Enter your Sign on user name and password.

Forgot password?
Sign In | Subscribe
Start learning today, and be successful in your academic & professional career. Start Today!
Use Chrome browser to play professor video
Raffi Hovasapian

Raffi Hovasapian

Vapor Pressure of Solutions

Slide Duration:

Table of Contents

I. Review
Naming Compounds

41m 24s

Intro
0:00
Periodic Table of Elements
0:15
Naming Compounds
3:13
Definition and Examples of Ions
3:14
Ionic (Symbol to Name): NaCl
5:23
Ionic (Name to Symbol): Calcium Oxide
7:58
Ionic - Polyatoms Anions: Examples
12:45
Ionic - Polyatoms Anions (Symbol to Name): KClO
14:50
Ionic - Polyatoms Anions (Name to Symbol): Potassium Phosphate
15:49
Ionic Compounds Involving Transition Metals (Symbol to Name): Co₂(CO₃)₃
20:48
Ionic Compounds Involving Transition Metals (Name to Symbol): Palladium 2 Acetate
22:44
Naming Covalent Compounds (Symbol to Name): CO
26:21
Naming Covalent Compounds (Name to Symbol): Nitrogen Trifluoride
27:34
Naming Covalent Compounds (Name to Symbol): Dichlorine Monoxide
27:57
Naming Acids Introduction
28:11
Naming Acids (Name to Symbol): Chlorous Acid
35:08
% Composition by Mass Example
37:38
Stoichiometry

37m 19s

Intro
0:00
Stoichiometry
0:25
Introduction to Stoichiometry
0:26
Example 1
5:03
Example 2
10:17
Example 3
15:09
Example 4
24:02
Example 5: Questions
28:11
Example 5: Part A - Limiting Reactant
30:30
Example 5: Part B
32:27
Example 5: Part C
35:00
II. Aqueous Reactions & Stoichiometry
Precipitation Reactions

31m 14s

Intro
0:00
Precipitation Reactions
0:53
Dissociation of ionic Compounds
0:54
Solubility Guidelines for ionic Compounds: Soluble Ionic Compounds
8:15
Solubility Guidelines for ionic Compounds: Insoluble ionic Compounds
12:56
Precipitation Reactions
14:08
Example 1: Mixing a Solution of BaCl₂ & K₂SO₄
21:21
Example 2: Mixing a Solution of Mg(NO₃)₂ & KI
26:10
Acid-Base Reactions

43m 21s

Intro
0:00
Acid-Base Reactions
1:00
Introduction to Acid: Monoprotic Acid and Polyprotic Acid
1:01
Introduction to Base
8:28
Neutralization
11:45
Example 1
16:17
Example 2
21:55
Molarity
24:50
Example 3
26:50
Example 4
30:01
Example 4: Limiting Reactant
37:51
Example 4: Reaction Part
40:01
Oxidation Reduction Reactions

47m 58s

Intro
0:00
Oxidation Reduction Reactions
0:26
Oxidation and Reduction Overview
0:27
How Can One Tell Whether Oxidation-Reduction has Taken Place?
7:13
Rules for Assigning Oxidation State: Number 1
11:22
Rules for Assigning Oxidation State: Number 2
12:46
Rules for Assigning Oxidation State: Number 3
13:25
Rules for Assigning Oxidation State: Number 4
14:50
Rules for Assigning Oxidation State: Number 5
15:41
Rules for Assigning Oxidation State: Number 6
17:00
Example 1: Determine the Oxidation State of Sulfur in the Following Compounds
18:20
Activity Series and Reduction Properties
25:32
Activity Series and Reduction Properties
25:33
Example 2: Write the Balance Molecular, Total Ionic, and Net Ionic Equations for Al + HCl
31:37
Example 3
34:25
Example 4
37:55
Stoichiometry Examples

31m 50s

Intro
0:00
Stoichiometry Example 1
0:36
Example 1: Question and Answer
0:37
Stoichiometry Example 2
6:57
Example 2: Questions
6:58
Example 2: Part A Solution
12:16
Example 2: Part B Solution
13:05
Example 2: Part C Solution
14:00
Example 2: Part D Solution
14:38
Stoichiometry Example 3
17:56
Example 3: Questions
17:57
Example 3: Part A Solution
19:51
Example 3: Part B Solution
21:43
Example 3: Part C Solution
26:46
III. Gases
Pressure, Gas Laws, & The Ideal Gas Equation

49m 40s

Intro
0:00
Pressure
0:22
Pressure Overview
0:23
Torricelli: Barometer
4:35
Measuring Gas Pressure in a Container
7:49
Boyle's Law
12:40
Example 1
16:56
Gas Laws
21:18
Gas Laws
21:19
Avogadro's Law
26:16
Example 2
31:47
Ideal Gas Equation
38:20
Standard Temperature and Pressure (STP)
38:21
Example 3
40:43
Partial Pressure, Mol Fraction, & Vapor Pressure

32m

Intro
0:00
Gases
0:27
Gases
0:28
Mole Fractions
5:52
Vapor Pressure
8:22
Example 1
13:25
Example 2
22:45
Kinetic Molecular Theory and Real Gases

31m 58s

Intro
0:00
Kinetic Molecular Theory and Real Gases
0:45
Kinetic Molecular Theory 1
0:46
Kinetic Molecular Theory 2
4:23
Kinetic Molecular Theory 3
5:42
Kinetic Molecular Theory 4
6:27
Equations
7:52
Effusion
11:15
Diffusion
13:30
Example 1
19:54
Example 2
23:23
Example 3
26:45
AP Practice for Gases

25m 34s

Intro
0:00
Example 1
0:34
Example 1
0:35
Example 2
6:15
Example 2: Part A
6:16
Example 2: Part B
8:46
Example 2: Part C
10:30
Example 2: Part D
11:15
Example 2: Part E
12:20
Example 2: Part F
13:22
Example 3
14:45
Example 3
14:46
Example 4
18:16
Example 4
18:17
Example 5
21:04
Example 5
21:05
IV. Thermochemistry
Energy, Heat, and Work

37m 32s

Intro
0:00
Thermochemistry
0:25
Temperature and Heat
0:26
Work
3:07
System, Surroundings, Exothermic Process, and Endothermic Process
8:19
Work & Gas: Expansion and Compression
16:30
Example 1
24:41
Example 2
27:47
Example 3
31:58
Enthalpy & Hess's Law

32m 34s

Intro
0:00
Thermochemistry
1:43
Defining Enthalpy & Hess's Law
1:44
Example 1
6:48
State Function
13:11
Example 2
17:15
Example 3
24:09
Standard Enthalpies of Formation

23m 9s

Intro
0:00
Thermochemistry
1:04
Standard Enthalpy of Formation: Definition & Equation
1:05
∆H of Formation
10:00
Example 1
11:22
Example 2
19:00
Calorimetry

39m 28s

Intro
0:00
Thermochemistry
0:21
Heat Capacity
0:22
Molar Heat Capacity
4:44
Constant Pressure Calorimetry
5:50
Example 1
12:24
Constant Volume Calorimetry
21:54
Example 2
24:40
Example 3
31:03
V. Kinetics
Reaction Rates and Rate Laws

36m 24s

Intro
0:00
Kinetics
2:18
Rate: 2 NO₂ (g) → 2NO (g) + O₂ (g)
2:19
Reaction Rates Graph
7:25
Time Interval & Average Rate
13:13
Instantaneous Rate
15:13
Rate of Reaction is Proportional to Some Power of the Reactant Concentrations
23:49
Example 1
27:19
Method of Initial Rates

