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Chapter 11
Properties of Solutions
Section 11.1
Solution Composition
Copyright © Cengage Learning. All rights reserved 2
Various Types of Solutions
Section 11.1
Solution Composition
Copyright © Cengage Learning. All rights reserved 3
Solution Composition
A
A
moles of solute
Molarity ( ) =
liters of solution
mass of solute
Mass (weight) percent = 100%
mass of solution
moles
Mole fraction ( ) =
total moles of solution
moles of solute
Molality ( ) =
kilogram of s


M
m
olvent
Section 11.1
Solution Composition
Copyright © Cengage Learning. All rights reserved 4
Molarity
moles of solute
Molarity ( ) =
liters of solution
M
Section 11.1
Solution Composition
Copyright © Cengage Learning. All rights reserved 5
You have 1.00 mol of sugar in 125.0 mL of
solution. Calculate the concentration in units of
molarity.
8.00 M
EXERCISE!
Section 11.1
Solution Composition
Copyright © Cengage Learning. All rights reserved 6
You have a 10.0 M sugar solution. What volume
of this solution do you need to have 2.00 mol of
sugar?
0.200 L
EXERCISE!
Section 11.1
Solution Composition
Copyright © Cengage Learning. All rights reserved 7
Consider separate solutions of NaOH and KCl
made by dissolving 100.0 g of each solute in 250.0
mL of solution. Calculate the concentration of
each solution in units of molarity.
10.0 M NaOH
5.37 M KCl
EXERCISE!
Section 11.1
Solution Composition
Copyright © Cengage Learning. All rights reserved 8
Mass Percent
mass of solute
Mass (weight)percent = 100%
mass of solution

Section 11.1
Solution Composition
Copyright © Cengage Learning. All rights reserved 9
What is the percent-by-mass concentration of
glucose in a solution made my dissolving 5.5 g of
glucose in 78.2 g of water?
6.6%
EXERCISE!
Section 11.1
Solution Composition
Copyright © Cengage Learning. All rights reserved 10
Mole Fraction
A
A
moles
Mole fraction ( ) =
total moles of solution

Section 11.1
Solution Composition
Copyright © Cengage Learning. All rights reserved 11
A solution of phosphoric acid was made by
dissolving 8.00 g of H3PO4 in 100.0 mL of water.
Calculate the mole fraction of H3PO4. (Assume
water has a density of 1.00 g/mL.)
0.0145
EXERCISE!
Section 11.1
Solution Composition
Copyright © Cengage Learning. All rights reserved 12
Molality
moles of solute
Molality ( ) =
kilogram of solvent
m
Section 11.1
Solution Composition
Copyright © Cengage Learning. All rights reserved 13
A solution of phosphoric acid was made by
dissolving 8.00 g of H3PO4 in 100.0 mL of water.
Calculate the molality of the solution. (Assume
water has a density of 1.00 g/mL.)
0.816 m
EXERCISE!
Section 11.2
The Energies of Solution Formation
Formation of a Liquid Solution
1. Separating the solute into its individual components
(expanding the solute).
2. Overcoming intermolecular forces in the solvent to
make room for the solute (expanding the solvent).
3. Allowing the solute and solvent to interact to form the
solution.
Copyright © Cengage Learning. All rights reserved 14
Section 11.2
The Energies of Solution Formation
Steps in the Dissolving Process
Copyright © Cengage Learning. All rights reserved 15
Section 11.2
The Energies of Solution Formation
Steps in the Dissolving Process
 Steps 1 and 2 require energy, since forces must be
overcome to expand the solute and solvent.
 Step 3 usually releases energy.
 Steps 1 and 2 are endothermic, and step 3 is often
exothermic.
Copyright © Cengage Learning. All rights reserved 16
Section 11.2
The Energies of Solution Formation
Enthalpy (Heat) of Solution
 Enthalpy change associated with the formation of the
solution is the sum of the ΔH values for the steps:
ΔHsoln = ΔH1 + ΔH2 + ΔH3
 ΔHsoln may have a positive sign (energy absorbed) or a
negative sign (energy released).
