Transcript Chapter 12 - Highline Community College

Chemistry: A Molecular Approach, 2nd Ed.
Nivaldo Tro
Chapter 12
Solutions
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright  2011 Pearson Education, Inc.
Thirsty Seawater
• Drinking seawater can cause you to dehydrate
• Seawater is a homogeneous mixture of salts with
water
• Seawater contains high concentrations of salts
higher than the salt content of your cells
• As seawater passes through your body, it pulls
water out of your cells; due mainly to nature’s
tendency toward spontaneous mixing
• This reduces your cells’ water level and usually
results in diarrhea as this extra liquid flows out
with the seawater
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Seawater
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Seawater
• Drinking seawater will dehydrate you and give
•
you diarrhea
The cell wall acts as a barrier to solute moving
so 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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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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Examples of Solutions
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Common Types of Solution
Solute
Phase
Gas
Solvent
Phase
Example
Gas
Air (mostly N2 & O2)
Liquid solutions
Gas
Liquid
Solid
Liquid
Liquid
Liquid
Soda (CO2 in H2O)
Vodka (C2H5OH in H2O)
Seawater (NaCl in H2O)
Solid solutions
Solid
Solid
Brass (Zn in Cu)
Solution Phase
Gaseous solutions
• 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
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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
toward mixing, and the types of
intermolecular attractive forces
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Spontaneous Mixing
When solutions with different solute concentrations
come in contact, they spontaneously mix to result in a
uniform distribution of solute throughout the solution
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Mixing and the Solution Process:
Entropy
• Most processes occur because the
end result has less potential energy
• But 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
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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Intermolecular Attractions
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Solution Interactions
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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 increase in entropy from mixing
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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 are attracted
to each other
• Polar molecules and ionic compounds will be
•
more soluble in polar solvents
Nonpolar molecules will be more soluble in
nonpolar solvents
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Classifying Solvents
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Example 12.1a: Predict whether the following
vitamin is soluble in fat or water
Water is a polar solvent.
Fat is mostly made of
nonpolar molecules.
The four OH groups
make the molecule
highly polar and it will
also H-bond to
water.
Vitamin C is water
soluble.
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Example 12.1b: Predict whether the following
vitamin is soluble in fat or water
Water is a polar solvent.
Fat is mostly made of
nonpolar molecules.
The two C=O groups are
polar, but their geometric
symmetry suggests their
pulls will cancel and the
molecule will be
nonpolar.
Vitamin K3
Vitamin K3 is fat soluble.
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Practice – Decide if the following are more
soluble in hexane, C6H14, or water
nonpolar molecule
more soluble in C6H14
polar molecule
more soluble in H2O
nonpolar part dominant
more soluble in C6H14
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Practice – Explain the solubility trends
seen in the table below
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Practice – Explain the solubility trends
seen in the table below
These alcohols all have a polar
OH part and a nonpolar CHn part.
As we go down the table the
nonpolar part gets larger, but the
amount of OH stays the same.
We therefore expect that the
solubility in water (polar solvent)
should decrease and the
solubility in hexane (nonpolar
solvent) should increase, and it
does.
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Heat of Solution
• When some compounds, such as NaOH,
dissolve in water, a lot of heat is released
the container gets hot
• When other compounds, such as NH4NO3,
dissolve in water, heat is absorbed from the
surroundings
the container gets cold
• Why is this?
