Submit Search
Upload
Mc 12 mac
•
Download as PPT, PDF
•
5 likes
•
3,770 views
D
dirksr
Follow
mixtures, solutions, concentration
Read less
Read more
Education
Report
Share
Report
Share
1 of 105
Download now
Recommended
Chap 15 section 2 titration
Chap 15 section 2 titration
dirksr
solvent extraction
solvent extraction
AADHIBHAGAWAN COLLEGE OF PHARMACY, THIRUVANNAMAI, TAMIL NADU
Chemistry of non aqueous solvents part-1
Chemistry of non aqueous solvents part-1
Sakina Bootwala
solubility & Pka presentation
solubility & Pka presentation
Mayuri Sawant
Chemistry project
Chemistry project
PRINTDESK by Dan
Solutions & colloids
Solutions & colloids
ysitko2
DETERMINATION OF THE EQUILIBRIUM CONSTANT FOR THE FORMATION OF TRI-IODIDE ION
DETERMINATION OF THE EQUILIBRIUM CONSTANT FOR THE FORMATION OF TRI-IODIDE ION
Augustine Adu
State of matter and properties of matter (Part-10)(Physicochemical properti...
State of matter and properties of matter (Part-10)(Physicochemical properti...
Ms. Pooja Bhandare
Recommended
Chap 15 section 2 titration
Chap 15 section 2 titration
dirksr
solvent extraction
solvent extraction
AADHIBHAGAWAN COLLEGE OF PHARMACY, THIRUVANNAMAI, TAMIL NADU
Chemistry of non aqueous solvents part-1
Chemistry of non aqueous solvents part-1
Sakina Bootwala
solubility & Pka presentation
solubility & Pka presentation
Mayuri Sawant
Chemistry project
Chemistry project
PRINTDESK by Dan
Solutions & colloids
Solutions & colloids
ysitko2
DETERMINATION OF THE EQUILIBRIUM CONSTANT FOR THE FORMATION OF TRI-IODIDE ION
DETERMINATION OF THE EQUILIBRIUM CONSTANT FOR THE FORMATION OF TRI-IODIDE ION
Augustine Adu
State of matter and properties of matter (Part-10)(Physicochemical properti...
State of matter and properties of matter (Part-10)(Physicochemical properti...
Ms. Pooja Bhandare
New chm 151_unit_13_power_points-su13_s
New chm 151_unit_13_power_points-su13_s
caneman1
Solvent extraction
Solvent extraction
vanessawhitehawk
Partition coefficient
Partition coefficient
Sagar Savale
pH stability profile
pH stability profile
SayaliPawar37
Solutions and Colligative properties
Solutions and Colligative properties
khali29
Solvent Extraction
Solvent Extraction
Suman Roy
Solutions
Solutions
prakrsinha
Volumetric Analysis by Dr. A. Amsavel
Volumetric Analysis by Dr. A. Amsavel
Dr. Amsavel A
Partition coefficient: To determine partition coefficient of benzoic acid bet...
Partition coefficient: To determine partition coefficient of benzoic acid bet...
SONALI PAWAR
pKa and log p determination
pKa and log p determination
Sanathoiba Singha
Chem unit 8 presentation
Chem unit 8 presentation
bobcatchemistry
Titration
Titration
Marvin Florendo
Chapter8
Chapter8
Brie Clarke
Titrimetric analysis
Titrimetric analysis
Dhanashree Kad
Analytical chemistry
Analytical chemistry
geetha T
Nyb F09 Unit 2 Slides 1 25
Nyb F09 Unit 2 Slides 1 25
Ben
Phph 1
Phph 1
Firas Aziz
Theory extract
Theory extract
marulituamanurung
Basic of titration
Basic of titration
Chia Siew Lian
Volumetric Analysis ( Titrimetric analysis) or Titration
Volumetric Analysis ( Titrimetric analysis) or Titration
Aman Kakne
Hstp ch04
Hstp ch04
Hatice Yarali
Ch 14 notes for upload
Ch 14 notes for upload
dirksr
More Related Content
What's hot
New chm 151_unit_13_power_points-su13_s
New chm 151_unit_13_power_points-su13_s
caneman1
Solvent extraction
Solvent extraction
vanessawhitehawk
Partition coefficient
Partition coefficient
Sagar Savale
pH stability profile
pH stability profile
SayaliPawar37
Solutions and Colligative properties
Solutions and Colligative properties
khali29
Solvent Extraction
Solvent Extraction
Suman Roy
Solutions
Solutions
prakrsinha
Volumetric Analysis by Dr. A. Amsavel
Volumetric Analysis by Dr. A. Amsavel
Dr. Amsavel A
Partition coefficient: To determine partition coefficient of benzoic acid bet...