Copyright © Cengage Learning. All rights reserved 17
Section 11.2
The Energies of Solution Formation
Enthalpy (Heat) of Solution
Copyright © Cengage Learning. All rights reserved 18
Section 11.2
The Energies of Solution Formation
Explain why water and oil (a long chain hydrocarbon)
do not mix. In your explanation, be sure to address
how ΔH plays a role.
Copyright © Cengage Learning. All rights reserved 19
CONCEPT CHECK!
Section 11.2
The Energies of Solution Formation
The Energy Terms for Various Types of Solutes and Solvents
Copyright © Cengage Learning. All rights reserved 20
ΔH1 ΔH2 ΔH3 ΔHsoln Outcome
Polar solute, polar solvent Large Large Large, negative Small Solution forms
Nonpolar solute, polar solvent Small Large Small Large, positive No solution
forms
Nonpolar solute, nonpolar
solvent
Small Small Small Small Solution forms
Polar solute, nonpolar solvent Large Small Small Large, positive No solution
forms
Section 11.2
The Energies of Solution Formation
In General
 One factor that favors a process is an increase in
probability of the state when the solute and solvent are
mixed.
 Processes that require large amounts of energy tend not
to occur.
 Overall, remember that “like dissolves like”.
Copyright © Cengage Learning. All rights reserved 21
Section 11.3
Factors Affecting Solubility
 Structure Effects:
 Polarity
 Pressure Effects:
 Henry’s law
 Temperature Effects:
 Affecting aqueous solutions
Copyright © Cengage Learning. All rights reserved 22
Section 11.3
Factors Affecting Solubility
Structure Effects
 Hydrophobic (water fearing)
 Non-polar substances
 Hydrophilic (water loving)
 Polar substances
Copyright © Cengage Learning. All rights reserved 23
Section 11.3
Factors Affecting Solubility
Pressure Effects
 Little effect on solubility of solids or liquids
 Henry’s law: C = kP
C = concentration of dissolved gas
k = constant
P = partial pressure of gas solute
above the solution
 Amount of gas dissolved in a solution is directly proportional to
the pressure of the gas above the solution.
Copyright © Cengage Learning. All rights reserved 24
Section 11.3
Factors Affecting Solubility
A Gaseous Solute
Copyright © Cengage Learning. All rights reserved 25
Section 11.3
Factors Affecting Solubility
Temperature Effects (for Aqueous Solutions)
 Although the solubility of most solids in water increases
with temperature, the solubilities of some substances
decrease with increasing temperature.
 Predicting temperature dependence of solubility is very
difficult.
 Solubility of a gas in solvent typically decreases with
increasing temperature.
Copyright © Cengage Learning. All rights reserved 26
Section 11.3
Factors Affecting Solubility
The Solubilities
of Several Solids
as a Function of
Temperature
Copyright © Cengage Learning. All rights reserved 27
Section 11.3
Factors Affecting Solubility
The Solubilities of
Several Gases in Water
Copyright © Cengage Learning. All rights reserved 28
Section 11.4
The Vapor Pressures of Solutions
An Aqueous Solution and Pure Water in a Closed Environment
Copyright © Cengage Learning. All rights reserved 29
Section 11.4
The Vapor Pressures of Solutions
Liquid/Vapor Equilibrium
Copyright © Cengage Learning. All rights reserved 30
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Section 11.4
The Vapor Pressures of Solutions
Vapor Pressure Lowering: Addition of a Solute
Copyright © Cengage Learning. All rights reserved 31
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Section 11.4
The Vapor Pressures of Solutions
Vapor Pressures of Solutions
 Nonvolatile solute lowers the vapor pressure of a
solvent.