30m 48s

Intro
0:00
Kinetics
0:33
Rate
0:34
Idea
2:24
Example 1: NH₄⁺ + NO₂⁻ → NO₂ (g) + 2 H₂O
5:36
Example 2: BrO₃⁻ + 5 Br⁻ + 6 H⁺ → 3 Br₂ + 3 H₂O
19:29
Integrated Rate Law & Reaction Half-Life

32m 17s

Intro
0:00
Kinetics
0:52
Integrated Rate Law
0:53
Example 1
6:26
Example 2
15:19
Half-life of a Reaction
20:40
Example 3: Part A
25:41
Example 3: Part B
28:01
Second Order & Zero-Order Rate Laws

26m 40s

Intro
0:00
Kinetics
0:22
Second Order
0:23
Example 1
6:08
Zero-Order
16:36
Summary for the Kinetics Associated with the Reaction
21:27
Activation Energy & Arrhenius Equation

40m 59s

Intro
0:00
Kinetics
0:53
Rate Constant
0:54
Collision Model
2:45
Activation Energy
5:11
Arrhenius Proposed
9:54
2 Requirements for a Successful Reaction
15:39
Rate Constant
17:53
Arrhenius Equation
19:51
Example 1
25:00
Activation Energy & the Values of K
32:12
Example 2
36:46
AP Practice for Kinetics

29m 8s

Intro
0:00
Kinetics
0:43
Example 1
0:44
Example 2
6:53
Example 3
8:58
Example 4
11:36
Example 5
16:36
Example 6: Part A
21:00
Example 6: Part B
25:09
VI. Equilibrium
Equilibrium, Part 1

46m

Intro
0:00
Equilibrium
1:32
Introduction to Equilibrium
1:33
Equilibrium Rules
14:00
Example 1: Part A
16:46
Example 1: Part B
18:48
Example 1: Part C
22:13
Example 1: Part D
24:55
Example 2: Part A
27:46
Example 2: Part B
31:22
Example 2: Part C
33:00
Reverse a Reaction
36:04
Example 3
37:24
Equilibrium, Part 2

40m 53s

Intro
0:00
Equilibrium
1:31
Equilibriums Involving Gases
1:32
General Equation
10:11
Example 1: Question
11:55
Example 1: Answer
13:43
Example 2: Question
19:08
Example 2: Answer
21:37
Example 3: Question
33:40
Example 3: Answer
35:24
Equilibrium: Reaction Quotient

45m 53s

Intro
0:00
Equilibrium
0:57
Reaction Quotient
0:58
If Q > K
5:37
If Q < K
6:52
If Q = K
7:45
Example 1: Part A
8:24
Example 1: Part B
13:11
Example 2: Question
20:04
Example 2: Answer
22:15
Example 3: Question
30:54
Example 3: Answer
32:52
Steps in Solving Equilibrium Problems
42:40
Equilibrium: Examples

31m 51s

Intro
0:00
Equilibrium
1:09
Example 1: Question
1:10
Example 1: Answer
4:15
Example 2: Question
13:04
Example 2: Answer
15:20
Example 3: Question
25:03
Example 3: Answer
26:32
Le Chatelier's principle & Equilibrium

40m 52s

Intro
0:00
Le Chatelier
1:05
Le Chatelier Principle
1:06
Concentration: Add 'x'
5:25
Concentration: Subtract 'x'
7:50
Example 1
9:44
Change in Pressure
12:53
Example 2
20:40
Temperature: Exothermic and Endothermic
24:33
Example 3
29:55
Example 4
35:30
VII. Acids & Bases
Acids and Bases

50m 11s

Intro
0:00
Acids and Bases
1:14
Bronsted-Lowry Acid-Base Model
1:28
Reaction of an Acid with Water
4:36
Acid Dissociation
10:51
Acid Strength
13:48
Example 1
21:22
Water as an Acid & a Base
25:25
Example 2: Part A
32:30
Example 2: Part B
34:47
Example 3: Part A
35:58
Example 3: Part B
39:33
pH Scale
41:12
Example 4
43:56
pH of Weak Acid Solutions

43m 52s

Intro
0:00
pH of Weak Acid Solutions
1:12
pH of Weak Acid Solutions
1:13
Example 1
6:26
Example 2
14:25
Example 3
24:23
Example 4
30:38
Percent Dissociation: Strong & Weak Bases

43m 4s

Intro
0:00
Bases
0:33
Percent Dissociation: Strong & Weak Bases
0:45
Example 1
6:23
Strong Base Dissociation
11:24
Example 2
13:02
Weak Acid and General Reaction
17:38
Example: NaOH → Na⁺ + OH⁻
20:30
Strong Base and Weak Base
23:49
Example 4
24:54
Example 5
33:51
Polyprotic Acids

35m 34s

Intro
0:00
Polyprotic Acids
1:04
Acids Dissociation
1:05
Example 1
4:51
Example 2
17:30
Example 3
31:11
Salts and Their Acid-Base Properties

41m 14s

Intro
0:00
Salts and Their Acid-Base Properties
0:11
Salts and Their Acid-Base Properties
0:15
Example 1
7:58
Example 2
14:00
Metal Ion and Acidic Solution
22:00
Example 3
28:35
NH₄F → NH₄⁺ + F⁻
34:05
Example 4
38:03
Common Ion Effect & Buffers

41m 58s

Intro
0:00
Common Ion Effect & Buffers
1:16
Covalent Oxides Produce Acidic Solutions in Water
1:36
Ionic Oxides Produce Basic Solutions in Water
4:15
Practice Example 1
6:10
Practice Example 2
9:00
Definition
12:27
Example 1: Part A
16:49
Example 1: Part B
19:54
Buffer Solution
25:10
Example of Some Buffers: HF and NaF
30:02
Example of Some Buffers: Acetic Acid & Potassium Acetate
31:34
Example of Some Buffers: CH₃NH₂ & CH₃NH₃Cl
33:54
Example 2: Buffer Solution
36:36
Buffer

32m 24s

Intro
0:00
Buffers
1:20
Buffer Solution
1:21
Adding Base
5:03
Adding Acid
7:14
Example 1: Question
9:48
Example 1: Recall
12:08
Example 1: Major Species Upon Addition of NaOH
16:10
Example 1: Equilibrium, ICE Chart, and Final Calculation
24:33
Example 1: Comparison
29:19
Buffers, Part II

40m 6s

Intro
0:00
Buffers
1:27
Example 1: Question
1:32
Example 1: ICE Chart
3:15
Example 1: Major Species Upon Addition of OH⁻, But Before Rxn
7:23
Example 1: Equilibrium, ICE Chart, and Final Calculation
12:51
Summary
17:21
Another Look at Buffering & the Henderson-Hasselbalch equation
19:00
Example 2
27:08
Example 3
32:01
Buffers, Part III

38m 43s

Intro
0:00
Buffers
0:25
Buffer Capacity Part 1
0:26
Example 1
4:10
Buffer Capacity Part 2
19:29
Example 2
25:12
Example 3
32:02
Titrations: Strong Acid and Strong Base

42m 42s

Intro
0:00
Titrations: Strong Acid and Strong Base
1:11
Definition of Titration
1:12
Sample Problem
3:33
Definition of Titration Curve or pH Curve
9:46
Scenario 1: Strong Acid- Strong Base Titration
11:00
Question
11:01
Part 1: No NaOH is Added
14:00
Part 2: 10.0 mL of NaOH is Added
15:50
Part 3: Another 10.0 mL of NaOH & 20.0 mL of NaOH are Added
22:19
Part 4: 50.0 mL of NaOH is Added
26:46
Part 5: 100.0 mL (Total) of NaOH is Added
27:26
Part 6: 150.0 mL (Total) of NaOH is Added
32:06
Part 7: 200.0 mL of NaOH is Added
35:07
Titrations Curve for Strong Acid and Strong Base
35:43
Titrations: Weak Acid and Strong Base