Partition coefficient: To determine partition coefficient of benzoic acid bet...
SONALI PAWAR
pKa and log p determination
pKa and log p determination
Sanathoiba Singha
Chem unit 8 presentation
Chem unit 8 presentation
bobcatchemistry
Titration
Titration
Marvin Florendo
Chapter8
Chapter8
Brie Clarke
Titrimetric analysis
Titrimetric analysis
Dhanashree Kad
Analytical chemistry
Analytical chemistry
geetha T
Nyb F09 Unit 2 Slides 1 25
Nyb F09 Unit 2 Slides 1 25
Ben
Phph 1
Phph 1
Firas Aziz
Theory extract
Theory extract
marulituamanurung
Basic of titration
Basic of titration
Chia Siew Lian
Volumetric Analysis ( Titrimetric analysis) or Titration
Volumetric Analysis ( Titrimetric analysis) or Titration
Aman Kakne
What's hot
(20)
New chm 151_unit_13_power_points-su13_s
New chm 151_unit_13_power_points-su13_s
Solvent extraction
Solvent extraction
Partition coefficient
Partition coefficient
pH stability profile
pH stability profile
Solutions and Colligative properties
Solutions and Colligative properties
Solvent Extraction
Solvent Extraction
Solutions
Solutions
Volumetric Analysis by Dr. A. Amsavel
Volumetric Analysis by Dr. A. Amsavel
Partition coefficient: To determine partition coefficient of benzoic acid bet...
Partition coefficient: To determine partition coefficient of benzoic acid bet...
pKa and log p determination
pKa and log p determination
Chem unit 8 presentation
Chem unit 8 presentation
Titration
Titration
Chapter8
Chapter8
Titrimetric analysis
Titrimetric analysis
Analytical chemistry
Analytical chemistry
Nyb F09 Unit 2 Slides 1 25
Nyb F09 Unit 2 Slides 1 25
Phph 1
Phph 1
Theory extract
Theory extract
Basic of titration
Basic of titration
Volumetric Analysis ( Titrimetric analysis) or Titration
Volumetric Analysis ( Titrimetric analysis) or Titration
Viewers also liked
Hstp ch04
Hstp ch04
Hatice Yarali
Ch 14 notes for upload
Ch 14 notes for upload
dirksr
Chapter 6.2
Chapter 6.2
Allison Barnette
Net ionic and colligative regular no math
Net ionic and colligative regular no math
dirksr
Chemistry
Chemistry
Fred Phillips
Biochemistry
Biochemistry
Fred Phillips
Populations
Populations
Fred Phillips
Modern Phylogenetic Taxonomy Notes
Modern Phylogenetic Taxonomy Notes
Fred Phillips
Energy And Enzymes Notes New
Energy And Enzymes Notes New
Fred Phillips
Meiosis Notes
Meiosis Notes
Fred Phillips
Ch. 3 Notes
Ch. 3 Notes
kbolinsky
Chapter 4 notes
Chapter 4 notes
kbolinsky
Humsns And The Environment
Humsns And The Environment
Fred Phillips
Community Ecology
Community Ecology
Fred Phillips
Taxonomy Biology Notes
Taxonomy Biology Notes
Fred Phillips
Cell
Cell
Fred Phillips
Evidence Of Evolution
Evidence Of Evolution
Fred Phillips
Mendel And Genetics Notes
Mendel And Genetics Notes
Fred Phillips
Introduction To Ecology
Introduction To Ecology
Fred Phillips
Genetics Notes
Genetics Notes
Fred Phillips
Viewers also liked
(20)
Hstp ch04