 Raoult’s Law:
Psoln = observed vapor pressure of solution
solv = mole fraction of solvent
= vapor pressure of pure solvent
Copyright © Cengage Learning. All rights reserved 32

soln solv solv
= 
P P
solv
P
Section 11.4
The Vapor Pressures of Solutions
A Solution Obeying Raoult’s Law
Copyright © Cengage Learning. All rights reserved 33
Section 11.4
The Vapor Pressures of Solutions
Nonideal Solutions
 Liquid-liquid solutions where both components are
volatile.
 Modified Raoult’s Law:
 Nonideal solutions behave ideally as the mole fractions
approach 0 and 1.
Copyright © Cengage Learning. All rights reserved 34
Total A A B B
= +
 
P P P
Section 11.4
The Vapor Pressures of Solutions
Vapor Pressure for a Solution of Two Volatile Liquids
Copyright © Cengage Learning. All rights reserved 35
Section 11.4
The Vapor Pressures of Solutions
Summary of the Behavior of Various Types of Solutions
Copyright © Cengage Learning. All rights reserved 36
Interactive Forces Between
Solute (A) and Solvent (B)
Particles
ΔHsoln
ΔT for Solution
Formation
Deviation
from
Raoult’s
Law
Example
A  A, B  B  A  B Zero Zero
None (ideal
solution)
Benzene-
toluene
A  A, B  B < A  B
Negative
(exothermic)
Positive Negative
Acetone-
water
A  A, B  B > A  B
Positive
(endothermic)
Negative Positive
Ethanol-
hexane
Section 11.4
The Vapor Pressures of Solutions
For each of the following solutions, would you expect it
to be relatively ideal (with respect to Raoult’s Law),
show a positive deviation, or show a negative deviation?
a) Hexane (C6H14) and chloroform (CHCl3)
b) Ethyl alcohol (C2H5OH) and water
c) Hexane (C6H14) and octane (C8H18)
Copyright © Cengage Learning. All rights reserved 37
CONCEPT CHECK!
Section 11.5
Boiling-Point Elevation and Freezing-Point
Depression
Colligative Properties
 Depend only on the number, not on the identity, of the
solute particles in an ideal solution:
 Boiling-point elevation
 Freezing-point depression
 Osmotic pressure
Copyright © Cengage Learning. All rights reserved 38
Section 11.5
Boiling-Point Elevation and Freezing-Point
Depression
Boiling-Point Elevation
 Nonvolatile solute elevates the boiling point of the
solvent.
 ΔT = Kbmsolute
ΔT = boiling-point elevation
Kb = molal boiling-point elevation constant
msolute = molality of solute
Copyright © Cengage Learning. All rights reserved 39
Section 11.5
Boiling-Point Elevation and Freezing-Point
Depression
Boiling Point Elevation: Liquid/Vapor Equilibrium
Copyright © Cengage Learning. All rights reserved 40
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Section 11.5
Boiling-Point Elevation and Freezing-Point
Depression
Boiling Point Elevation: Addition of a Solute
Copyright © Cengage Learning. All rights reserved 41
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Section 11.5
Boiling-Point Elevation and Freezing-Point
Depression
Boiling Point Elevation: Solution/Vapor Equilibrium
Copyright © Cengage Learning. All rights reserved 42
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Section 11.5
Boiling-Point Elevation and Freezing-Point
Depression
Freezing-Point Depression
 When a solute is dissolved in a solvent, the freezing
point of the solution is lower than that of the pure
solvent.
 ΔT = Kfmsolute
ΔT = freezing-point depression
Kf = molal freezing-point depression constant
msolute = molality of solute
Copyright © Cengage Learning. All rights reserved 43
Section 11.5
Boiling-Point Elevation and Freezing-Point
Depression
Freezing Point Depression: Solid/Liquid Equilibrium
Copyright © Cengage Learning. All rights reserved 44
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Section 11.5
Boiling-Point Elevation and Freezing-Point
Depression
Freezing Point Depression: Addition of a Solute
Copyright © Cengage Learning. All rights reserved 45
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Section 11.5
Boiling-Point Elevation and Freezing-Point
Depression
Freezing Point Depression: Solid/Solution Equilibrium
Copyright © Cengage Learning. All rights reserved 46
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Section 11.5
Boiling-Point Elevation and Freezing-Point
Depression
Changes in Boiling Point and Freezing Point of Water
Copyright © Cengage Learning. All rights reserved 47
Section 11.5
Boiling-Point Elevation and Freezing-Point
Depression
A solution was prepared by dissolving 25.00 g of
glucose in 200.0 g water. The molar mass of
glucose is 180.16 g/mol. What is the boiling point
of the resulting solution (in °C)? Glucose is a
molecular solid that is present as individual
molecules in solution.