42m 3s

Intro
0:00
Titrations: Weak Acid and Strong Base
0:43
Question
0:44
Part 1: No NaOH is Added
1:54
Part 2: 10.0 mL of NaOH is Added
5:17
Part 3: 25.0 mL of NaOH is Added
14:01
Part 4: 40.0 mL of NaOH is Added
21:55
Part 5: 50.0 mL (Total) of NaOH is Added
22:25
Part 6: 60.0 mL (Total) of NaOH is Added
31:36
Part 7: 75.0 mL (Total) of NaOH is Added
35:44
Titration Curve
36:09
Titration Examples & Acid-Base Indicators

52m 3s

Intro
0:00
Examples and Indicators
0:25
Example 1: Question
0:26
Example 1: Solution
2:03
Example 2: Question
12:33
Example 2: Solution
14:52
Example 3: Question
23:45
Example 3: Solution
25:09
Acid/Base Indicator Overview
34:45
Acid/Base Indicator Example
37:40
Acid/Base Indicator General Result
47:11
Choosing Acid/Base Indicator
49:12
VIII. Solubility
Solubility Equilibria

36m 25s

Intro
0:00
Solubility Equilibria
0:48
Solubility Equilibria Overview
0:49
Solubility Product Constant
4:24
Definition of Solubility
9:10
Definition of Solubility Product
11:28
Example 1
14:09
Example 2
20:19
Example 3
27:30
Relative Solubilities
31:04
Solubility Equilibria, Part II

42m 6s

Intro
0:00
Solubility Equilibria
0:46
Common Ion Effect
0:47
Example 1
3:14
pH & Solubility
13:00
Example of pH & Solubility
15:25
Example 2
23:06
Precipitation & Definition of the Ion Product
26:48
If Q > Ksp
29:31
If Q < Ksp
30:27
Example 3
32:58
Solubility Equilibria, Part III

43m 9s

Intro
0:00
Solubility Equilibria
0:55
Example 1: Question
0:56
Example 1: Step 1 - Check to See if Anything Precipitates
2:52
Example 1: Step 2 - Stoichiometry
10:47
Example 1: Step 3 - Equilibrium
16:34
Example 2: Selective Precipitation (Question)
21:02
Example 2: Solution
23:41
Classical Qualitative Analysis
29:44
Groups: 1-5
38:44
IX. Complex Ions
Complex Ion Equilibria

43m 38s

Intro
0:00
Complex Ion Equilibria
0:32
Complex Ion
0:34
Ligan Examples
1:51
Ligand Definition
3:12
Coordination
6:28
Example 1
8:08
Example 2
19:13
Complex Ions & Solubility

31m 30s

Intro
0:00
Complex Ions and Solubility
0:23
Recall: Classical Qualitative Analysis
0:24
Example 1
6:10
Example 2
16:16
Dissolving a Water-Insoluble Ionic Compound: Method 1
23:38
Dissolving a Water-Insoluble Ionic Compound: Method 2
28:13
X. Chemical Thermodynamics
Spontaneity, Entropy, & Free Energy, Part I

56m 28s

Intro
0:00
Spontaneity, Entropy, Free Energy
2:25
Energy Overview
2:26
Equation: ∆E = q + w
4:30
State Function/ State Property
8:35
Equation: w = -P∆V
12:00
Enthalpy: H = E + PV
14:50
Enthalpy is a State Property
17:33
Exothermic and Endothermic Reactions
19:20
First Law of Thermodynamic
22:28
Entropy
25:48
Spontaneous Process
33:53
Second Law of Thermodynamic
36:51
More on Entropy
42:23
Example
43:55
Spontaneity, Entropy, & Free Energy, Part II

39m 55s

Intro
0:00
Spontaneity, Entropy, Free Energy
1:30
∆S of Universe = ∆S of System + ∆S of Surrounding
1:31
Convention
3:32
Examining a System
5:36
Thermodynamic Property: Sign of ∆S
16:52
Thermodynamic Property: Magnitude of ∆S
18:45
Deriving Equation: ∆S of Surrounding = -∆H / T
20:25
Example 1
25:51
Free Energy Equations
29:22
Spontaneity, Entropy, & Free Energy, Part III

30m 10s

Intro
0:00
Spontaneity, Entropy, Free Energy
0:11
Example 1
2:38
Key Concept of Example 1
14:06
Example 2
15:56
Units for ∆H, ∆G, and S
20:56
∆S of Surrounding & ∆S of System
22:00
Reaction Example
24:17
Example 3
26:52
Spontaneity, Entropy, & Free Energy, Part IV

30m 7s

Intro
0:00
Spontaneity, Entropy, Free Energy
0:29
Standard Free Energy of Formation
0:58
Example 1
4:34
Reaction Under Non-standard Conditions
13:23
Example 2
16:26
∆G = Negative
22:12
∆G = 0
24:38
Diagram Example of ∆G
26:43
Spontaneity, Entropy, & Free Energy, Part V

44m 56s

Intro
0:00
Spontaneity, Entropy, Free Energy
0:56
Equations: ∆G of Reaction, ∆G°, and K
0:57
Example 1: Question
6:50
Example 1: Part A
9:49
Example 1: Part B
15:28
Example 2
17:33
Example 3
23:31
lnK = (- ∆H° ÷ R) ( 1 ÷ T) + ( ∆S° ÷ R)
31:36
Maximum Work
35:57
XI. Electrochemistry
Oxidation-Reduction & Balancing

39m 23s

Intro
0:00
Oxidation-Reduction and Balancing
2:06
Definition of Electrochemistry
2:07
Oxidation and Reduction Review
3:05
Example 1: Assigning Oxidation State
10:15
Example 2: Is the Following a Redox Reaction?
18:06
Example 3: Step 1 - Write the Oxidation & Reduction Half Reactions
22:46
Example 3: Step 2 - Balance the Reaction
26:44
Example 3: Step 3 - Multiply
30:11
Example 3: Step 4 - Add
32:07
Example 3: Step 5 - Check
33:29
Galvanic Cells

43m 9s

Intro
0:00
Galvanic Cells
0:39
Example 1: Balance the Following Under Basic Conditions
0:40
Example 1: Steps to Balance Reaction Under Basic Conditions
3:25
Example 1: Solution
5:23
Example 2: Balance the Following Reaction
13:56
Galvanic Cells
18:15
Example 3: Galvanic Cells
28:19
Example 4: Galvanic Cells
35:12
Cell Potential

48m 41s

Intro
0:00
Cell Potential
2:08
Definition of Cell Potential
2:17
Symbol and Unit
5:50
Standard Reduction Potential
10:16
Example Figure 1
13:08
Example Figure 2
19:00
All Reduction Potentials are Written as Reduction
23:10
Cell Potential: Important Fact 1
26:49
Cell Potential: Important Fact 2
27:32
Cell Potential: Important Fact 3
28:54
Cell Potential: Important Fact 4
30:05
Example Problem 1
32:29
Example Problem 2
38:38
Potential, Work, & Free Energy

41m 23s

Intro
0:00
Potential, Work, Free Energy
0:42
Descriptions of Galvanic Cell
0:43
Line Notation
5:33
Example 1
6:26
Example 2
11:15
Example 3
15:18
Equation: Volt
22:20
Equations: Cell Potential, Work, and Charge
28:30
Maximum Cell Potential is Related to the Free Energy of the Cell Reaction
35:09
Example 4
37:42
Cell Potential & Concentration

34m 19s

Intro
0:00
Cell Potential & Concentration
0:29
Example 1: Question
0:30
Example 1: Nernst Equation
4:43
Example 1: Solution
7:01
Cell Potential & Concentration
11:27
Example 2
16:38
Manipulating the Nernst Equation
25:15
Example 3
28:43
Electrolysis