Hstp ch04
Ch 14 notes for upload
Ch 14 notes for upload
Chapter 6.2
Chapter 6.2
Net ionic and colligative regular no math
Net ionic and colligative regular no math
Chemistry
Chemistry
Biochemistry
Biochemistry
Populations
Populations
Modern Phylogenetic Taxonomy Notes
Modern Phylogenetic Taxonomy Notes
Energy And Enzymes Notes New
Energy And Enzymes Notes New
Meiosis Notes
Meiosis Notes
Ch. 3 Notes
Ch. 3 Notes
Chapter 4 notes
Chapter 4 notes
Humsns And The Environment
Humsns And The Environment
Community Ecology
Community Ecology
Taxonomy Biology Notes
Taxonomy Biology Notes
Cell
Cell
Evidence Of Evolution
Evidence Of Evolution
Mendel And Genetics Notes
Mendel And Genetics Notes
Introduction To Ecology
Introduction To Ecology
Genetics Notes
Genetics Notes
Similar to Mc 12 mac
mc_ch12.ppt
mc_ch12.ppt
SamarjeetKumar18
chapter11.ppt
chapter11.ppt
KimberlyAnnePagdanga1
Colligative properties
Colligative properties
dirksr
Solubility of drugs
Solubility of drugs
SwatiMittal62
Solubility of drugs
Solubility of drugs
Swati Mittal
Solubility lec 1
Solubility lec 1
am_mudhafar
Chapter8 120319075446-phpapp01
Chapter8 120319075446-phpapp01
Cleophas Rwemera
Chem 1 unit 11 presentation
Chem 1 unit 11 presentation
bobcatchemistry
Solubility of drugs
Solubility of drugs
PriyaManeDeshmukh
Chem 1 unit 11 presentation
Chem 1 unit 11 presentation
bobcatchemistry
Chem 1 unit 11 presentation
Chem 1 unit 11 presentation
bobcatchemistry
Chem 1 unit 11 presentation
Chem 1 unit 11 presentation
bobcatchemistry
Chemunit11presentation 120308075246-phpapp01
Chemunit11presentation 120308075246-phpapp01
Cleophas Rwemera
httpsdrive.google.comfiled1q4wdFDrp2E2nHW8xrn5AfnNugXMsE6Fmviewusp=share_link...
httpsdrive.google.comfiled1q4wdFDrp2E2nHW8xrn5AfnNugXMsE6Fmviewusp=share_link...
SatyendraMishra77
Buffer solution ( ).pdf
Buffer solution ( ).pdf
TanvirMahmud54
Unit 2
Unit 2
alekey08
Chapter 13
Chapter 13
ewalenta
Properties of solutions
Properties of solutions
Channa Karunathilaka
chapter_13.ppt
chapter_13.ppt
Vinod Raorane
Chapter 13
Chapter 13
ChemistryFun
Similar to Mc 12 mac
(20)
mc_ch12.ppt
mc_ch12.ppt
chapter11.ppt
chapter11.ppt
Colligative properties
Colligative properties
Solubility of drugs
Solubility of drugs
Solubility of drugs
Solubility of drugs
Solubility lec 1
Solubility lec 1
Chapter8 120319075446-phpapp01
Chapter8 120319075446-phpapp01
Chem 1 unit 11 presentation
Chem 1 unit 11 presentation
Solubility of drugs
Solubility of drugs
Chem 1 unit 11 presentation
Chem 1 unit 11 presentation
Chem 1 unit 11 presentation
Chem 1 unit 11 presentation
Chem 1 unit 11 presentation
Chem 1 unit 11 presentation
Chemunit11presentation 120308075246-phpapp01
Chemunit11presentation 120308075246-phpapp01
httpsdrive.google.comfiled1q4wdFDrp2E2nHW8xrn5AfnNugXMsE6Fmviewusp=share_link...
httpsdrive.google.comfiled1q4wdFDrp2E2nHW8xrn5AfnNugXMsE6Fmviewusp=share_link...