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Energetics of Solution Formation: the
Enthalpy of Solution
• To make a solution you must
1. overcome all attractions between the solute
particles; therefore DHsolute is endothermic
2. overcome some attractions between solvent
molecules; therefore DHsolvent is endothermic
3. form new attractions between solute particles and
solvent molecules; therefore DHmix is exothermic
• The overall DH for making a solution depends on the
relative sizes of the DH for these three 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
attractions,
releasingattractions
energy
2. Add
energy
in solute–solvent
to overcome some
solvent–solvent
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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
released
in
than
the energy
released
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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Ion-Dipole Interactions
• When ions dissolve in water they become
hydrated
each ion is surrounded by water molecules
• The formation of these ion-dipole attractions
causes the heat of hydration to
be very exothermic
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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+lattice
−DH
DHenergy
hydration
lattice energy
hydration
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Heat of Hydration
DHsolution = DHhydration− DHlattice energy
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Comparing Heat of Solution to
Heat of Hydration
• Because 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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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
Solve:
Check: the unit is correct, 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 after some solute is dissolved, 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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Solution Equilibrium
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Solubility Limit
• A solution that has the solute and solvent in
dynamic equilibrium is said to be saturated
 if you add more solute it will not dissolve
 the saturation concentration depends on the
temperature
 and pressure of gases
• A solution that has less solute than saturation is
said to be unsaturated
 more solute will dissolve at this temperature
• 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 a Crystal of NaC2H3O2 to a
Supersaturated Solution
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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
removing impurities from a solid
compound
• 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
then 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
• Gases generally have lower solubility in water
than ionic or polar covalent solids because
most are nonpolar molecules
gases with high solubility usually are actually
reacting with water
• 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 the Henry’s
Law Constant
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Relationship between Partial Pressure
and Solubility of a Gas
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persrst
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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: S = [CO2] = 0.12 M,
Find: P of CO2, atm
Conceptual
[CO2]
Plan:
P
Relationships: S = k P, k = 3.4 x 10−2 M/atm
H
H
Solve:
Check: the unit is correct, 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 = 760torr, 1 mol =17.04 g
Solve:
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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 moles
of CaCl2 in each liter of solution
1 L = 1.0 moles 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 moles Ca2+, 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, kilogram, 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
Solve:
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
Solve:
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 x 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: 0.2771
17.2 g Cmol
O22H
, 0.500
kg H2kg
O,H515
sol’n
0.515
L
2H6C
6O2, 0.500
2O, mL
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
Solve:
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 x 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 x 103 mL of solution
Given: 2.00
34.0 mol
g NH
NH
mL
L sol’n
sol’n
3, 2000
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
Solve:
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: 17.2 g C2H6O2, 0.500 kg H2O, 515 mL sol’n
Find: m
Conceptual g C2H6O2
Plan:
mol C2H6O2
kg H2O
m
Relationships: m = mol/kg, 1 mol C2H6O2 = 62.07 g
Solve:
Check: the unit is correct, the magnitude is
reasonable
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – Calculate the molality of a solution made by
dissolving 34.0 g of NH3 in 2.00 x 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 x 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 molNH3=17.04 g, 1kg=1000 g, 1.00g=1 mL
Solve:
Check: the unit is correct, the magnitude is
reasonable
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – Calculate the molality of a solution made by
dissolving 34.0 g of NH3 in 2.00 x 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 x 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
m
Relationships: m=mol/kg, 1 molNH3=17.04 g, 1kg=1000 g
Solve:
Check: the unit is correct, the magnitude is
reasonable
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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
Solve:
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 x 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 x 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 sol’n
g H2O
%
Relationships: % = g/g x 100%, 1.00 g=1 mL
Solve:
Check: the unit is correct, the magnitude is
reasonable
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – Calculate the parts per million of a solution made by
dissolving 0.340 g of NH3 in 2.00 x 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 x 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 sol’n
g H2O
ppm
Relationships: ppm = g/g x 106, 1.00 g=1 mL
Solve:
Check: the unit is correct, the magnitude is
reasonable
Tro: Chemistry: A Molecular Approach, 2/e
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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
Solve:
Check: the unit is correct, the magnitude is
reasonable
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – Calculate the mole fraction of a solution made
by dissolving 34.0 g of NH3 in 2.00 x 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 x 103 mL of water
Given: 2.00
34.0 mol
g NHNH
mLmol
H2O
H2O, 113.1 tot mol
3, 2000
3, 111.1
Find: cc
Conceptual
Plan:
g NH3
mL H2O
Relationships:
mol NH3
g H2O
mol H2O
c
c=mol/mol, 1 mol NH3=17.04 g, 1mol H2O =18.02 g, 1.00 g =1 mL
Solve:
Check: the unit is correct, the magnitude is
reasonable
Tro: Chemistry: A Molecular Approach, 2/e
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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 = 62.07g, 1 mol H2O=18.02 g
Solve:
Check: the unit is correct, the magnitude is
reasonable
Tro: Chemistry: A Molecular Approach, 2/e
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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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Tro: Chemistry: A Molecular Approach, 2/e
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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
mol C6H12O6
mL
L sol’n
M
Relationships: M =mol/L, 1mol C6H12O6=180.16g, 1mL=0.001L, 1mL=1.03g
Solve:
Check: the unit is correct, the magnitude is
reasonable
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – Calculate the molality of 16.2 M H2SO4(aq)
(MMH2SO4 = 98.08 g/mol, dsol’n = 1.80 g/mL)
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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
g sol’n
mol H2SO4
g H2O
kg H2O
m
Relationships: m=mol/kg, 1molH2SO4=98.08g, 1kg=1000g, 1.80g=1mL
Solve:
Check: the unit is correct, the magnitude is reasonable
Tro: Chemistry: A Molecular Approach, 2/e
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Colligative Properties
• Colligative properties are properties whose
value depends only on the number of solute
particles, and not on what they are
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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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,
is reAdditionequilibrium
of a nonvolatile
established,
butpure
withthe
a smaller
solvent
solute The
reduces
rate of
number
of vapordecreasing
–
establishes
amolecules
liquid  vapor
vaporization,
the
therefore the
vapor
equilibrium
amount
ofpressure
vapor will
be lower
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Thirsty Solutions Revisited
• 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
Tro: Chemistry: A Molecular Approach, 2/e
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Thirsty Solutions
Beakers
with equal
When
equilibrium
is
liquid levelsthe
of liquid
pure
established,
solvent
and
a
level
in the
solution
solution
are placed
beaker
is higher
than in
a bell
jar. Solvent
the
solution
level in
molecules
evaporate
the
pure solvent
from each
and fill
beaker
– theone
thirsty
the bellgrabs
jar, and
solution
establishing
an
holds
solvent vapor
equilibrium
with the
more
effectively
liquids in the
beakers.
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Raoult’s Law
• The vapor pressure of a volatile solvent above
a solution is equal to its normal vapor pressure,
P°, multiplied by its mole fraction in the solution
Psolvent in solution = csolvent∙P°
because 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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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 H2 O
mol H2O
cH2O
PH2O
Relationships: P°H2O = 23.8 torr, 1mol C12H22O11 = 342.30g, 1mol H2O = 18.02g
Solve:
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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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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 H2 O
mol H2O
CH2O
PH2O
Relationships: P°H2O = 92.5torr, 1mol C6H12O6 = 180.2g, 1mol H2O = 18.02g
Solve:
because glucose is nonvolatile, the total vapor pressure = vapor pressure H2O
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Vapor Pressure Lowering
• The vapor pressure of a solvent in a solution is
•
•
always lower than the vapor pressure of the
pure solvent
The vapor pressure of the solution is directly
proportional to the amount of the solvent in the
solution
The difference between the vapor pressure of
the pure solvent and the vapor pressure of the
solvent in solution is called the vapor pressure
lowering
DP = Pºsolvent − Psolution = csolute  Pºsolvent
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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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Example 12.7: Calculate the component and total
vapor pressure of a solution prepared by mixing
3.95 g of CS2 with 2.43 g of C3H6O
Given:
Find:
Conceptual
Plan:
0.05187molCS
