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Page 1: Chapter 11 Properties of Solutions. Chapter 11 Table of Contents Copyright © Cengage Learning. All rights reserved 2 11.1 Solution Composition 11.2 The

Chapter 11

Properties of Solutions

Page 2: Chapter 11 Properties of Solutions. Chapter 11 Table of Contents Copyright © Cengage Learning. All rights reserved 2 11.1 Solution Composition 11.2 The

Chapter 11

Table of Contents

Copyright © Cengage Learning. All rights reserved 2

11.1 Solution Composition

11.2 The Energies of Solution Formation

11.3 Factors Affecting Solubility

11.4 The Vapor Pressures of Solutions

11.5 Boiling-Point Elevation and Freezing-Point Depression

11.6 Osmotic Pressure

11.7 Colligative Properties of Electrolyte Solutions

11.8 Colloids

Page 3: Chapter 11 Properties of Solutions. Chapter 11 Table of Contents Copyright © Cengage Learning. All rights reserved 2 11.1 Solution Composition 11.2 The

Section 11.1

Solution Composition

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Various Types of Solutions

ExampleState of Solution State of Solute

State of Solvent

Air, natural gas Gas Gas Gas

Vodka, antifreeze Liquid Liquid Liquid

Brass Solid Solid Solid

Carbonated water (soda) Liquid Gas Liquid

Seawater, sugar solution Liquid Solid Liquid

Hydrogen in platinum Solid Gas Solid

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Section 11.1

Solution Composition

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Solution Composition

AA

moles of soluteMolarity ( ) =

liters of solution

mass of soluteMass (weight) percent = 100%

mass of solution

molesMole fraction ( ) =

total moles of solution

moles of soluteMolality ( ) =

kilogram of s

M

molvent

Page 5: Chapter 11 Properties of Solutions. Chapter 11 Table of Contents Copyright © Cengage Learning. All rights reserved 2 11.1 Solution Composition 11.2 The

Section 11.1

Solution Composition

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Molarity

moles of soluteMolarity ( ) =

liters of solution M

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Section 11.1

Solution Composition

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Exercise

You have 1.00 mol of sugar in 125.0 mL of solution. Calculate the concentration in units of molarity.

8.00 M

Page 7: Chapter 11 Properties of Solutions. Chapter 11 Table of Contents Copyright © Cengage Learning. All rights reserved 2 11.1 Solution Composition 11.2 The

Section 11.1

Solution Composition

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Exercise

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

Page 8: Chapter 11 Properties of Solutions. Chapter 11 Table of Contents Copyright © Cengage Learning. All rights reserved 2 11.1 Solution Composition 11.2 The

Section 11.1

Solution Composition

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Exercise

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

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Section 11.1

Solution Composition

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Mass Percent

mass of soluteMass (weight) percent = 100%

mass of solution

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Section 11.1

Solution Composition

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Exercise

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%

Page 11: Chapter 11 Properties of Solutions. Chapter 11 Table of Contents Copyright © Cengage Learning. All rights reserved 2 11.1 Solution Composition 11.2 The

Section 11.1

Solution Composition

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Mole Fraction

AA

molesMole fraction ( ) =

total moles of solution

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Section 11.1

Solution Composition

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Exercise

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

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Section 11.1

Solution Composition

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Molality

moles of soluteMolality ( ) =

kilogram of solvent m

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Section 11.1

Solution Composition

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Exercise

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

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Section 11.2

Atomic Masses

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The Energies of Solution Formation

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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.

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Section 11.2

Atomic Masses

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The Energies of Solution Formation

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Steps in the Dissolving Process

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Section 11.2

Atomic Masses

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The Energies of Solution Formation

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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.

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Section 11.2

Atomic Masses

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The Energies of Solution Formation

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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).

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Section 11.2

Atomic Masses

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The Energies of Solution Formation

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Enthalpy (Heat) of Solution

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Section 11.2

Atomic Masses

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The Energies of Solution Formation

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Concept Check

Explain why water and oil (a long chain hydrocarbon) do not mix. In your explanation, be sure to address how ΔH plays a role.

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Section 11.2

Atomic Masses

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The Energies of Solution Formation

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The Energy Terms for Various Types of Solutes and Solvents

H1 H2 H3 Hsoln 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

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Section 11.2

Atomic Masses

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The Energies of Solution Formation

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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”.

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Section 11.3

The Mole Factors Affecting Solubility

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• Structural Effects: Polarity

• Pressure Effects: Henry’s law

• Temperature Effects: Affecting aqueous solutions

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Section 11.3

The Mole Factors Affecting Solubility

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Pressure Effects

• Henry’s law: C = kPC = 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.

