Dissolve 57.5 g of CaCl2 in enough water to total
250.0 mL.
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Solution Concentrations
Mole Fraction, XA
• The mole fraction is the fraction of the moles
•
•
•
of one component in the total moles of all the
components of the solution.
total of all the mole fractions in a solution = 1
unitless
The mole percentage is the percentage of the
moles of one component in the total moles of all
the components of the solution.
 = mole fraction × 100%
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Example 12.4a What is the molarity of a solution
prepared by mixing 17.2 g of C2H6O2 with 0.500 kg
of H2O to make 515 mL of solution?
Given: 17.2
0.2771
C2H
0.515
L
g Cmol
, 0.500
kg H2kg
O, H
515
sol’n
6O2, 0.500
2O,mL
2H6O
Find: M
Conceptual g C2H6O2
Plan:
mol C2H6O2
mL sol’n
L sol’n
M
Relationships: M = mol/L, 1 mol C2H6O2 = 62.07 g, 1 mL = 0.001 L
Solution:
Check:
The unit is correct; the magnitude is
reasonable.
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Practice—Calculate the molarity of a solution made by
dissolving 34.0 g of NH3 in 2.00 × 103 mL of solution.
(MMNH3 = 17.04 g/mol)
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Practice—Calculate the molarity of a solution made by
dissolving 34.0 g of NH3 in 2.00 × 103 mL of solution.
Given: 34.0
2.00
g NH
mol
2000
mL sol’n
L sol’n
3, NH
3, 2.00
Find: M M
Conceptual
Plan:
g NH3
mol NH3
mL sol’n
L sol’n
M
Relationships: M = mol/L, 1 mol NH3 = 17.04 g, 1 mL = 0.001 L
Solution:
Check:
The unit is correct; the magnitude is
reasonable.
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Example 12.4b What is the molality of a solution
prepared by mixing 17.2 g of C2H6O2 with 0.500 kg
of H2O to make 515 mL of solution?
Given: 0.2771
17.2 g C
mol
0.500
kg kg
H2H
O,2O,
515
515
mL
mLsol’n
sol’n
2H
2, 0.500
2HC
6O
2,6O
Find: m
Conceptual g C2H6O2
Plan:
mol C2H6O2
kg H2O
m
Relationships: m = mol/kg, 1 mol C2H6O2 = 62.07 g
Solution:
Check:
The unit is correct; the magnitude is
reasonable.
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Practice—Calculate the molality of a solution made by
dissolving 34.0 g of NH3 in 2.00 × 103 mL of water.
(MMNH3 = 17.04 g/mol, dH2O = 1.00 g/mL)
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Practice—Calculate the molality of a solution made by
dissolving 34.0 g of NH3 in 2.00 × 103 mL of water.
Given: 34.0
2.00 gmol
kgHH
NHNH
mL
3, 2.00
2O
3, 2000
2O
Find: m
Conceptual
Plan:
g NH3
mL H2O
mol NH3
g H2O
kg H2O
m
Relationships: m = mol/kg, 1 mol NH3 = 17.04 g, 1 kg = 1000 g, 1.00 g = 1 mL
Solution:
Check:
The unit is correct; the magnitude is
reasonable.
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Practice—Calculate the molality of a solution made by
dissolving 34.0 g of NH3 in 2.00 × 103 g of solution.
(MMNH3 = 17.04 g/mol)
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Practice—Calculate the molality of a solution made by
dissolving 34.0 g of NH3 in 2.00 × 103 g of solution.
Given:
Find:
Conceptual
Plan:
2.00
34.0 mol
g NHNH
g kg
solution
H 2O
3, 2000
3, 1.97
m
g NH3
g sol’n
mol NH3
g H2O
kg H2O
Relationships: m = mol/kg, 1 mol NH3 = 17.04 g, 1 kg = 1000 g
Solution:
Check:
The unit is correct; the magnitude is
reasonable.
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m
Example 12.4c What is the percent by mass of a
solution prepared by mixing 17.2 g of C2H6O2 with
0.500 kg of H2O to make 515 mL of solution?
Given: 17.2 g C2H6O2, 0.500 kg H2O, 515 mL sol’n
Find: %(m/m)
Conceptual
Plan:
g C2H6O2
g sol’n
g solvent
%
Relationships: 1 kg = 1000 g
Solution:
Check:
The unit is correct; the magnitude is
reasonable.