100.35 °C
Copyright © Cengage Learning. All rights reserved 48
EXERCISE!
Section 11.5
Boiling-Point Elevation and Freezing-Point
Depression
You take 20.0 g of a sucrose (C12H22O11) and NaCl
mixture and dissolve it in 1.0 L of water. The freezing
point of this solution is found to be -0.426°C.
Assuming ideal behavior, calculate the mass percent
composition of the original mixture, and the mole
fraction of sucrose in the original mixture.
72.8% sucrose and 27.2% sodium chloride;
mole fraction of the sucrose is 0.313
Copyright © Cengage Learning. All rights reserved 49
EXERCISE!
Section 11.5
Boiling-Point Elevation and Freezing-Point
Depression
A plant cell has a natural concentration of
0.25 m. You immerse it in an aqueous solution with a
freezing point of –0.246°C. Will the
cell explode, shrivel, or do nothing?
Copyright © Cengage Learning. All rights reserved 50
EXERCISE!
Section 11.6
Osmotic Pressure
 Osmosis – flow of solvent into the solution through a
semipermeable membrane.
 = MRT
= osmotic pressure (atm)
M = molarity of the solution
R = gas law constant
T = temperature (Kelvin)
Copyright © Cengage Learning. All rights reserved 51


Section 11.6
Osmotic Pressure
Copyright © Cengage Learning. All rights reserved 52
Section 11.6
Osmotic Pressure
Osmosis
Copyright © Cengage Learning. All rights reserved 53
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Section 11.6
Osmotic Pressure
Copyright © Cengage Learning. All rights reserved 54
Section 11.6
Osmotic Pressure
When 33.4 mg of a compound is dissolved in
10.0 mL of water at 25°C, the solution has an
osmotic pressure of 558 torr. Calculate the
molar mass of this compound.
111 g/mol
Copyright © Cengage Learning. All rights reserved 55
EXERCISE!
Section 11.7
Colligative Properties of Electrolyte Solutions
van’t Hoff Factor, i
 The relationship between the moles of solute dissolved
and the moles of particles in solution is usually
expressed as:
Copyright © Cengage Learning. All rights reserved 56
moles of particles in solution
=
moles of solute dissolved
i
Section 11.7
Colligative Properties of Electrolyte Solutions
Ion Pairing
 At a given instant a small percentage of the sodium and
chloride ions are paired and thus count as a single
particle.
Copyright © Cengage Learning. All rights reserved 57
Section 11.7
Colligative Properties of Electrolyte Solutions
Examples
 The expected value for i can be determined for a salt by
noting the number of ions per formula unit (assuming
complete dissociation and that ion pairing does not
occur).
 NaCl i = 2
 KNO3 i = 2
 Na3PO4 i = 4
Copyright © Cengage Learning. All rights reserved 58
Section 11.7
Colligative Properties of Electrolyte Solutions
Ion Pairing
 Ion pairing is most important in concentrated solutions.
 As the solution becomes more dilute, the ions are
farther apart and less ion pairing occurs.
 Ion pairing occurs to some extent in all electrolyte
solutions.
 Ion pairing is most important for highly charged ions.
Copyright © Cengage Learning. All rights reserved 59
Section 11.7
Colligative Properties of Electrolyte Solutions
Modified Equations
Copyright © Cengage Learning. All rights reserved 60
=
T imK
=
 iMRT
Section 11.8
Colloids
 A suspension of tiny particles in some medium.
 Tyndall effect – scattering of light by particles.