33m 21s

Intro
0:00
Electrolysis
3:16
Electrolysis: Part 1
3:17
Electrolysis: Part 2
5:25
Galvanic Cell Example
7:13
Nickel Cadmium Battery
12:18
Ampere
16:00
Example 1
20:47
Example 2
25:47
XII. Light
Light

44m 45s

Intro
0:00
Light
2:14
Introduction to Light
2:15
Frequency, Speed, and Wavelength of Waves
3:58
Units and Equations
7:37
Electromagnetic Spectrum
12:13
Example 1: Calculate the Frequency
17:41
E = hν
21:30
Example 2: Increment of Energy
25:12
Photon Energy of Light
28:56
Wave and Particle
31:46
Example 3: Wavelength of an Electron
34:46
XIII. Quantum Mechanics
Quantum Mechanics & Electron Orbitals

54m

Intro
0:00
Quantum Mechanics & Electron Orbitals
0:51
Quantum Mechanics & Electron Orbitals Overview
0:52
Electron Orbital and Energy Levels for the Hydrogen Atom
8:47
Example 1
13:41
Quantum Mechanics: Schrodinger Equation
19:19
Quantum Numbers Overview
31:10
Principal Quantum Numbers
33:28
Angular Momentum Numbers
34:55
Magnetic Quantum Numbers
36:35
Spin Quantum Numbers
37:46
Primary Level, Sublevels, and Sub-Sub-Levels
39:42
Example
42:17
Orbital & Quantum Numbers
49:32
Electron Configurations & Diagrams

34m 4s

Intro
0:00
Electron Configurations & Diagrams
1:08
Electronic Structure of Ground State Atom
1:09
Order of Electron Filling
3:50
Electron Configurations & Diagrams: H
8:41
Electron Configurations & Diagrams: He
9:12
Electron Configurations & Diagrams: Li
9:47
Electron Configurations & Diagrams: Be
11:17
Electron Configurations & Diagrams: B
12:05
Electron Configurations & Diagrams: C
13:03
Electron Configurations & Diagrams: N
14:55
Electron Configurations & Diagrams: O
15:24
Electron Configurations & Diagrams: F
16:25
Electron Configurations & Diagrams: Ne
17:00
Electron Configurations & Diagrams: S
18:08
Electron Configurations & Diagrams: Fe
20:08
Introduction to Valence Electrons
23:04
Valence Electrons of Oxygen
23:44
Valence Electrons of Iron
24:02
Valence Electrons of Arsenic
24:30
Valence Electrons: Exceptions
25:36
The Periodic Table
27:52
XIV. Intermolecular Forces
Vapor Pressure & Changes of State

52m 43s

Intro
0:00
Vapor Pressure and Changes of State
2:26
Intermolecular Forces Overview
2:27
Hydrogen Bonding
5:23
Heat of Vaporization
9:58
Vapor Pressure: Definition and Example
11:04
Vapor Pressures is Mostly a Function of Intermolecular Forces
17:41
Vapor Pressure Increases with Temperature
20:52
Vapor Pressure vs. Temperature: Graph and Equation
22:55
Clausius-Clapeyron Equation
31:55
Example 1
32:13
Heating Curve
35:40
Heat of Fusion
41:31
Example 2
43:45
Phase Diagrams & Solutions

31m 17s

Intro
0:00
Phase Diagrams and Solutions
0:22
Definition of a Phase Diagram
0:50
Phase Diagram Part 1: H₂O
1:54
Phase Diagram Part 2: CO₂
9:59
Solutions: Solute & Solvent
16:12
Ways of Discussing Solution Composition: Mass Percent or Weight Percent
18:46
Ways of Discussing Solution Composition: Molarity
20:07
Ways of Discussing Solution Composition: Mole Fraction
20:48
Ways of Discussing Solution Composition: Molality
21:41
Example 1: Question
22:06
Example 1: Mass Percent
24:32
Example 1: Molarity
25:53
Example 1: Mole Fraction
28:09
Example 1: Molality
29:36
Vapor Pressure of Solutions

37m 23s

Intro
0:00
Vapor Pressure of Solutions
2:07
Vapor Pressure & Raoult's Law
2:08
Example 1
5:21
When Ionic Compounds Dissolve
10:51
Example 2
12:38
Non-Ideal Solutions
17:42
Negative Deviation
24:23
Positive Deviation
29:19
Example 3
31:40
Colligatives Properties

34m 11s

Intro
0:00
Colligative Properties
1:07
Boiling Point Elevation
1:08
Example 1: Question
5:19
Example 1: Solution
6:52
Freezing Point Depression
12:01
Example 2: Question
14:46
Example 2: Solution
16:34
Osmotic Pressure
20:20
Example 3: Question
28:00
Example 3: Solution
30:16
XV. Bonding
Bonding & Lewis Structure

48m 39s

Intro
0:00
Bonding & Lewis Structure
2:23
Covalent Bond
2:24
Single Bond, Double Bond, and Triple Bond
4:11
Bond Length & Intermolecular Distance
5:51
Definition of Electronegativity
8:42
Bond Polarity
11:48
Bond Energy
20:04
Example 1
24:31
Definition of Lewis Structure
31:54
Steps in Forming a Lewis Structure
33:26
Lewis Structure Example: H₂
36:53
Lewis Structure Example: CH₄
37:33
Lewis Structure Example: NO⁺
38:43
Lewis Structure Example: PCl₅
41:12
Lewis Structure Example: ICl₄⁻
43:05
Lewis Structure Example: BeCl₂
45:07
Resonance & Formal Charge

36m 59s

Intro
0:00
Resonance and Formal Charge
0:09
Resonance Structures of NO₃⁻
0:25
Resonance Structures of NO₂⁻
12:28
Resonance Structures of HCO₂⁻
16:28
Formal Charge
19:40
Formal Charge Example: SO₄²⁻
21:32
Formal Charge Example: CO₂
31:33
Formal Charge Example: HCN
32:44
Formal Charge Example: CN⁻
33:34
Formal Charge Example: 0₃
34:43
Shapes of Molecules

41m 21s

Intro
0:00
Shapes of Molecules
0:35
VSEPR
0:36
Steps in Determining Shapes of Molecules
6:18
Linear
11:38
Trigonal Planar
11:55
Tetrahedral
12:45
Trigonal Bipyramidal
13:23
Octahedral
14:29
Table: Shapes of Molecules
15:40
Example: CO₂
21:11
Example: NO₃⁻
24:01
Example: H₂O
27:00
Example: NH₃
29:48
Example: PCl₃⁻
32:18
Example: IF₄⁺
34:38
Example: KrF₄
37:57
Hybrid Orbitals

40m 17s

Intro
0:00
Hybrid Orbitals
0:13
Introduction to Hybrid Orbitals
0:14
Electron Orbitals for CH₄
5:02
sp³ Hybridization
10:52
Example: sp³ Hybridization
12:06
sp² Hybridization
14:21
Example: sp² Hybridization
16:11
σ Bond
19:10
π Bond
20:07
sp Hybridization & Example
22:00
dsp³ Hybridization & Example
27:36
d²sp³ Hybridization & Example
30:36
Example: Predict the Hybridization and Describe the Molecular Geometry of CO
32:31
Example: Predict the Hybridization and Describe the Molecular Geometry of BF₄⁻
35:17
Example: Predict the Hybridization and Describe the Molecular Geometry of XeF₂
37:09
XVI. AP Practice Exam
AP Practice Exam: Multiple Choice, Part I