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Practice—Calculate the percent by mass of a solution made by
dissolving 34.0 g of NH3 in 2.00 × 103 mL of water.
(MMNH3 = 17.04 g/mol, dH2O = 1.00 g/mL)
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Practice—Calculate the percent by mass of a solution made by
dissolving 34.0 g of NH3 in 2.00 × 103 mL of water.
Given:
Find:
34.0 g NH3, 2000 g
mL
H2HO,
2O2034 g sol’n
%(m/m)
Conceptual
Plan:
g NH3
mL H2O
g H2O
g sol’n
Relationships: % = g/g × 100%, 1.00 g = 1 mL
Solution:
Check:
The unit is correct; the magnitude is
reasonable.
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%
Practice—Calculate the parts per million of a solution made by
dissolving 0.340 g of NH3 in 2.00 × 103 mL of water.
(MMNH3 = 17.04 g/mol, dH2O = 1.00 g/mL)
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Practice—Calculate the parts per million of a solution made by
dissolving 0.340 g of NH3 in 2.00 × 103 mL of water.
Given:
Find:
0.340 g NH3, 2000 g
mL
H2HO,
2O2000 g sol’n
ppm
Conceptual
Plan:
g NH3
mL H2O
g H2O
g sol’n
ppm
Relationships: ppm = g/g × 106, 1.00 g = 1 mL
Solution:
Check:
The unit is correct; the magnitude is
reasonable.
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Example 12.4d What is the mole fraction of a
solution prepared by mixing 17.2 g of C2H6O2 with
0.500 kg of H2O to make 515 mL of solution?
Given: 17.2 g C2H6O2, 0.500 kg H2O, 515 mL sol’n
Find: C
Conceptual g C2H6O2
Plan:
mol C2H6O2
g H2O
mol H2O
C
Relationships: C = molA/moltot, 1 mol C2H6O2 = 62.07 g, 1 mol H2O = 18.02 g
Solution:
Check:
The unit is correct; the magnitude is
reasonable.
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Practice—Calculate the mole fraction of a solution made
by dissolving 34.0 g of NH3 in 2.00 × 103 mL of water.
(MMNH3 = 17.04 g/mol, dH2O = 1.00 g/mL)
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Practice—Calculate the mole fraction of a solution made
by dissolving 34.0 g of NH3 in 2.00 × 103 mL of water.
Given:
Find:
Conceptual
Plan:
2.00
34.0 mol
g NHNH
mLmol
H2OH2O, 113.1 tot mol
3, 2000
3, 111.1
C
g NH3
mL H2O
mol NH3
g H2O
mol H2O
C
Relationships: C= mol/mol, 1 mol NH3 = 17.04 g, 1 mol H2O = 18.02 g, 1.00 g = 1 mL
Solution:
Check:
The unit is correct; the magnitude is
reasonable.
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84
Example 12.4d What is the mole percent of a
solution prepared by mixing 17.2 g of C2H6O2 with
0.500 kg of H2O to make 515 mL of solution?
Given: 17.2 g C2H6O2, 0.500 kg H2O, 515 mL sol’n
Find: C%
Conceptual g C2H6O2
Plan:
mol C2H6O2
g H2O
mol H2O
C%
Relationships: C = molA/moltot, 1 mol C2H6O2 = 62.07 g, 1 mol H2O = 18.02 g
Solution:
Check:
The unit is correct; the magnitude is
reasonable.
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85
Converting Concentration Units
1. Write the given concentration as a ratio.
2. Separate the numerator and denominator.
 Separate into the solute part and solution part.
3. Convert the solute part into the required
unit.
4. Convert the solution part into the required
unit.
5. Use the definitions to calculate the new
concentration units.
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Example 12.5 What is the molarity of 6.55% by
mass glucose (C6H12O6) solution?
Given: 6.55
g C6mol
H12O
sol’n L
6.55%(m/m)
C662,H100
O26,g0.09709
0.03636
12
6O
Find: M
Conceptual g C6H12O6
Plan:
g sol’n
Relationships:
mol C6H12O6
L sol’n
mL
M
M = mol/L, 1 mol C6H12O6 = 180.16 g, 1 mL = 0.001 L, 1 mL = 1.03 g
Solution:
Check:
The unit is correct; the magnitude is
reasonable.