 Suspended particles are single large molecules or
aggregates of molecules or ions ranging in size from 1 to
1000 nm.
Copyright © Cengage Learning. All rights reserved 61
Section 11.8
Colloids
Types of Colloids
Copyright © Cengage Learning. All rights reserved 62
Section 11.8
Colloids
Coagulation
 Destruction of a colloid.
 Usually accomplished either by heating or by adding an
electrolyte.
Copyright © Cengage Learning. All rights reserved 63

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chapter11.ppt

  • 2. Section 11.1 Solution Composition Copyright © Cengage Learning. All rights reserved 2 Various Types of Solutions
  • 3. Section 11.1 Solution Composition Copyright © Cengage Learning. All rights reserved 3 Solution Composition A A moles of solute Molarity ( ) = liters of solution mass of solute Mass (weight) percent = 100% mass of solution moles Mole fraction ( ) = total moles of solution moles of solute Molality ( ) = kilogram of s   M m olvent
  • 4. Section 11.1 Solution Composition Copyright © Cengage Learning. All rights reserved 4 Molarity moles of solute Molarity ( ) = liters of solution M
  • 5. Section 11.1 Solution Composition Copyright © Cengage Learning. All rights reserved 5 You have 1.00 mol of sugar in 125.0 mL of solution. Calculate the concentration in units of molarity. 8.00 M EXERCISE!
  • 6. Section 11.1 Solution Composition Copyright © Cengage Learning. All rights reserved 6 You have a 10.0 M sugar solution. What volume of this solution do you need to have 2.00 mol of sugar? 0.200 L EXERCISE!
  • 7. Section 11.1 Solution Composition Copyright © Cengage Learning. All rights reserved 7 Consider separate solutions of NaOH and KCl made by dissolving 100.0 g of each solute in 250.0 mL of solution. Calculate the concentration of each solution in units of molarity. 10.0 M NaOH 5.37 M KCl EXERCISE!
  • 8. Section 11.1 Solution Composition Copyright © Cengage Learning. All rights reserved 8 Mass Percent mass of solute Mass (weight)percent = 100% mass of solution 
  • 9. Section 11.1 Solution Composition Copyright © Cengage Learning. All rights reserved 9 What is the percent-by-mass concentration of glucose in a solution made my dissolving 5.5 g of glucose in 78.2 g of water? 6.6% EXERCISE!
  • 10. Section 11.1 Solution Composition Copyright © Cengage Learning. All rights reserved 10 Mole Fraction A A moles Mole fraction ( ) = total moles of solution 
  • 11. Section 11.1 Solution Composition Copyright © Cengage Learning. All rights reserved 11 A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in 100.0 mL of water. Calculate the mole fraction of H3PO4. (Assume water has a density of 1.00 g/mL.) 0.0145 EXERCISE!
  • 12. Section 11.1 Solution Composition Copyright © Cengage Learning. All rights reserved 12 Molality moles of solute Molality ( ) = kilogram of solvent m
  • 13. Section 11.1 Solution Composition Copyright © Cengage Learning. All rights reserved 13 A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in 100.0 mL of water. Calculate the molality of the solution. (Assume water has a density of 1.00 g/mL.) 0.816 m EXERCISE!