52m 34s

Intro
0:00
Multiple Choice
1:21
Multiple Choice 1
1:22
Multiple Choice 2
2:23
Multiple Choice 3
3:38
Multiple Choice 4
4:34
Multiple Choice 5
5:16
Multiple Choice 6
5:41
Multiple Choice 7
6:20
Multiple Choice 8
7:03
Multiple Choice 9
7:31
Multiple Choice 10
9:03
Multiple Choice 11
11:52
Multiple Choice 12
13:16
Multiple Choice 13
13:56
Multiple Choice 14
14:52
Multiple Choice 15
15:43
Multiple Choice 16
16:20
Multiple Choice 17
16:55
Multiple Choice 18
17:22
Multiple Choice 19
18:59
Multiple Choice 20
20:24
Multiple Choice 21
22:20
Multiple Choice 22
23:29
Multiple Choice 23
24:30
Multiple Choice 24
25:24
Multiple Choice 25
26:21
Multiple Choice 26
29:06
Multiple Choice 27
30:42
Multiple Choice 28
33:28
Multiple Choice 29
34:38
Multiple Choice 30
35:37
Multiple Choice 31
37:31
Multiple Choice 32
38:28
Multiple Choice 33
39:50
Multiple Choice 34
42:57
Multiple Choice 35
44:18
Multiple Choice 36
45:52
Multiple Choice 37
48:02
Multiple Choice 38
49:25
Multiple Choice 39
49:43
Multiple Choice 40
50:16
Multiple Choice 41
50:49
AP Practice Exam: Multiple Choice, Part II

32m 15s

Intro
0:00
Multiple Choice
0:12
Multiple Choice 42
0:13
Multiple Choice 43
0:33
Multiple Choice 44
1:16
Multiple Choice 45
2:36
Multiple Choice 46
5:22
Multiple Choice 47
6:35
Multiple Choice 48
8:02
Multiple Choice 49
10:05
Multiple Choice 50
10:26
Multiple Choice 51
11:07
Multiple Choice 52
12:01
Multiple Choice 53
12:55
Multiple Choice 54
16:12
Multiple Choice 55
18:11
Multiple Choice 56
19:45
Multiple Choice 57
20:15
Multiple Choice 58
23:28
Multiple Choice 59
24:27
Multiple Choice 60
26:45
Multiple Choice 61
29:15
AP Practice Exam: Multiple Choice, Part III

32m 50s

Intro
0:00
Multiple Choice
0:16
Multiple Choice 62
0:17
Multiple Choice 63
1:57
Multiple Choice 64
6:16
Multiple Choice 65
8:05
Multiple Choice 66
9:18
Multiple Choice 67
10:38
Multiple Choice 68
12:51
Multiple Choice 69
14:32
Multiple Choice 70
17:35
Multiple Choice 71
22:44
Multiple Choice 72
24:27
Multiple Choice 73
27:46
Multiple Choice 74
29:39
Multiple Choice 75
30:23
AP Practice Exam: Free response Part I

47m 22s

Intro
0:00
Free Response
0:15
Free Response 1: Part A
0:16
Free Response 1: Part B
4:15
Free Response 1: Part C
5:47
Free Response 1: Part D
9:20
Free Response 1: Part E. i
10:58
Free Response 1: Part E. ii
16:45
Free Response 1: Part E. iii
26:03
Free Response 2: Part A. i
31:01
Free Response 2: Part A. ii
33:38
Free Response 2: Part A. iii
35:20
Free Response 2: Part B. i
37:38
Free Response 2: Part B. ii
39:30
Free Response 2: Part B. iii
44:44
AP Practice Exam: Free Response Part II

43m 5s

Intro
0:00
Free Response
0:12
Free Response 3: Part A
0:13
Free Response 3: Part B
6:25
Free Response 3: Part C. i
11:33
Free Response 3: Part C. ii
12:02
Free Response 3: Part D
14:30
Free Response 4: Part A
21:03
Free Response 4: Part B
22:59
Free Response 4: Part C
24:33
Free Response 4: Part D
27:22
Free Response 4: Part E
28:43
Free Response 4: Part F
29:35
Free Response 4: Part G
30:15
Free Response 4: Part H
30:48
Free Response 5: Diagram
32:00
Free Response 5: Part A
34:14
Free Response 5: Part B
36:07
Free Response 5: Part C
37:45
Free Response 5: Part D
39:00
Free Response 5: Part E
40:26
AP Practice Exam: Free Response Part III

28m 36s

Intro
0:00
Free Response
0:43
Free Response 6: Part A. i
0:44
Free Response 6: Part A. ii
3:08
Free Response 6: Part A. iii
5:02
Free Response 6: Part B. i
7:11
Free Response 6: Part B. ii
9:40
Free Response 7: Part A
11:14
Free Response 7: Part B
13:45
Free Response 7: Part C
15:43
Free Response 7: Part D
16:54
Free Response 8: Part A. i
19:15
Free Response 8: Part A. ii
21:16
Free Response 8: Part B. i
23:51
Free Response 8: Part B. ii
25:07
Loading...
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of AP Chemistry
  • Discussion

  • Download Lecture Slides

  • Table of Contents

  • Transcription

  • Related Books & Services

Lecture Comments (10)

3 answers

Last reply by: Professor Hovasapian
Fri Nov 18, 2016 8:43 PM

Post by Kaye Lim on October 24, 2016

I have 2 questions:

1/ The vapor pressure of water at 25*C (P* water solvent)= 23.76 torr. Is this fixed value the measurement for 1 mol or 18g of water at 25*C? If not, then what is the amount of water for this P*solvent value?

2/ Why the density of solid water is 1, but liquid water is less than 1? Lets say for 1 mol of water (18g), if this amount of water is ice, doesn't it occupy more volume than the same 18g of water in liquid based on the fact that H2O expands going from liquid to solid? Then why would density of ice bigger if the Volume on the denominator is bigger for ice?

Thank you!

1 answer

Last reply by: Professor Hovasapian
Thu Dec 17, 2015 1:46 AM

Post by Gaurav Kumar on December 14, 2015

Hi Professor,

I have one more question about the last example. Did you use the measured value of 265 torr in your work? Wouldn't you have to compare 265 torr with 319 torr to see whether or not it is an ideal solution or not?


Thank you!  

1 answer

Last reply by: Professor Hovasapian
Thu Dec 17, 2015 1:38 AM

Post by Gaurav Kumar on December 14, 2015

Hi Professor,

i am a little confused on what you said about how we need to take in account for the total number of free particles when ionic compounds dissolve. Can you please clarify what a free particle is and what it means to take them into account.

Thank you so much!

1 answer

Last reply by: Professor Hovasapian
Thu Dec 17, 2015 1:32 AM

Post by Tammy T on December 13, 2015

Hello Prof. H!

For the Raoult's law equation, I was given a somewhat different equation with extra variables in it.
The equation given in my class is Pi = ai.P*i = yi xi P*i
i stands for component i
ai = activity of component i or also "effective concentration"
yi = Raoult's Law activity coefficient for component i
xi = mol fraction of component i

I understand the equation given in your lecture, but how come the equation given in my class is so different. I was wonder if you could please run through the equation above to explain the meaning for those extra variables and why the need for those when everything was fine using just your given Raoult's law.

Your lectures help me understand chemistry so much. Thank you!

