How to Use This Presentation

Slides:



Advertisements
Similar presentations
Chapter 13 Preview Multiple Choice Short Answer
Advertisements

Chapter 13- Unit 2 Colligative Properties - are properties of solutions that depend on the number of molecules in a given volume of solvent and not on.
Ions in aqueous Solutions And Colligative Properties
Ions in Aqueous Solutions and Colligative Properties
Chapter 13 Preview Lesson Starter Objectives Dissociation Ionization
1 Ions in Aqueous Solutions and Colligative Properties Chapter 14 Chemistry chapter 14.
Ions in Aqueous Solutions and Colligative Properties
Chapter 13: Ions in Aqueous Solutions and Colligative Properties.
Chapter 13 Section 1 Dissociation, Ionization, Electrolytes & Net Ionic Equations.
Section 1 Compounds in Aqueous Solution
Colligative Properties. Properties that depend upon the concentration of solute particles are called colligative properties. Generally these properties.
Chapter 13 Table of Contents Section 1 Compounds in Aqueous Solutions
Ions in Aqueous Solutions and Colligative Properties
Chapter #14 Ions in Aqueous Solutions and Colligative Properties.
Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter Presentation Transparencies Lesson Starters Standardized.
Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu How to Use This Presentation To View the presentation as a slideshow.
Chapter 14 Ions in Aqueous Solutions & Colligative Properties.
Modern Chemistry Chapter 13
Ions in Aqueous Solutions and Colligative Properties Chapter 13 Preview Image Bank Dissociation of NaCl General Solubility Guidelines Soluble and Insoluble.
Chapter 13 Ions in Aqueous Solutions and Colligative Properties.
Aqueous Solutions Compound in Aqueous Solutions.  Objectives  Write equations for the dissolution of soluble ionic compounds in water  Predict whether.
Unit 8 Solution Chemistry
Ions in Aqueous Solutions and Colligative Properties.
Colligative Properties
Colligative Properties
Preview Lesson Starter Objectives Dissociation Ionization Strong and Weak Electrolytes Chapter 13.
© Houghton Mifflin Harcourt Publishing Company Preview Lesson Starter Objectives Dissociation Ionization Strong and Weak Electrolytes Chapter 13.
Preview Lesson Starter Objectives Dissociation Ionization Strong and Weak Electrolytes Chapter 13.
Ions in Aqueous Solutions
Chapter 13 Preview Lesson Starter Objectives
COLLIGATIVE PROPERTIES
Chapters 12-13: Solutions Modern Chemistry.
Colligative Properties
COLLIGATIVE PROPERTIES
Colligative properties
V. Colligative Properties
Heterogeneous vs. Homogeneous
Colligative Properties
Drill – 4/11/08 What two factors determine if a substance is in the solid, liquid, or gas phase? How would you define boiling point? Melting point?
Modern Chemistry Chapter 13
12.6 NOTES Colligative Properties
Colligative Properties HONORS CHEMISTRY
GENERAL PROPERTIES OF SOLUTIONS
AQUEOUS SOLUTIONS HOLT MODERN CHEMISTRY.
Ions in Solution Chapter 14.
Ch 13 Ions in Aq Soln and Colligative Properties 13
Colligative Properties of Solutions
Section 1 Types of Mixtures
Section 1 Compounds in Aqueous Solution
Chapter 13 Preview Lesson Starter Objectives Dissociation Ionization
Colligative Properties
* 07/16/96 SOLUTIONS *.
Colligative Properties of Solutions
Chapter 13-2 Colligative Properties of Solutions
Colligative Properties of Solutions
Section 1 Compounds in Aqueous Solution
Colligative Properties
Section 1 Compounds in Aqueous Solution
Chapter 13 – Aqueous Solutions
CHEMISTRY Matter and Change
CHEMISTRY Matter and Change
Chapter 12: Solutions Mrs. Taylor HAHS H. Chem 1B.
Modern Chemistry Chapter 13
How to Use This Presentation
Heterogeneous vs. Homogeneous
Colligative Properties of Solutions (chapter 16)
4B Chapter 13 Notes.
Chapter 13 Section 5: Colligative Properties
Compounds in Aqueous Solution
Chapter 13 Solutions.
Presentation transcript:

How to Use This Presentation To View the presentation as a slideshow with effects select “View” on the menu bar and click on “Slide Show.” To advance through the presentation, click the right-arrow key or the space bar. From the resources slide, click on any resource to see a presentation for that resource. From the Chapter menu screen click on any lesson to go directly to that lesson’s presentation. You may exit the slide show at any time by pressing the Esc key.