P
=148torr
PºCS2
=515torr,
0.04184molC
H66O,
O,
Pº
3.95 g CS2, Pº
==285torr,
515 torr,0.04184molC
2.43 g C3H633H
O,
PºPC3H6O
==332torr
332 torr
2,CS2
C3H6O
CS2
C3H6O
P
PCS2,,PPC3H6O, P, total
P
CS2
C3H6O
total
g CS2
mol CS2
g C3H6O
mol C3H6O
c
P
Relationships: 1mol CS2 = 76.15 g, 1mol C3H6O = 58.0 g
Solve:
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – Calculate the total vapor pressure of an ideal
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 an ideal
solution made by mixing 0.500 mol of ether (C4H10O)
with 0.250 mol of ethanol (C2H6O) at 20°C
Given:
Find:
Conceptual
Plan:
0.500 mol C4H10O, Pº
PC4H10O
=293torr,
torr,0.250
0.250mol
molCC
PºPethanol
=14.9
44.6 torr
torr
ether=440
2H
2H
6O,
6O,
C2H6O=
PPtotal
total
mol C4H10O
mol C2H6O
c
P
Relationships:
Solve:
Tro: Chemistry: A Molecular Approach, 2/e
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Ideal vs. Nonideal Solution
• In ideal solutions, the made solute–solvent
interactions are equal to the sum of the broken
solute–solute and solvent–solvent interactions
ideal solutions follow Raoult’s Law
• Effectively, the solute is diluting the solvent
• If the solute–solvent interactions are stronger
or weaker than the broken interactions the
solution is nonideal
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Vapor Pressure of a
Nonideal Solution
• When the solute–solvent interactions are stronger
than the solute–solute  solvent–solvent, the total
vapor pressure of the solution will be less than
predicted by Raoult’s Law
 because the vapor pressures of the solute and solvent
are lower than ideal
• When the solute–solvent interactions are weaker
than the solute–solute  solvent–solvent, the total
vapor pressure of the solution will be more than
predicted by Raoult’s Law
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Deviations from Raoult’s Law
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Example 12.7, cont’d: The experimentally measured total
vapor pressure of the solution is 645 torr. Is the solution
ideal? If not, what can you say about the relative
strength of carbon disulfide–acetone interactions?
Given:
Find:
Ptotal(expt) = 645 torr, Ptotal(ideal) = 443 torr
is the solution ideal?, interaction strength if not ideal
Solve:
Ptotal(expt) = 645 torr > Ptotal(ideal) = 443 torr
The solution is not ideal and shows positive deviations from
Raoult’s law. Therefore, carbon disulfide–acetone
interactions must be weaker than acetone–acetone and
carbon disulfide–carbon disulfide interactions.
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Other Colligative Properties Related to
Vapor Pressure Lowering
• Vapor pressure lowering occurs at all
•
•
temperatures
This results in the temperature required to boil
the solution being higher than the boiling point
of the pure solvent
This also results in the temperature required to
freeze the solution being lower than the
freezing point of the pure solvent
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Freezing Salt Water
• Pure water freezes at 0 ºC. At this temperature,
•
•
•
ice and liquid water are in dynamic equilibrium.
Adding salt disrupts the equilibrium. The salt
particles dissolve in the water, but do not attach
easily to the solid ice.
When an aqueous solution containing a dissolved
solid solute freezes slowly, the ice that forms does
not normally contain much of the solute.
To return the system to equilibrium, the
temperature must be lowered sufficiently to make
the water molecules slow down enough so that
more can attach themselves to the ice.
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Freezing Point Depression
• The freezing point of a solution is lower than the
freezing point of the pure solvent
 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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Kf
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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
Solve:
Check: the unit is correct, the freezing point lower
than the normal freezing point makes sense
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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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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
Plan:
Relationships:
FPsol’n
DTf
DTf =FPH2O−FPsol’n
m
mol
MM
m  kgH2O= mol MM=g/mol
DTf=mKf, m =mol/kg, MM=g/mol, Kf=1.86 °C/m, FPH2O=0.00 °C
Solve:
Tro: Chemistry: A Molecular Approach, 2/e
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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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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
Solve:
Tro: Chemistry: A Molecular Approach, 2/e
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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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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
Solve:
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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 semi-permeable membrane allows
solvent to flow through it, but not solute
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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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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= 0.08206atm∙L/mol∙K
M = mol/L, 1 mL = 0.001 L, MM=g/mol, 1atm =760 torr
Solve:
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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 x 10−3 atm at 25 °C. Calculate the
molar mass of lysozyme.