Vapor Pressure of Solutions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Vapor Pressure of Solutions 2:07
    • Vapor Pressure & Raoult's Law
    • Example 1
    • When Ionic Compounds Dissolve
    • Example 2
    • Non-Ideal Solutions
    • Negative Deviation
    • Positive Deviation
    • Example 3

Transcription: Vapor Pressure of Solutions

Hello, and welcome back to Educator.com; welcome back to AP Chemistry.0000

Today, we are going to be talking about the vapor pressure of solutions.0003

Now, we have talked about vapor pressure before, and that is basically: at any given temperature, there is always going to be some of the molecules on the surface of a liquid that have escaped into the gas phase.0008

So, in a closed container, there is always going to be some gas above the particular liquid--like, if I have water in a closed container, yes, you see the liquid water there; but on top of the liquid water, there is a little bit of water vapor, at every temperature--not just at 100 degrees Celsius.0018

100 degrees Celsius is the boiling temperature of water; that is when every single molecule of liquid water has enough energy to go and stay in the vapor state.0036

But, in every liquid, there is always some amount of gas above it.0045

Well, again, pressure is just a measure of the number of particles in a particular closed container; so we call that thing the vapor pressure of the solvent.0051

Well, today we are going to be talking about the vapor pressure of solutions; if you take a non-volatile solute (like if you take salt and dissolve it in some water), that actually ends up changing the vapor pressure of the solvent.0059

Now, you can call it different things, and in the process of discussing it, sometimes I'm going to say "vapor pressure of solute"; sometimes I'm going to say "vapor pressure of solution"; we'll know what we are talking about if we are talking about pure solvent or some solution.0071

But, in essence, when I put a solute (like salt) in water and dissolve it, now the water molecules are actually competing with other particles (namely, the sodium ion and the chloride ion) to make their way up to the surface, so they can actually escape into the gas phase.0086

Well, because these ions are actually running interference, not as many water molecules make it to the surface; therefore, not as many jump up into the vapor phase; therefore, the vapor pressure (which is a measure of how much is on top of the liquid at any given time) is going to be lower, in general.0104

Let's go ahead and get started; that is just a quick description of what it is that we are going to do.0122

OK, so let's write down: the vapor pressure of a...no, not "the vapor pressure of a"...what am I saying here?--I can't write that.0127

The presence (I'm getting ahead of myself) of a non-volatile solute (and non-volatile means, generally, it is not a liquid; it's a solid of some sort--a salt or sugar--some covalent compound that forms a crystal) lowers the vapor pressure of the solvent.0148

And again, if you need to do a quick review of vapor pressure, just go back some lessons, where we actually talk about what vapor pressure is.0184

We have actually also talked about it twice, if you remember--when we did gases, when we were collecting a gas above a liquid and using that technique of displacement to measure the volume of a gas, we talked about vapor pressure then.0191

So, we have actually discussed it a couple of times.0203

All right, vapor pressure of a solvent...now, this is our important equation: it's something called Raoult's Law, and it goes something like this.0207

You know what, I'm always worried when I use different variables; I actually like to write my variables out; I don't really care for letters, myself, personally--because there are so many letters in science that you collect, after a certain number of classes, that they tend to get confusing.0218

Well, you know what, let me just go ahead and leave it as I have.0236

The vapor pressure of the solution is equal to the mole fraction of the solvent (it's very, very important to identify, because now we have solvent and solute; we have to specify what we are talking about), times the initial vapor pressure of the solvent.0240

So, Raoult's Law says that, if I have a particular solution, and there is some solute that has been put into solvent to create the solution, the vapor pressure that I end up measuring under ideal conditions--it will be: I take the normal vapor pressure of the pure solvent, and I multiply it by the mole fraction of the solvent.0258

So now, the mole fraction is no longer 1; there is actually solvent and solute in the solution; therefore, the mole fraction is going to be less than 1.0283

Remember what mole fraction is: the moles of the A over the total moles.0295

So, in this case, it's going to be the moles of solvent over moles of solute plus moles of solvent.0299

It is going to be less than 1, so your pressure is actually going to be less than the normal pressure, which makes sense, because again, molecules--the solvent molecules--are interfering with the solvent molecules' ability to get to the surface so they can jump up into the vapor phase.0305

Let's do an example.0321

We'll do Example 1: OK, now, what is the expected vapor pressure (I'll just call it V/P) at 25 degrees Celsius, for a sucrose solution of 120.0 grams of sucrose, dissolved in (oops, it's not 572; I want...) 527.0 milliliters of water?0326

OK, now we're going to have some information that we need here: At 25 degrees Celsius, the density of water (that's a little Greek delta) equals 0.9971 grams per milliliter.0374

The vapor pressure, the normal vapor pressure, of H2O (the solvent) equals 23.76 torr.0393

That means that, at 25 degrees Celsius (just really briefly), if I have a glass of water that is covered, at 25 degrees Celsius, the amount of vapor--water vapor--above the liquid in the glass actually accounts for 23.76 torricelli.0407

Not a lot, but it is some pressure; it is some gas particles that are bouncing around, hitting the walls of the container, causing a pressure; that is what vapor pressure is.0425

It is the measure of how much vapor is actually above a liquid at a given temperature.0434

Sorry to keep repeating myself, but I find that repetition sort of enforces comfort with these things.0439

Now, sucrose (which is just normal table sugar--so we're just making a sugar solution) is 342.3 grams per mole; so these are the things that we are going to need.0447

OK, so Raoult's Law: this is the law that we are going to use; let me go ahead and circle this in red.0460

That is what we want to use; so we need to find--we have this one already; that is the 23.76 torr; we need to find the mole fraction of water, the solvent.0465

We go...so we are going to need to find, first of all, the moles of sucrose; well, the moles of sucrose--we said we had 120 grams, times 1 mole (nice stoichiometric conversion), over 342.3 grams; that gives us 0.3506 mol.0478

You know, let me write my numbers a little bit bigger; I think they might be a little tiny here: 0.3506 moles of sucrose.0500

OK, now we need the moles of H2O.0509

Well, H2O...we have 527 milliliters, so we need to convert that to grams first, so that is why we need the density.0513

We have 0.9971 grams per milliliter, times (grams goes down here; mol goes up there) 18 grams in 1 mole; so it's going to be 527, times .9971, divided by 18; and we get 29.193 moles.0521

29.193 mol: therefore, the mole fraction of H2O is equal to the number of moles of water, divided by the total moles (which is 29.193+0.3506).0543

When I do that, I get 0.9881; so it's going to be less than 1; it's a fraction--it's a part over the whole.0565

If we wanted the mole fraction of sucrose, well, we would say .3506 in the numerator; the denominator is the total moles; I hope that is OK.0574

OK, so now, when we plug this into the equation, the vapor pressure of the solution is going to be 0.9881 (which is the mole fraction of water), times the normal vapor pressure at that temperature (23.76), and we end up with 23.47 torricelli (or torr, or millimeters of mercury...if you want to convert that to atmospheres, psi, kilopascal, whatever you need; it's just a unit of pressure).0583

Notice, 23.47 is not a lot less than 23.76, but it is less.0615

That is what is important: when you add a solute to a solvent (in other words, when you create a solution--a non-volatile solute), your vapor pressure will lower.0621

What that means is that fewer molecules of liquid will hang around in the vapor state on top of the liquid, because fewer molecules will be able to get to the surface, because the molecules of solute are getting in their way, so they can't jump up as much.0633

OK, now, this was for a covalent compound: sugar is a covalent compound--it dissolves, but when it dissolves, it doesn't actually break up into its individual atoms.0650

It just dissolves: the crystal comes apart--the molecules separate.0661

Ionic compounds are different; so, when ionic compounds dissolve, they dissociate; in other words, they separate into their ions; so we must account for the total number of dissolved particles--that is what is important: the total number of free particles.0665

For example, if I had some magnesium chloride that I dropped into water, well, I have: 1 mole of magnesium chloride produces 1 mole of magnesium ions, plus 2 moles of chloride ions.0713

So, I have to account for...whatever the number of moles is that I get of that, it is actually 3 times that many particles.0727

1 mole of magnesium chloride produces 3 moles of free-flowing particles--of free particles--so it's actually 3 that is what I use when I calculate the mole fraction, OK?0737

That is it; that is the only thing that is different--you have to differentiate between ionic compounds and covalent compounds by accounting for the number of free particles that are floating around.0751