  • 14. Section 11.2 The Energies of Solution Formation Formation of a Liquid Solution 1. Separating the solute into its individual components (expanding the solute). 2. Overcoming intermolecular forces in the solvent to make room for the solute (expanding the solvent). 3. Allowing the solute and solvent to interact to form the solution. Copyright © Cengage Learning. All rights reserved 14
  • 15. Section 11.2 The Energies of Solution Formation Steps in the Dissolving Process Copyright © Cengage Learning. All rights reserved 15
  • 16. Section 11.2 The Energies of Solution Formation Steps in the Dissolving Process  Steps 1 and 2 require energy, since forces must be overcome to expand the solute and solvent.  Step 3 usually releases energy.  Steps 1 and 2 are endothermic, and step 3 is often exothermic. Copyright © Cengage Learning. All rights reserved 16
  • 17. Section 11.2 The Energies of Solution Formation Enthalpy (Heat) of Solution  Enthalpy change associated with the formation of the solution is the sum of the ΔH values for the steps: ΔHsoln = ΔH1 + ΔH2 + ΔH3  ΔHsoln may have a positive sign (energy absorbed) or a negative sign (energy released). Copyright © Cengage Learning. All rights reserved 17
  • 18. Section 11.2 The Energies of Solution Formation Enthalpy (Heat) of Solution Copyright © Cengage Learning. All rights reserved 18
  • 19. Section 11.2 The Energies of Solution Formation Explain why water and oil (a long chain hydrocarbon) do not mix. In your explanation, be sure to address how ΔH plays a role. Copyright © Cengage Learning. All rights reserved 19 CONCEPT CHECK!
  • 20. Section 11.2 The Energies of Solution Formation The Energy Terms for Various Types of Solutes and Solvents Copyright © Cengage Learning. All rights reserved 20 ΔH1 ΔH2 ΔH3 ΔHsoln Outcome Polar solute, polar solvent Large Large Large, negative Small Solution forms Nonpolar solute, polar solvent Small Large Small Large, positive No solution forms Nonpolar solute, nonpolar solvent Small Small Small Small Solution forms Polar solute, nonpolar solvent Large Small Small Large, positive No solution forms
  • 21. Section 11.2 The Energies of Solution Formation In General  One factor that favors a process is an increase in probability of the state when the solute and solvent are mixed.  Processes that require large amounts of energy tend not to occur.  Overall, remember that “like dissolves like”. Copyright © Cengage Learning. All rights reserved 21
  • 22. Section 11.3 Factors Affecting Solubility  Structure Effects:  Polarity  Pressure Effects:  Henry’s law  Temperature Effects:  Affecting aqueous solutions Copyright © Cengage Learning. All rights reserved 22
  • 23. Section 11.3 Factors Affecting Solubility Structure Effects  Hydrophobic (water fearing)  Non-polar substances  Hydrophilic (water loving)  Polar substances Copyright © Cengage Learning. All rights reserved 23
  • 24. Section 11.3 Factors Affecting Solubility Pressure Effects  Little effect on solubility of solids or liquids  Henry’s law: C = kP C = concentration of dissolved gas k = constant P = partial pressure of gas solute above the solution  Amount of gas dissolved in a solution is directly proportional to the pressure of the gas above the solution. Copyright © Cengage Learning. All rights reserved 24
  • 25. Section 11.3 Factors Affecting Solubility A Gaseous Solute Copyright © Cengage Learning. All rights reserved 25
  • 26. Section 11.3 Factors Affecting Solubility Temperature Effects (for Aqueous Solutions)  Although the solubility of most solids in water increases with temperature, the solubilities of some substances decrease with increasing temperature.  Predicting temperature dependence of solubility is very difficult.  Solubility of a gas in solvent typically decreases with increasing temperature. Copyright © Cengage Learning. All rights reserved 26
  • 27. Section 11.3 Factors Affecting Solubility The Solubilities of Several Solids as a Function of Temperature Copyright © Cengage Learning. All rights reserved 27
  • 28. Section 11.3 Factors Affecting Solubility The Solubilities of Several Gases in Water Copyright © Cengage Learning. All rights reserved 28
  • 29. Section 11.4 The Vapor Pressures of Solutions An Aqueous Solution and Pure Water in a Closed Environment Copyright © Cengage Learning. All rights reserved 29
  • 30. Section 11.4 The Vapor Pressures of Solutions Liquid/Vapor Equilibrium Copyright © Cengage Learning. All rights reserved 30 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE
  • 31. Section 11.4 The Vapor Pressures of Solutions Vapor Pressure Lowering: Addition of a Solute Copyright © Cengage Learning. All rights reserved 31 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE