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Practice—Calculate the molality of a 16.2 M H2SO4
solution.
(MMH2SO4 = 98.08 g/mol, dsol’n = 1.80 g/mL)
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88
Practice—Calculate the molality of a 16.2 M
H2SO4 solution.
Given: 16.2 mol
sol’n
M H2HSO
0.210Lkg
H2O
2SO
4 4,, 1.00
Find: m
Conceptual
Plan:
g H2SO4
L
mL
mol H2SO4
g H2 O
g sol’n
kg H2O
m
Relationships: m = mol/kg, 1 mol H2SO4 = 98.08 g, 1 kg = 1000 g, 1.80 g = 1 mL
Solution:
Check:
The unit is correct; the magnitude is reasonable.
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89
Colligative Properties
• Colligative properties are properties whose
value depends only on the number of solute
particles, and not on what they are.
The value of the property depends on the
concentration of the solution.
• The difference in the value of the property
between the solution and the pure substance
is generally related to the different attractive
forces and solute particles occupying solvent
molecules positions.
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90
Vapor Pressure of Solutions
• The vapor pressure of a solvent above a
solution is lower than the vapor pressure
of the pure solvent.
The solute particles replace some of the
solvent molecules at the surface.
Eventually, equilibrium is reAddition
ofpure
a nonvolatile
established,
but
with
a smaller
The
solvent
solute
the rate
of
number
ofreduces
vaporamolecules;
establishes
liquid
 vapor
vaporization,
decreasing
the
therefore,
theequilibrium.
vapor
pressure
amount
of vapor.
will
be lower.
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91
Thirsty Solutions
• A concentrated solution will draw solvent
molecules toward it due to the natural drive
for materials in nature to mix.
• Similarly, a concentrated solution will draw
pure solvent vapor into it due to this
tendency to mix.
• The result is reduction in vapor pressure.
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92
Thirsty Solutions
Beakers
with equalis
When equilibrium
liquid
levels of
established,
thepure
liquid
solvent
andsolution
a
level in the
solution
placed
in
beaker isare
higher
than
a
bell
jar. Solvent
the
solution
level in
molecules
evaporate
the pure solvent
from
each one
and fill
beaker—the
thirsty
the
bell jar,
solution
grabs and
establishing
holds solventan
vapor
equilibrium
with the
more effectively.
liquids in the
beakers.
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93
Raoult’s Law
• The vapor pressure of a volatile solvent
above a solution is equal to its mole fraction
of its normal vapor pressure, P°.
Psolvent in solution = csolvent∙P°
Since the mole fraction is always less than 1,
the vapor pressure of the solvent in solution will
always be less than the vapor pressure of the
pure solvent.
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94
Example 12.6 Calculate the vapor pressure of
water in a solution prepared by mixing 99.5 g of
C12H22O11 with 300.0 mL of H2O.
Given:
Find:
99.5 g C12H22O11, 300.0 mL H2O
PH2O
Conceptual g C12H22O11
mol C12H22O11
Plan:
mL H2O
g H2O
mol H2O
CH2O
PH2O
Relationships: P°H2O = 23.8 torr, 1 mol C12H22O11 = 342.30 g, 1 mol H2O = 18.02 g
Solution:
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95
Practice—Calculate the total vapor pressure of a
solution made by dissolving 25.0 g of glucose
(C6H12O6) in 215 g of water at 50 °C.
(MMC6H12O6 = 180.2 g/mol, MMH2O = 18.02 g/mol
Vapor Pressure of H2O @ 50 °C = 92.5 torr)
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96
Practice—Calculate the total vapor pressure of a
solution made by dissolving 25.0 g of glucose
(C6H12O6) in 215 g of water at 50 °C.
Given:
Find:
25.0 g C6H12O6, 215 g H2O
PH2O
Conceptual g C6H12O6
Plan:
mol C6H12O6
g H2O
mol H2O
CH2O
PH2O
Relationships: P°H2O = 92.5 torr, 1 mol C6H12O6 = 180.2 g, 1 mol H2O = 18.02 g
Solution:
Since glucose is nonvolatile, the total vapor pressure = vapor pressure H2O.
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97
Raoult’s Law for Volatile Solute
• When both the solvent and the solute can evaporate,
•
both molecules will be found in the vapor phase.