Buffer solution ( ).pdf
Buffer solution ( ).pdf
Unit 2
Unit 2
Chapter 13
Chapter 13
Properties of solutions
Properties of solutions
chapter_13.ppt
chapter_13.ppt
Chapter 13
Chapter 13
More from dirksr
Chapter 7 review jeopardy nomenclature only
Chapter 7 review jeopardy nomenclature only
dirksr
Chapter 7 review: Formulas & Moles
Chapter 7 review: Formulas & Moles
dirksr
Chapter 5 review jeopardy reg modern chem
Chapter 5 review jeopardy reg modern chem
dirksr
Chapter 5 review jeopardy
Chapter 5 review jeopardy
dirksr
Ch 2 jeopardy review
Ch 2 jeopardy review
dirksr
Chapter 1 review
Chapter 1 review
dirksr
More from dirksr
(6)
Chapter 7 review jeopardy nomenclature only
Chapter 7 review jeopardy nomenclature only
Chapter 7 review: Formulas & Moles
Chapter 7 review: Formulas & Moles
Chapter 5 review jeopardy reg modern chem
Chapter 5 review jeopardy reg modern chem
Chapter 5 review jeopardy
Chapter 5 review jeopardy
Ch 2 jeopardy review
Ch 2 jeopardy review
Chapter 1 review
Chapter 1 review
Recently uploaded
ESSENTIAL of (CS/IT/IS) class 06 (database)
ESSENTIAL of (CS/IT/IS) class 06 (database)
Dr. Mazin Mohamed alkathiri
Proudly South Africa powerpoint Thorisha.pptx
Proudly South Africa powerpoint Thorisha.pptx
thorishapillay1
Roles & Responsibilities in Pharmacovigilance
Roles & Responsibilities in Pharmacovigilance
SamikshaHamane
Alper Gobel In Media Res Media Component
Alper Gobel In Media Res Media Component
InMediaRes1
Hierarchy of management that covers different levels of management
Hierarchy of management that covers different levels of management
mkooblal
Organic Name Reactions for the students and aspirants of Chemistry12th.pptx
Organic Name Reactions for the students and aspirants of Chemistry12th.pptx
VS Mahajan Coaching Centre
Solving Puzzles Benefits Everyone (English).pptx
Solving Puzzles Benefits Everyone (English).pptx
OH TEIK BIN
Capitol Tech U Doctoral Presentation - April 2024.pptx
Capitol Tech U Doctoral Presentation - April 2024.pptx
CapitolTechU
DATA STRUCTURE AND ALGORITHM for beginners
DATA STRUCTURE AND ALGORITHM for beginners
Sabitha Banu
Gas measurement O2,Co2,& ph) 04/2024.pptx
Gas measurement O2,Co2,& ph) 04/2024.pptx
Dr.Ibrahim Hassaan
EPANDING THE CONTENT OF AN OUTLINE using notes.pptx
EPANDING THE CONTENT OF AN OUTLINE using notes.pptx
RaymartEstabillo3
MICROBIOLOGY biochemical test detailed.pptx
MICROBIOLOGY biochemical test detailed.pptx
abhijeetpadhi001
CELL CYCLE Division Science 8 quarter IV.pptx
CELL CYCLE Division Science 8 quarter IV.pptx
JiesonDelaCerna
Earth Day Presentation wow hello nice great
Earth Day Presentation wow hello nice great
YousafMalik24
MARGINALIZATION (Different learners in Marginalized Group
MARGINALIZATION (Different learners in Marginalized Group
JonathanParaisoCruz
Full Stack Web Development Course for Beginners
Full Stack Web Development Course for Beginners
Sabitha Banu
Computed Fields and api Depends in the Odoo 17
Computed Fields and api Depends in the Odoo 17
Celine George
OS-operating systems- ch04 (Threads) ...
OS-operating systems- ch04 (Threads) ...