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Section 11.3

The Mole Factors Affecting Solubility

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A Gaseous Solute

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Section 11.3

The Mole Factors Affecting Solubility

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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.

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Section 11.3

The Mole Factors Affecting Solubility

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The Solubilities of Several Solids as a Function of Temperature

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Section 11.3

The Mole Factors Affecting Solubility

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The Solubilities of Several Gases in Water

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Section 11.4

The Vapor Pressures of Solutions

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An Aqueous Solution and Pure Water in a Closed Environment

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Section 11.4

The Vapor Pressures of Solutions

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Liquid/Vapor Equilibrium

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Section 11.4

The Vapor Pressures of Solutions

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Vapor Pressure Lowering: Addition of a Solute

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Section 11.4

The Vapor Pressures of Solutions

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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

soln solv solv = P P

solvP

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Section 11.4

The Vapor Pressures of Solutions

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A Solution Obeying Raoult’s Law

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Section 11.4

The Vapor Pressures of Solutions

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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.

Total A A B B = + P P P

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Section 11.4

The Vapor Pressures of Solutions

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Vapor Pressure for a Solution of Two Volatile Liquids

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Section 11.4

The Vapor Pressures of Solutions

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Summary of the Behavior of Various Types of Solutions

Interactive Forces Between Solute (A) and Solvent (B)

ParticlesHsoln

T for Solution Formation

Deviation from

Raoult’s Law

Example

A A, B B A B Zero ZeroNone (ideal

solution)Benzene-toluene

A A, B B < A BNegative

(exothermic)Positive Negative

Acetone-water

A A, B B > A BPositive

(endothermic)Negative Positive

Ethanol-hexane

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Section 11.4

The Vapor Pressures of Solutions

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Concept Check

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)

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Section 11.5

Boiling-Point Elevation and Freezing-Point Depression

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• 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

Colligative Properties

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Section 11.5

Boiling-Point Elevation and Freezing-Point Depression

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• 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

Boiling-Point Elevation

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Section 11.5

Boiling-Point Elevation and Freezing-Point Depression

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Boiling Point Elevation: Liquid/Vapor Equilibrium

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Section 11.5

Boiling-Point Elevation and Freezing-Point Depression

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Boiling Point Elevation: Addition of a Solute

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Section 11.5

Boiling-Point Elevation and Freezing-Point Depression

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Boiling Point Elevation: Solution/Vapor Equilibrium

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Section 11.5

Boiling-Point Elevation and Freezing-Point Depression

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• 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

Freezing-Point Depression

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Section 11.5

Boiling-Point Elevation and Freezing-Point Depression

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Freezing Point Depression: Solid/Liquid Equilibrium

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Section 11.5

Boiling-Point Elevation and Freezing-Point Depression

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Freezing Point Depression: Addition of a Solute

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Section 11.5

Boiling-Point Elevation and Freezing-Point Depression

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Freezing Point Depression: Solid/Solution Equilibrium

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Section 11.5

Boiling-Point Elevation and Freezing-Point Depression

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Changes in Boiling Point and Freezing Point of Water

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Section 11.5

Boiling-Point Elevation and Freezing-Point Depression

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Exercise

A solution was prepared by dissolving 25.00 g 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

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Section 11.5

Boiling-Point Elevation and Freezing-Point Depression

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Exercise

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

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Section 11.5

Boiling-Point Elevation and Freezing-Point Depression

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Exercise

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?

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Section 11.6

Osmotic Pressure

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• 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)

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Section 11.6

Osmotic Pressure

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Section 11.6

Osmotic Pressure

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Osmosis

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Section 11.6

Osmotic Pressure

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Section 11.6

Osmotic Pressure

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Exercise

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

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Section 11.7

Colligative Properties of Electrolyte Solutions

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• The relationship between the moles of solute dissolved and the moles of particles in solution is usually expressed as:

van’t Hoff Factor, i

moles of particles in solution =

moles of solute dissolvedi

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Section 11.7

Colligative Properties of Electrolyte Solutions

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Ion Pairing

• At a given instant a small percentage of the sodium and chloride ions are paired and thus count as a single particle.

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Section 11.7

Colligative Properties of Electrolyte Solutions

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• 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

Examples

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Section 11.7

Colligative Properties of Electrolyte Solutions

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• 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.

Ion Pairing

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Section 11.7

Colligative Properties of Electrolyte Solutions

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Modified Equations

= T imK

= iMRT

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Section 11.8

Colloids

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• 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.

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Section 11.8

Colloids

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Types of Colloids

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Section 11.8

Colloids

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• Destruction of a colloid.• Usually accomplished either by heating or

by adding an electrolyte.

Coagulation