Standardized Test Prep Resources Chapter Presentation Lesson Starters Transparencies Sample Problems Visual Concepts Standardized Test Prep

Chapter 13 Table of Contents Section 1 Compounds in Aqueous Solutions Ions in Aqueous Solutions and Colligative Properties Chapter 13 Table of Contents Section 1 Compounds in Aqueous Solutions Section 2 Colligative Properties

Chapter 13 Lesson Starter Section 1 Compounds in Aqueous Solution Chapter 13 Lesson Starter Compare the composition and arrangement of particles in the solid crystals of CuSO4•5H2O with those in the solution.

Section 1 Compounds in Aqueous Solution Chapter 13 Objectives Write equations for the dissolution of soluble ionic compounds in water. Predict whether a precipitate will form when solutions of soluble ionic compounds are combined, and write net ionic equations for precipitation reactions. Compare dissociation of ionic compounds with ionization of molecular compounds.

Chapter 13 Objectives, continued Section 1 Compounds in Aqueous Solution Chapter 13 Objectives, continued Draw the structure of the hydronium ion, and explain why it is used to represent the hydrogen ion in solution. Distinguish between strong electrolytes and weak electrolytes.

Section 1 Compounds in Aqueous Solution Chapter 13 Dissociation Dissociation is separation of ions that occurs when an ionic compound dissolves. 1 mol 1 mol 1 mol 1 mol 1 mol 2 mol

Chapter 13 Dissociation of NaCl Section 1 Compounds in Aqueous Solution Chapter 13 Dissociation of NaCl

Dissociation, continued Section 1 Compounds in Aqueous Solution Chapter 13 Dissociation, continued Sample Problem A Write the equation for the dissolution of aluminum sulfate, Al2(SO4)3 , in water. How many moles of aluminum ions and sulfate ions are produced by dissolving 1 mol of aluminum sulfate? What is the total number of moles of ions produced by dissolving 1 mol of aluminum sulfate?

Dissociation, continued Section 1 Compounds in Aqueous Solution Chapter 13 Dissociation, continued Sample Problem A Solution Given: amount of solute = 1 mol Al2(SO4)3 solvent identity = water Unknown: a. moles of aluminum ions and sulfate ions b. total number of moles of solute ions produced Solution:

Dissociation, continued Section 1 Compounds in Aqueous Solution Chapter 13 Dissociation, continued Precipitation Reactions Although no ionic compound is completely insoluble, compounds of very low solubility can be considered insoluble for most practical purposes.

Section 2 Colligative Properties of Solutions Chapter 13 Objectives List four colligative properties, and explain why they are classified as colligative properties. Calculate freezing-point depression, boiling-point elevation, and solution molality of nonelectrolyte solutions. Calculate the expected changes in freezing point and boiling point of an electrolyte solution. Discuss causes of the differences between expected and experimentally observed colligative properties of electrolyte solutions.

Colligative Properties of Solutions Section 2 Colligative Properties of Solutions Chapter 13 Colligative Properties of Solutions Properties that depend on the concentration of solute particles but not on their identity are called colligative properties. Vapor-Pressure Lowering Freezing-Point Depression Boiling-Point Elevation Osmotic Pressure

Vapor-Pressure Lowering Section 2 Colligative Properties of Solutions Chapter 13 Vapor-Pressure Lowering A nonvolatile substance is one that has little tendency to become a gas under existing conditions. The boiling point and freezing point of a solution differ from those of the pure solvent. A nonvolatile solute raises the boiling point and lowers the freezing point.