Let's do an example.0761

This is going to be Example #2.0765

OK, we have: 40.0 grams of solid potassium sulfate is dissolved (let's actually write it so we can read it; that might be nice) in 180.0 grams of water.0770

So, this time, they went ahead and they gave you the mass of the water directly, instead of volume--not a big deal.0796

OK, what is the vapor pressure of solution?0802

OK, well, again, we are going to use Raoult's Law; so, since I don't have it on this page, let me write it one more time.0812

The vapor pressure of solution is equal to the mole fraction of solvent, times the initial vapor pressure of solvent.0819

OK, well, we have the vapor pressure of solvent; that is going to end up being...it looks like this is going to be...I'm sorry, I didn't specify: this is 25 degrees Celsius, so we are going to use the 23.46; that is water.0830

The initial of the solvent is going to be 23.76 torr; so that is that one.0842

We need to find the mole fraction of the solvent; OK.0856

Well, let's find the number of moles of water; so moles of H2O: we have 180.0 grams; 1 mole is 18 grams, 18.0 grams; so we are looking at 10 moles of water.0859

OK, now let's find the number of moles of K2SO4, potassium sulfate.0876

Well, we have 40 grams of the potassium sulfate; and we want to do 1 mole of it; when we check the molar mass using a periodic table, we get 174.2 grams (I hope I did the arithmetic correctly--if not, not a problem--it's just a number), equals 0.2296 moles.0882

OK, now here is where we have to be careful: K2SO4...when K2SO4 dissociates, it produces 2 moles of potassium ions, plus 1 mole of sulfate ion.0905

1 mole of potassium sulfate produces 3 moles of particles--3 moles of free particles.0919

That is what matters: the identity of the species don't matter; these are colligative properties--these are...what matters is the number of particles that are floating around in solution.0930

The more number of particles you have floating around in solution, the more they get in the way of the solvent doing what it needs to do, which is escape into the vapor phase.0940

3 moles of free particles are formed.0949

So, when we do our mole fraction, our mole fraction is going to look like this.0954

The mole fraction of H2O is equal to...well, 10 moles of H2 over 10 plus three times the .2296 (right? .2296 is the number of moles of potassium sulfate, times 3 because, for every mole of potassium sulfate, three moles of free particles are going to be floating around in that solution).0962

So, you get: 0.9356; therefore, the vapor pressure of solution is equal to 0.9356 (let me make the numbers a little bit bigger; I don't know why--I usually have a habit of doing big numbers, and lately I know I have been doing small numbers), times 23.76 torricelli, which gives us 22.23 torr.0986

There you go: the only difference--handle it exactly the same way, but when you are dealing with an ionic compound, make sure you account for the dissociation of the ion and the number of free particles that are floating around--very important.1023

OK, so the identity of the dissolved species doesn't matter, just how many; it doesn't matter if it's sucrose or potassium sulfate or whatever.1035

It is how many particles, not the identity--just like gases, remember?--the identity of the gas doesn't matter--1 mole of gas at standard temperature and pressure always occupies 22.4 liters.1043

All gases--it doesn't matter what the identity of the gas is; what matters is the number of particles.1055

OK, so now, let's discuss non-ideal solutions.1063

What we have been discussing are ideal solutions; now, we will discuss non-ideal solutions.1066

This is analogous to discussing the non-ideal gas law versus the ideal gas law: the ideal gas law can be...actually, for most purposes, it's just fine; and Raoult's Law, for most purposes, is actually just fine.1075

But this is an experimental science--it is chemistry--so when we do the experiment, it doesn't always fit with the theory, with the numbers, and we need to actually explain why that is the case.1088

So, let's explain why.1099

OK, so: so far (well, you know what, I'm not going to write this part out), what we have assumed is that the solute is non-volatile--that is some solid crystal that you drop in the water; you mix it up; you dissolve it; it's non-volatile.1102

In other words, the solute itself that you put in the solvent doesn't have a vapor pressure of its own.1115

But, if you mix a liquid and a liquid (like methanol in water, or hexane in benzene, or sodium acetate in pentane--something like that), it is a liquid-liquid.1122

Well, all liquids have a certain vapor pressure; so now, when you mix a liquid and a liquid, it is still just a solution; you have some liquid that you are putting into another liquid, some solute that you are putting into a solvent.1142

Now, both of those liquids--the molecules can escape into the vapor phase.1155

The vapor pressure of solution is now a sum of the mole fractions of each, times the vapor pressures of each.1162

Let's see--so non-ideal solutions: for liquid-liquid solutions, where both are volatile, Raoult's Law becomes: the vapor pressure of solution is equal to the mole fraction of A, times the vapor pressure of A (pure A), plus the mole fraction of B, times the vapor pressure of pure B.1171

That is it: we are just sort of combining the Raoult's Law for the individual liquids, but now we are mixing both liquids, so we have to account for both of their partial pressures.1222

The total pressure of the solution--the total number of molecules of vapor--some of the vapor molecules are going to be of A; some of the molecules are going to be of B; each one of them contributes to the total pressure (the total number of particles above the liquid at a given temperature).1233

OK, so let me draw this out so you see what is going on.1248

What I am going to draw is an ideal situation: so under ideal conditions, this is what we would expect; this is the kind of behavior we would expect.1255

Let me go back to blue.1263

Now, I'm going to go from here; I'm going to go up to here; then, I'm going to go here; I'm going to go down this way, and I'll...the mole fraction of A increases as we move in this direction; the mole fraction of B increases as we move in this direction.1265

Therefore, at this point, this is pure B; and at this point, we have pure A, right?1285

So, if we start with absolutely no A and pure B, well, that is what this represents.1295

As we add more A, less B, more A, less B...this here represents the mixture, if you will: a certain amount of B, a certain amount of A; a certain amount of B, a certain amount of A; a certain amount of B, a certain amount of A; that is all this is.1301

OK, and they are represented by mole fraction: as mole fraction goes from 0 to 1...well, for pure A, when it's 1, it is pure A; when the mole fraction of B is 1, it's pure B.1320

There is any other combination; so if you want to draw vertical lines, that just represents how much B we have and how much A we have in a given solution.1331

Well, the vapor pressure is going to look like this.1339

It is going to be this number, plus that number.1349

Again, you are just adding graphs; that is all that is happening here.1352

It is this number plus that number, and you are going to get this number; so this actually, down here, doesn't really matter all that much; what we want you to realize is that the vapor pressure of the solution is the sum of the mole fractions of each, times the standard vapor pressure of that liquid alone.1356

That is what this is saying: you are just adding two functions to get a final function.1376

You have been doing this for years in your precalculus and your algebra classes, so it shouldn't be a problem.1380

Now, this is ideal behavior right here.1385

In other words, all we have done is take 2 things that are volatile and added them together.1394

This is what should happen--what we expect to happen, theoretically.1399

OK, when the two liquids are very much alike, this happens--when the two liquids are very much alike.1403

What we mean by "very much alike": we mean they have similar properties and similar structures--"very much alike."1415

So, for example, if you had something like water and ethanol: ethanol is just regular alcohol--the alcohol that you drink; the structures of those two are actually very, very similar, and because they are very, very similar, they are actually going to behave in an ideal fashion.1422

Now, we will talk about how, actually, it deviates from ideal behavior; but any time you put two liquids together--when you mix two liquids to make a solution--if the two liquids have a lot of things in common (a lot of properties, and are very much alike structurally), what you are going to get is ideal behavior.1441

In other words, your vapor pressure of solution is going to equal this right here.1459

Now, let's talk about deviation from ideal behavior.1464

There are two ways to deviate from ideal behavior: there is something called negative deviation, and what that means is: when the two liquids have a special affinity for each other, such as hydrogen bonding (and more often than not, it is going to be hydrogen bonding that sort of creates the negative deviation), the Psolution measured is lower than expected by Raoult's Law.1468