The total vapor pressure above the solution will be
the sum of the vapor pressures of the solute and
solvent.
 for an ideal solution
Ptotal = Psolute + Psolvent
• The solvent decreases the solute vapor pressure in
the same way the solute decreased the solvent’s.
Psolute = csolute∙P°solute and Psolvent = csolvent∙P°solvent
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98
Example 12.7 Calculate the component and total
vapor pressure in a solution prepared by mixing
3.95 g of CS2 with 2.43 g of C3H6O.
0.05187
molCS
CS2,2,PPº
===285
515torr,
torr,
0.04184
0.04184
CH36HO,
O,O,PC3H6O
PºC3H6O
= 148
= 332
torrtorr
3.95 gmol
CS
CS2
CS2
515
torr,
2.43 mol
gmol
CC33H
Given: 0.05187
2, Pº
CS2
6 6 Pº
C3H6O = 332 torr
PC3H6O
Ptotal
CS2, ,P
C3H6O, ,P
total
Find: PPPCS2
CS2, PC3H6O, Ptotal
Conceptual
Plan:
g CS2
mol CS2
g C3 H 6 O
mol C3H6O
Relationships: 1 mol CS2 = 76.15 g, 1 mol C3H6O = 58.0 g
Solution:
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99
C
P
Practice—Calculate the total vapor pressure of a
solution made by mixing 0.500 mol of ether (C4H10O)
with 0.250 mol of ethanol (C2H6O) at 20 °C.
(Vapor Pressure of ether @ 20 °C = 440 torr
Vapor Pressure of ethanol @ 20 °C = 44.6 torr)
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Practice—Calculate the total vapor pressure of a
solution made by mixing 0.500 mol of ether (C4H10O)
with 0.250 mol of ethanol (C2H6O) at 20 °C.
293torr,
torr,0.250
0.250mol
molCC
Pethanol
= 332
14.9torr
torr
0.500 mol
mol C
C44H
H10
O, P
PºC4H10O
2H
6O,Pº
C2H6O=
10O,
ether ==440
2H
6O,
Given: 0.500
Find: PPtotal
total
Conceptual
Plan:
mol C4H10O
C
mol C2H6O
Relationships:
Solution:
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101
P
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104
Freezing Point Depression
• The freezing point of a solution is lower than the
freezing point of the pure solvent.
 for a nonvolatile solute
 Therefore, the melting point of the solid solution is lower.
• The difference between the freezing point of the
solution and freezing point of the pure solvent is
directly proportional to the molal concentration of solute
particles.
(FPsolvent – FPsolution) = DTf = m∙Kf
• The proportionality constant is called the Freezing
Point Depression Constant, Kf.
 The value of Kf depends on the solvent.
 The units of Kf are °C/m.
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105
Kf
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106
Example 12.8 What is the freezing point of a 1.7 m
aqueous ethylene glycol solution, C2H6O2?
Given: 1.7 m C2H6O2(aq)
Find: Tf, °C
Conceptual
Plan:
DTf
m
Relationships: DTf = m∙Kf, Kf for H2O = 1.86 °C/m, FPH2O = 0.00 °C
Solution:
Check:
The unit is correct; the freezing point lower
than the normal freezing point makes sense.
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107
Practice—Calculate the molar mass of a
compound if a solution of 12.0 g dissolved in
80.0 g of water freezes at −1.94 °C.
(Kf water = 1.86 °C/m)
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108
Practice—Calculate the molar mass of a compound if a
solution of 12.0 g dissolved in 80.0 g of water freezes
at −1.94 °C.
Given: masssolute = 12.0 g, massH2O = 80.0 g, FPsol’n = −1.94 °C
Find: Tf, °C
Conceptual FP
DTf
sol’n
Plan: DT = FP − FP
f
H2O
sol’n
Relationships:
m
mol
MM
m  kgH2O = mol MM = g/mol
DTf = m Kf, m = mol/kg, MM = g/mol, Kf = 1.86 °C/m, FPH2O = 0.00 °C
Solution:
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109
Boiling Point Elevation
• The boiling point of a solution is higher than the
boiling point of the pure solvent.
 for a nonvolatile solute
• The difference between the boiling point of the
•
solution and boiling point of the pure solvent is
directly proportional to the molal concentration of
solute particles.