Comparing Volatile and Nonvolatile Liquids Visual Concepts Chapter 13 Comparing Volatile and Nonvolatile Liquids Click below to watch the Visual Concept. Visual Concept

Section 2 Colligative Properties of Solutions Chapter 13

Vapor Pressures of Pure Water and a Water Solution Section 2 Colligative Properties of Solutions Chapter 13 Vapor Pressures of Pure Water and a Water Solution

Freezing-Point Depression Section 2 Colligative Properties of Solutions Chapter 13 Freezing-Point Depression The freezing-point depression, ∆tf , is the difference between the freezing points of the pure solvent and a solution of a nonelectrolyte in that solvent, and it is directly proportional to the molal concentration of the solution. The molal freezing-point constant (Kf ) is the freezing-point depression of the solvent in a 1-molal solution of a nonvolatile, nonelectrolyte solute. ∆tf = Kfm

Freezing-Point Depression, continued Section 2 Colligative Properties of Solutions Chapter 13 Freezing-Point Depression, continued Molal Freezing-Point and Boiling-Point Constants

Freezing-Point Depression Visual Concepts Chapter 13 Freezing-Point Depression Click below to watch the Visual Concept. Visual Concept

Freezing-Point Depression, continued Section 2 Colligative Properties of Solutions Chapter 13 Freezing-Point Depression, continued Sample Problem C What is the freezing-point depression of water in a solution of 17.1 g of sucrose, C12H22O11, in 200. g of water? What is the actual freezing point of the solution?

Freezing-Point Depression, continued Section 2 Colligative Properties of Solutions Chapter 13 Freezing-Point Depression, continued Sample Problem C Solution Given: solute mass and chemical formula = 17.1 g C12H22O11 solvent mass and identity = 200. g water Unknown: a. freezing-point depression b. freezing point of the solution Solution:

Freezing-Point Depression, continued Section 2 Colligative Properties of Solutions Chapter 13 Freezing-Point Depression, continued Sample Problem C Solution, continued Solution: ∆tf = Kfm f.p. solution = f.p. solvent + ∆tf

Freezing-Point Depression, continued Section 2 Colligative Properties of Solutions Chapter 13 Freezing-Point Depression, continued Sample Problem C Solution, continued Solution:

Freezing-Point Depression, continued Section 2 Colligative Properties of Solutions Chapter 13 Freezing-Point Depression, continued Sample Problem C Solution, continued Solution: ∆tf = Kfm ∆tf = 0.250 m × (1.86°C/m) = 0.465°C f.p. solution = f.p. solvent + ∆tf f.p. solution = 0.000°C + ( 0.465°C) = 0.465°C

Boiling-Point Elevation Section 2 Colligative Properties of Solutions Chapter 13 Boiling-Point Elevation The boiling-point elevation, ∆tb, is the difference between the boiling points of the pure solvent and a nonelectrolyte solution of that solvent, and it is directly proportional to the molal concentration of the solution. The molal boiling-point constant (Kb) is the boiling-point elevation of the solvent in a 1-molal solution of a nonvolatile, nonelectrolyte solute. ∆tb = Kbm

Boiling-Point Elevation and the Presence of Solutes Visual Concepts Chapter 13 Boiling-Point Elevation and the Presence of Solutes Click below to watch the Visual Concept. Visual Concept

Boiling-Point Elevation, continued Section 2 Colligative Properties of Solutions Chapter 13 Boiling-Point Elevation, continued Sample Problem E What is the boiling-point elevation of a solution made from 20.1 g of a nonelectrolyte solute and 400.0 g of water? The molar mass of the solute is 62.0 g.

Boiling-Point Elevation, continued Section 2 Colligative Properties of Solutions Chapter 13 Boiling-Point Elevation, continued Sample Problem E Solution Given: solute mass = 20.1 g solute molar mass = 62.0 g solvent mass and identity = 400.0 g of water Unknown: boiling-point elevation Solution:

Boiling-Point Elevation, continued Section 2 Colligative Properties of Solutions Chapter 13 Boiling-Point Elevation, continued Sample Problem E Solution, continued Solution: ∆tb = Kbm

Boiling-Point Elevation, continued Section 2 Colligative Properties of Solutions Chapter 13 Boiling-Point Elevation, continued Sample Problem E Solution, continued Solution: ∆tb = 0.51°C/m  0.810 m = 0.41°C