  • 32. Section 11.4 The Vapor Pressures of Solutions Vapor Pressures of Solutions  Nonvolatile solute lowers the vapor pressure of a solvent.  Raoult’s Law: Psoln = observed vapor pressure of solution solv = mole fraction of solvent = vapor pressure of pure solvent Copyright © Cengage Learning. All rights reserved 32  soln solv solv =  P P solv P
  • 33. Section 11.4 The Vapor Pressures of Solutions A Solution Obeying Raoult’s Law Copyright © Cengage Learning. All rights reserved 33
  • 34. Section 11.4 The Vapor Pressures of Solutions Nonideal Solutions  Liquid-liquid solutions where both components are volatile.  Modified Raoult’s Law:  Nonideal solutions behave ideally as the mole fractions approach 0 and 1. Copyright © Cengage Learning. All rights reserved 34 Total A A B B = +   P P P
  • 35. Section 11.4 The Vapor Pressures of Solutions Vapor Pressure for a Solution of Two Volatile Liquids Copyright © Cengage Learning. All rights reserved 35
  • 36. Section 11.4 The Vapor Pressures of Solutions Summary of the Behavior of Various Types of Solutions Copyright © Cengage Learning. All rights reserved 36 Interactive Forces Between Solute (A) and Solvent (B) Particles ΔHsoln ΔT for Solution Formation Deviation from Raoult’s Law Example A  A, B  B  A  B Zero Zero None (ideal solution) Benzene- toluene A  A, B  B < A  B Negative (exothermic) Positive Negative Acetone- water A  A, B  B > A  B Positive (endothermic) Negative Positive Ethanol- hexane
  • 37. Section 11.4 The Vapor Pressures of Solutions For each of the following solutions, would you expect it to be relatively ideal (with respect to Raoult’s Law), show a positive deviation, or show a negative deviation? a) Hexane (C6H14) and chloroform (CHCl3) b) Ethyl alcohol (C2H5OH) and water c) Hexane (C6H14) and octane (C8H18) Copyright © Cengage Learning. All rights reserved 37 CONCEPT CHECK!
  • 38. Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Colligative Properties  Depend only on the number, not on the identity, of the solute particles in an ideal solution:  Boiling-point elevation  Freezing-point depression  Osmotic pressure Copyright © Cengage Learning. All rights reserved 38
  • 39. Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Boiling-Point Elevation  Nonvolatile solute elevates the boiling point of the solvent.  ΔT = Kbmsolute ΔT = boiling-point elevation Kb = molal boiling-point elevation constant msolute = molality of solute Copyright © Cengage Learning. All rights reserved 39
  • 40. Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Boiling Point Elevation: Liquid/Vapor Equilibrium Copyright © Cengage Learning. All rights reserved 40 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE
  • 41. Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Boiling Point Elevation: Addition of a Solute Copyright © Cengage Learning. All rights reserved 41 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE
  • 42. Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Boiling Point Elevation: Solution/Vapor Equilibrium Copyright © Cengage Learning. All rights reserved 42 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE
  • 43. Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Freezing-Point Depression  When a solute is dissolved in a solvent, the freezing point of the solution is lower than that of the pure solvent.  ΔT = Kfmsolute ΔT = freezing-point depression Kf = molal freezing-point depression constant msolute = molality of solute Copyright © Cengage Learning. All rights reserved 43
  • 44. Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Freezing Point Depression: Solid/Liquid Equilibrium Copyright © Cengage Learning. All rights reserved 44 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE
  • 45. Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Freezing Point Depression: Addition of a Solute Copyright © Cengage Learning. All rights reserved 45 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE
  • 46. Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Freezing Point Depression: Solid/Solution Equilibrium Copyright © Cengage Learning. All rights reserved 46 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE
  • 47. Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Changes in Boiling Point and Freezing Point of Water Copyright © Cengage Learning. All rights reserved 47
  • 48. Section 11.5 Boiling-Point Elevation and Freezing-Point Depression A solution was prepared by dissolving 25.00 g of glucose in 200.0 g water. The molar mass of glucose is 180.16 g/mol. What is the boiling point of the resulting solution (in °C)? Glucose is a molecular solid that is present as individual molecules in solution. 100.35 °C Copyright © Cengage Learning. All rights reserved 48 EXERCISE!