(BPsolution – BPsolvent) = DTb = m∙Kb
The proportionality constant is called the Boiling
Point Elevation Constant, Kb.
 The value of Kb depends on the solvent.
 The units of Kb are °C/m.
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110
Example 12.9 How many grams of ethylene glycol,
C2H6O2, must be added to 1.0 kg H2O to give a
solution that boils at 105 °C?
Given: 1.0 kg H2O, Tb = 105 °C
Find: mass C2H6O2, g
DTb
Conceptual
Plan:
kg H2O
m
mol C2H6O2
g C2H6O2
Relationships: DTb = m∙Kb, Kb H2O = 0.512 °C/m, BPH2O = 100.0 °C
MMC2H6O2 = 62.07 g/mol, 1 kg = 1000 g
Solution:
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111
Practice—Calculate the boiling point of a
solution made by dissolving 1.00 g of
glycerin, C3H8O3, in 54.0 g of water.
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112
Practice—Calculate the boiling point of a solution
made by dissolving 1.00 g of glycerin, C3H8O3, in
54.0 g of water.
Given: 1.00 g C3H8O3, 54.0 g H2O
Find: Tb, sol’n, °C
Conceptual g C H O , kg H O
3 8 3
2
Plan:
Relationships:
m
DTb
m = mol/kg, DTb = m∙Kb, Kb for H2O = 0.512 °C/m,
BPH2O = 100.0 °C, 1 mol C3H8O3 = 92.09 g
Solution:
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113
Osmosis
• Osmosis is the flow of solvent from a
solution of low concentration into a
solution of high concentration.
• The solutions may be separated by a
semipermeable membrane.
• A semipermeable membrane allows
solvent to flow through it, but not solute.
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114
Osmotic Pressure
• The amount of pressure needed to keep
osmotic flow from taking place is called the
osmotic pressure.
• The osmotic pressure, P, is directly
proportional to the molarity of the solute
particles.
R = 0.08206 (atm∙L)/(mol∙K)
P = MRT
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115
Tro, Principles of Chemistry: A Molecular Approach
116
Example 12.10 What is the molar mass of a protein
if 5.87 mg per 10 mL gives an osmotic pressure of
2.45 torr at 25 °C?
Given: 5.87 mg/10 mL, P = 2.45 torr, T = 25 °C
Find: molar mass, g/mol
P,T
Conceptual
Plan:
mL
M
L
mol protein
Relationships: P = MRT, T(K) = T(°C) + 273.15, R = 0.08206 atm∙L/mol∙K
M = mol/L, 1 mL = 0.001 L, MM = g/mol, 1 atm = 760 torr
Solution:
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117
Practice—Lysozyme is an enzyme used to cleave
cell walls. A solution made by dissolving 0.0750 g
of lysozyme in 100.0 mL results in an osmotic
pressure of 1.32 × 10−3 atm at 25 °C. Calculate the
molar mass of lysozyme.
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118
Practice—What is the molar mass of lysozyme if
0.0750 g per 100.0 mL gives an osmotic pressure of
1.32 × 10−3 atm at 25 °C?
Given: 0.0750 g/100 mL, P = 1.32 × 10−3 atm, T = 25 °C
Find: molar mass, g/mol
P,T
Conceptual
Plan:
mL
M
L
mol protein
Relationships: P = MRT, T(K) = T(°C) + 273.15, R = 0.08206 atm∙L/mol∙K
M = mol/L, 1 mL = 0.001 L, MM = g/mol, 1 atm = 760 torr
Solution:
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119
van’t Hoff Factors
• Ionic compounds produce multiple solute
particles for each formula unit.
• The van’t Hoff factor, i, is the ratio of
moles of solute particles to moles of formula
units dissolved.
• Measured van’t Hoff factors are often lower
than you might expect due to ion pairing in
solution.
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120
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121
Example 12.11 What is the van’t Hoff factor if 0.050 m
CaCl2(aq) has a freezing point of −0.27 ºC?
Given: 0.050 m CaCl2(aq), Tf = −0.27 °C
Find: i
Conceptual
m, DTf
i
Plan:
Relationships: DTf = i∙m∙Kf, Kf for H2O = 1.86 °C/m, FPH2O = 0.00 °C
Solution:
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122