Dr. Mazin Mohamed alkathiri
Framing an Appropriate Research Question 6b9b26d93da94caf993c038d9efcdedb.pdf
Framing an Appropriate Research Question 6b9b26d93da94caf993c038d9efcdedb.pdf
UjwalaBharambe
Introduction to ArtificiaI Intelligence in Higher Education
Introduction to ArtificiaI Intelligence in Higher Education
pboyjonauth
Recently uploaded
(20)
ESSENTIAL of (CS/IT/IS) class 06 (database)
ESSENTIAL of (CS/IT/IS) class 06 (database)
Proudly South Africa powerpoint Thorisha.pptx
Proudly South Africa powerpoint Thorisha.pptx
Roles & Responsibilities in Pharmacovigilance
Roles & Responsibilities in Pharmacovigilance
Alper Gobel In Media Res Media Component
Alper Gobel In Media Res Media Component
Hierarchy of management that covers different levels of management
Hierarchy of management that covers different levels of management
Organic Name Reactions for the students and aspirants of Chemistry12th.pptx
Organic Name Reactions for the students and aspirants of Chemistry12th.pptx
Solving Puzzles Benefits Everyone (English).pptx
Solving Puzzles Benefits Everyone (English).pptx
Capitol Tech U Doctoral Presentation - April 2024.pptx
Capitol Tech U Doctoral Presentation - April 2024.pptx
DATA STRUCTURE AND ALGORITHM for beginners
DATA STRUCTURE AND ALGORITHM for beginners
Gas measurement O2,Co2,& ph) 04/2024.pptx
Gas measurement O2,Co2,& ph) 04/2024.pptx
EPANDING THE CONTENT OF AN OUTLINE using notes.pptx
EPANDING THE CONTENT OF AN OUTLINE using notes.pptx
MICROBIOLOGY biochemical test detailed.pptx
MICROBIOLOGY biochemical test detailed.pptx
CELL CYCLE Division Science 8 quarter IV.pptx
CELL CYCLE Division Science 8 quarter IV.pptx
Earth Day Presentation wow hello nice great
Earth Day Presentation wow hello nice great
MARGINALIZATION (Different learners in Marginalized Group
MARGINALIZATION (Different learners in Marginalized Group
Full Stack Web Development Course for Beginners
Full Stack Web Development Course for Beginners
Computed Fields and api Depends in the Odoo 17
Computed Fields and api Depends in the Odoo 17
OS-operating systems- ch04 (Threads) ...
OS-operating systems- ch04 (Threads) ...
Framing an Appropriate Research Question 6b9b26d93da94caf993c038d9efcdedb.pdf
Framing an Appropriate Research Question 6b9b26d93da94caf993c038d9efcdedb.pdf
Introduction to ArtificiaI Intelligence in Higher Education
Introduction to ArtificiaI Intelligence in Higher Education
Mc 12 mac
1.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Table of Contents Chapter 12 Solutions Section 1 Types of Mixtures Section 2 The Solution Process Section 3 Concentration of Solutions
2.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Objectives • Distinguish between electrolytes and nonelectrolytes. • List three different solute-solvent combinations. • Compare the properties of suspensions, colloids, and solutions. • Distinguish between electrolytes and nonelectrolytes. Chapter 12 Section 1 Types of Mixtures
3.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Solutions • You know from experience that sugar dissolves in water. Sugar is described as “soluble in water.” By soluble we mean capable of being dissolved. • When sugar dissolves, all its molecules become uniformly distributed among the water molecules. The solid sugar is no longer visible. • Such a mixture is called a solution. A solution is a homogeneous mixture of two or more substances in a single phase. Chapter 12 Section 1 Types of Mixtures
4.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Click below to watch the Visual Concept. Visual Concept Chapter 12 Solutions
5.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Solutions, continued • The dissolving medium in a solution is called the solvent, and the substance dissolved in a solution is called the solute. • Solutions may exist as gases, liquids, or solids. There are many possible solute-solvent combinations between gases, liquids, and solids. • example: Alloys are solid solutions in which the atoms of two or more metals are uniformly mixed. • Brass is made from zinc and copper. • Sterling silver is made from silver and copper. Chapter 12 Section 1 Types of Mixtures
6.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Click below to watch the Visual Concept. Visual Concept Chapter 12 Solutes, Solvents, and Solutions
7.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Chapter 12 Types of Solutions
8.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Particle Models for Gold and Gold Alloy Chapter 12 Section 1 Types of Mixtures
9.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Suspensions • If the particles in a solvent are so large that they settle out unless the mixture is constantly stirred or agitated, the mixture is called a suspension. • For example, a jar of muddy water consists of soil particles suspended in water. The soil particles will eventually all collect on the bottom of the jar, because the soil particles are denser than the solvent, water. • Particles over 1000 nm in diameter—1000 times as large as atoms, molecules or ions—form suspensions. Chapter 12 Section 1 Types of Mixtures
10.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Click below to watch the Visual Concept. Visual Concept Chapter 12 Suspensions
11.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Colloids • Particles that are intermediate in size between those in solutions and suspensions form mixtures known as colloidal dispersions, or simply colloids. • The particles in a colloid are small enough to be suspended throughout the solvent by the constant movement of the surrounding molecules. • Colloidal particles make up the dispersed phase, and water is the dispersing medium. • example: Mayonnaise is a colloid. • It is an emulsion of oil droplets in water. Chapter 12 Section 1 Types of Mixtures