  • 49. Section 11.5 Boiling-Point Elevation and Freezing-Point Depression You take 20.0 g of a sucrose (C12H22O11) and NaCl mixture and dissolve it in 1.0 L of water. The freezing point of this solution is found to be -0.426°C. Assuming ideal behavior, calculate the mass percent composition of the original mixture, and the mole fraction of sucrose in the original mixture. 72.8% sucrose and 27.2% sodium chloride; mole fraction of the sucrose is 0.313 Copyright © Cengage Learning. All rights reserved 49 EXERCISE!
  • 50. Section 11.5 Boiling-Point Elevation and Freezing-Point Depression A plant cell has a natural concentration of 0.25 m. You immerse it in an aqueous solution with a freezing point of –0.246°C. Will the cell explode, shrivel, or do nothing? Copyright © Cengage Learning. All rights reserved 50 EXERCISE!
  • 51. Section 11.6 Osmotic Pressure  Osmosis – flow of solvent into the solution through a semipermeable membrane.  = MRT = osmotic pressure (atm) M = molarity of the solution R = gas law constant T = temperature (Kelvin) Copyright © Cengage Learning. All rights reserved 51  
  • 52. Section 11.6 Osmotic Pressure Copyright © Cengage Learning. All rights reserved 52
  • 53. Section 11.6 Osmotic Pressure Osmosis Copyright © Cengage Learning. All rights reserved 53 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE
  • 54. Section 11.6 Osmotic Pressure Copyright © Cengage Learning. All rights reserved 54
  • 55. Section 11.6 Osmotic Pressure When 33.4 mg of a compound is dissolved in 10.0 mL of water at 25°C, the solution has an osmotic pressure of 558 torr. Calculate the molar mass of this compound. 111 g/mol Copyright © Cengage Learning. All rights reserved 55 EXERCISE!
  • 56. Section 11.7 Colligative Properties of Electrolyte Solutions van’t Hoff Factor, i  The relationship between the moles of solute dissolved and the moles of particles in solution is usually expressed as: Copyright © Cengage Learning. All rights reserved 56 moles of particles in solution = moles of solute dissolved i
  • 57. Section 11.7 Colligative Properties of Electrolyte Solutions Ion Pairing  At a given instant a small percentage of the sodium and chloride ions are paired and thus count as a single particle. Copyright © Cengage Learning. All rights reserved 57
  • 58. Section 11.7 Colligative Properties of Electrolyte Solutions Examples  The expected value for i can be determined for a salt by noting the number of ions per formula unit (assuming complete dissociation and that ion pairing does not occur).  NaCl i = 2  KNO3 i = 2  Na3PO4 i = 4 Copyright © Cengage Learning. All rights reserved 58
  • 59. Section 11.7 Colligative Properties of Electrolyte Solutions Ion Pairing  Ion pairing is most important in concentrated solutions.  As the solution becomes more dilute, the ions are farther apart and less ion pairing occurs.  Ion pairing occurs to some extent in all electrolyte solutions.  Ion pairing is most important for highly charged ions. Copyright © Cengage Learning. All rights reserved 59
  • 60. Section 11.7 Colligative Properties of Electrolyte Solutions Modified Equations Copyright © Cengage Learning. All rights reserved 60 = T imK =  iMRT
  • 61. Section 11.8 Colloids  A suspension of tiny particles in some medium.  Tyndall effect – scattering of light by particles.  Suspended particles are single large molecules or aggregates of molecules or ions ranging in size from 1 to 1000 nm. Copyright © Cengage Learning. All rights reserved 61
  • 62. Section 11.8 Colloids Types of Colloids Copyright © Cengage Learning. All rights reserved 62
  • 63. Section 11.8 Colloids Coagulation  Destruction of a colloid.  Usually accomplished either by heating or by adding an electrolyte. Copyright © Cengage Learning. All rights reserved 63