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Practice – What is the molar mass of lysozyme if
0.0750 g per 100.0 mL gives an osmotic pressure of
1.32x 10−3 atm at 25 °C?
Given: 0.0750 g/100 mL, P = 1.32 x 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= 0.08206atm∙L/mol∙K
M = mol/L, 1 mL = 0.001 L, MM=g/mol, 1atm =760 torr
Solve:
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van’t Hoff Factors
• Ionic compounds produce multiple solute
particles for each formula unit
• The theoretical van’t Hoff factor, i, is the ratio
of moles of solute particles to moles of formula
units dissolved
• The measured van’t Hoff factors are generally
less than the theoretical due to ion pairing in
solution
therefore the measured van’t Hoff factor often
causes the DT to be lower than you might expect
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Example 12.11: What is the measured van’t Hoff factor
if 0.050 m CaCl2(aq) has 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
Solve:
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Example 12.12: A solution contains 0.102 mol
Ca(NO3)2 and 0.927 mol H2O. Calculate the vapor
pressure of the solution at 55 ºC
Given: 0.102 mol Ca(NO3)2(aq), 0.927 mol H2O, T = 55 °C,
P°H2O = 118.1 torr (at 55 °C)
Find: Psolution
Conceptual
c, PH2O
Plan:
Relationships: Psolution = i∙cH2O∙PºH2O
Psolution
Solve:
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Practice – Calculate the theoretical boiling
point of a solution made by dissolving 10.0 g
of NaCl (MM 58.44) in 54.0 g of water
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Practice – Calculate the theoretical boiling point of a solution
made by dissolving 10.0 g of NaCl in 54.0 g of H2O
Given: 10.0 g NaCl, 54.0 g H2O
Find: Tb, sol’n, °C
Conceptual
Plan:
Relationships:
g NaCl, kg H2O
m
DTb
m = mol/kg, DTb = imKb Kb for H2O = 0.512 °C/m,
BPH2O = 100.0°C, 1 mol C3H8O3 = 92.09 g
Solve:
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A An
hyperosmotic
hyposmotic
isosmoticsolution
solution
solution
has
has
has
athe
lower
a higher
same
osmotic
osmotic
pressure
pressure
than
as the
the solution
solution inside
inside the
the cell
cell –– as
as aa result
result there
there is
is no
a
net
netflow
flowofofwater
waterinto
out
into
of
the
or
the
cell,
out
cell,
of
causing
causing
the cellit to
it to
swell
shrivel
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Mixtures
• Solutions = homogeneous
• Suspensions = heterogeneous, separate on
•
standing
Colloids = heterogeneous, do not separate on
standing
particles can coagulate
cannot pass through semi-permeable membrane
hydrophilic
stabilized by attraction for solvent (water)
hydrophobic
stabilized by charged surface repulsions
• Show the Tyndall Effect and Brownian motion
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Mixtures
Solution
Colloid
Suspenion
Particle size
< 1 nm
1-1000 nm
> 1000 nm
Gravity effect
none
none
settle out
Centrifuge effect
none
settle out
settle out
Ordinary filters
pass
through
pass through
retained
Ultrafilters
pass
through
retained
retained
Colligative properties
large
small
none
Visibility of particle w/ lite
microscope
invisible
barely visible
visible
Visibility of particle w/ elec.
mcrscp.
invisible
visible
visible
Light scattering
none
Tyndall Effect
scatter all light
Brownian movement
none visible
visible
none
Electrical charge
maybe
many
negligible
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Brownian Motion
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Types of Colloidal Suspensions
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Soaps
• Triglycerides can be broken down into fatty acid salts
and glycerol by treatment with a strong hydroxide
solution
• Fatty acid salts have a very polar “head” because it is
ionic and a very nonpolar “tail” because it is all C and
H
 hydrophilic head and hydrophobic tail
• This unique structure allows the fatty acid salts,
called soaps, to help oily substances be attracted to
water
 micelle formation
 emulsification
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Soap
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Soap
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