So, the equation that I just gave--that will give you an expected vapor pressure, but because the two liquids have a special affinity for each other (for example, water and ethanol--there is hydrogen bonding that takes place between those two)--because of that, the vapor pressure that we measure is going to be less than what we calculate by this law.1560

That is negative deviation, and what this looks like is the following.1582

I'm going to draw the same graph that I had...well, not graph--the same diagram that I had on the previous page.1588

And so, let me go this way, except--notice that I have reversed the colors this time.1595

I have one going that way; so this is...that would be ideal behavior; negative deviation looks like this.1600

This is lower; this is lower than expected; therefore, this is going to be lower than expected.1615

Let me go over that one in black.1625

Don't worry so much about these; this is the one that I am concerned about.1633

Ideal behavior would be right there: this is what the equation that I just gave you would predict.1636

However, because these things actually have a greater affinity for each other (in this case, hydrogen bonding--for example, hydrogen bonding), the vapor pressure that we measure is less than what we expect or calculate from this equation.1641

This is called negative deviation, and again, it makes perfect sense.1655

If you have two liquids that you would expect to just sort of treat...one liquid mixes with another liquid, so you just add up their mole fractions times their standard vapor pressures, and you should get a number; that number should be somewhere on that line, depending on the relative amounts of each liquid that you use.1659

However, because of the hydrogen bonding, this ethanol and this water actually are attracted to each other more; therefore, they will not let go of each other all that much so that they can go off into the vapor phase.1678

You actually need more energy to make that happen; but at a given temperature, the energy is fixed; therefore, the vapor pressure will actually be lower than you expect.1689

That is all that it is saying; it is saying that, when you measure a vapor pressure that is lower than you expect from calculation, that means it's a negative deviation--that means the two liquids have an extra-special affinity for each other.1700

That means they want to be together--they like being together.1712

The opposite of this is positive deviation.1716

Oh, let me give you an example of this one; so, for example, if I mix acetone (which has a molecular structure like this--wow, you know what, I like black--I think I'll stick with that) and water--if you mix acetone and water, well, there is some serious hydrogen bonding (let me draw it another way, so you can actually see the hydrogen bonding): this is going to be H; this is going to be O; this is going to be H.1718

There is hydrogen bonding; because of that hydrogen bonding, the acetone and the water are actually more attracted to each other; therefore, the vapor pressure we measure of their mixture is less than what we would expect: negative deviation.1746

OK, now let's do positive deviation.1759

Positive deviation is the opposite: it is when the two liquids' interactions are weaker than the solvents alone--weaker than the solvent and solute alone.1765

So, for example, if I had hexane--pure hexane--the hexane molecules interact with each other.1802

If I had water, the water molecules interact with each other.1808

If I put hexane and water together, hexane and water are not very much alike; in fact, they do not like to be together (you know this: oil and water--they don't mix).1811

Hexane is just a smaller version of oil, if you will--it's a hydrocarbon, not polar...and we will talk about what polar means; but water and hexane are not very much alike; they don't like to be together.1819

Therefore, they actually, literally, want to be apart; because they want to be apart, it is more likely that they will push away from each other and actually go into the vapor phase sooner, because they don't like being together.1831

Whereas, up here, these molecules that have a special affinity do like being together; so a positive deviation (if we stick to this graph, I'm going to do the positive deviation in blue)--what the Raoult's Law equation says for a given concentration--I should be here; but when I measure the vapor pressure, there is actually more vapor above the liquid.1845

What that means is that the molecules, mixed together--they don't want to be together, so they actually will go out of their way to escape into the vapor phase to avoid being so close together in the liquid form.1876

The vapor pressure is actually up here: the measured is higher than what I expect.1888

This is called positive deviation.1893

I'll let you think about that a little bit, and I will finish off with an example here.1896

Let's do this in black; so we have Example 3.1901

5.8 grams of acetone is mixed with 12 grams of CHCl3; OK, now, at 35 degrees Celsius, the measured vapor pressure of solution is equal to 265 torr.1909

OK, is this an ideal solution?1949

In other words, does it actually obey Raoult's Law?1958

Well, let's use Raoult's Law, calculate what we expect, compare it to what we measured (which is 265 torr), and see if it is lower or higher than that.1961

If it's lower than that, it's negative deviation; if it's higher than that, it's positive deviation; in either case, it's a non-ideal solution.1969

If it is close to 265 (you know, in the 260 to 270 range), it's an ideal solution.1976

OK, so our ideal is: solution equals the mole fraction of acetone (I'll just call it A...yes, that is fine), times the normal vapor pressure of acetone, plus the mole fraction of (I'll call this...I'll just do CHCl3...here we go again: chemistry with all of its symbolism, all over the place).1982

OK, so let's calculate the number of moles of CHCl3.2014

We have 12 grams, and 1 mole of that is actually equal to 119.4, equals 0.1005 mol.2021

And then, moles of C3H6O (that is the acetone): we have 5.8 grams, times 1 mole, which is 58.1 grams; OK, and we get 0.0998 mol.2037

Therefore, the mole fraction of acetone is equal to 0.0998, divided by 0.1005 plus 0.0998; you end up with 0.5.2056

And the mole fraction of the CHCl3 is equal to 0.1005, divided by 0.1005, plus 0.0998; and you actually end up with, again, .5.2076

So you see that the mole fraction of each is .5.2093

Now, our P solution is equal to the mole fraction of A (0.5), times its vapor pressure of the pure solvent at that temperature.2097

In the case of acetone at 35 degrees Celsius, it is 293 torr.2115

That is information that is actually given to you, or you can look it up.2120

...plus 0.5, which is the mole fraction of the CHCl3, and its vapor pressure (35 degrees Celsius, pure solvent) is 345 torr--that is going to equal 319 torr.2124

So, what I would expect under normal circumstances, without accounting for affinity or disaffinity, is 319 torr.2140

Well, what I measured was 293 torr; clearly, 293 torr is less than 319 torr, so this is negative deviation.2148

Negative deviation means that the acetone and the CHCl3 have some sort of an extra affinity to each other.2158

Well, let's see what that affinity is.2165

As it turns out, it is going to be exactly what you think: hydrogen bonding.2167

This is acetone; this is the H; this extra bonding--the hydrogen bonding--this intermolecular force between the acetone and the CHCl3--causes them to actually stick together more than they normally would.2173

Because they stick together more than they normally would, there are fewer molecules jumping out of solution into the vapor phase.2195

Because there are fewer molecules in the vapor phase, the vapor pressure that I measured is lower than what I expected.2204

That is what is going on; so that is it--standard Raoult's Law; I think the equation is pretty simple, and hopefully you understand what it is that is going on.2211

It is just: when you mix things, because now you don't have a pure solvent--you have other particles floating around in solution--they are actually getting in the way of the solvent particles from reaching the surface in order to break out into the vapor phase.2221

That is all that is going on: if you understand that physically, you will understand everything else.2234

Thank you for joining us here at Educator.com.2239

We'll see you next time; goodbye.2241

Educator®

Please sign in for full access to this lesson.

Sign-InORCreate Account

Enter your Sign-on user name and password.

Forgot password?

Start Learning Now

Our free lessons will get you started (Adobe Flash® required).
Get immediate access to our entire library.

Sign up for Educator.com

Membership Overview

  • Unlimited access to our entire library of courses.
  • Search and jump to exactly what you want to learn.
  • *Ask questions and get answers from the community and our teachers!
  • Practice questions with step-by-step solutions.
  • Download lesson files for programming and software training practice.
  • Track your course viewing progress.
  • Download lecture slides for taking notes.