12.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Colloids, continued Tyndall Effect • Many colloids look similar to solutions because their particles cannot be seen. • The Tyndall effect occurs when light is scattered by colloidal particles dispersed in a transparent medium. • example: a headlight beam is visible from the side on a foggy night • The Tyndall effect can be used to distinguish between a solution and a colloid. Chapter 12 Section 1 Types of Mixtures
13.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Colloids Chapter 12
14.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Emulsions Chapter 12
15.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Properties of Solutions, Colloids, and Suspensions Chapter 12 Section 1 Types of Mixtures
16.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Solutes: Electrolytes Versus Nonelectrolytes • A substance that dissolves in water to give a solution that conducts electric current is called an electrolyte. • Any soluble ionic compound, such as sodium chloride, NaCl, is an electrolyte. • The positive and negative ions separate from each other in solution and are free to move, making it possible for an electric current to pass through the solution. Chapter 12 Section 1 Types of Mixtures
17.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Solutes: Electrolytes Versus Nonelectrolytes, continued • Strong electrolytes completely dissociate when they are dissolved in water. • Weak electrolytes only partially dissociate when dissolved in water. Chapter 12 Section 1 Types of Mixtures
18.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Solutes: Electrolytes Versus Nonelectrolytes, continued • A substance that dissolves in water to give a solution that does not conduct electric current is called a nonelectrolyte. • Sugar is an example of a nonelectrolyte. • Neutral solute molecules do not contain mobile charged particles, so a solution of a nonelectrolyte cannot conduct electric current. Chapter 12 Section 1 Types of Mixtures
19.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Electrical Conductivity of Solutions Chapter 12 Section 1 Types of Mixtures
20.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Objectives • List and explain three factors that affect the rate at which a solid solute dissolves in a liquid solvent. • Explain solution equilibrium, and distinguish among saturated, unsaturated, and supersaturated solutions. • Explain the meaning of “like dissolves like” in terms of polar and nonpolar substances. Chapter 12 Section 2 The Solution Process
21.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Objectives, continued • List the three interactions that contribute to the enthalpy of a solution, and explain how they combine to cause dissolution to be exothermic or endothermic. • Compare the effects of temperature and pressure on solubility. Chapter 12 Section 2 The Solution Process
22.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu • Because the dissolution process occurs at the surface of the solute, it can be speeded up if the surface area of the solute is increased. • Stirring or shaking helps to disperse solute particles and increase contact between the solvent and solute surface. This speeds up the dissolving process. Chapter 12 Section 2 The Solution Process Factors Affecting the Rate of Dissolution • At higher temperatures, collisions between solvent molecules and solvent are more frequent and of higher energy. This helps to disperse solute molecules among the solvent molecules, and speed up the dissolving process.
23.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Click below to watch the Visual Concept. Visual Concept Factors Affecting the Rate of Dissolution Chapter 12
24.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Solubility • If you add spoonful after spoonful of sugar to tea, eventually no more sugar will dissolve. • This illustrates the fact that for every combination of solvent with a solid solute at a given temperature, there is a limit to the amount of solid that can be dissolved. • The point at which this limit is reached for any solute- solvent combination depends on the nature of the solute, the nature of the solvent, and the temperature. Chapter 12 Section 2 The Solution Process
25.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 12 Section 2 The Solution Process Particle Model for Soluble and Insoluble Substances
26.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 12 Section 2 The Solution Process Particle Model for Soluble and Insoluble Substances
27.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Solubility, continued • When a solute is first added to a solvent, solute molecules leave the solid surface and move about at random in the solvent. • As more solute is added, more collisions occur between dissolved solute particles. Some of the solute molecules return to the crystal. • When maximum solubility is reached, molecules are returning to the solid form at the same rate at which they are going into solution. Chapter 12 Section 2 The Solution Process
28.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Solubility, continued • Solution equilibrium is the physical state in which the opposing processes of dissolution and crystallization of a solute occur at the same rates. Chapter 12 Section 2 The Solution Process
29.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Click below to watch the Visual Concept. Visual Concept Solution Equilibrium Chapter 12
30.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Solubility, continued Saturated Versus Unsaturated Solutions • A solution that contains the maximum amount of dissolved solute is described as a saturated solution. • If more solute is added to a saturated solution, it falls to the bottom of the container and does not dissolve. • This is because an equilibrium has been established between ions leaving and entering the solid phase. • A solution that contains less solute than a saturated solution under the same conditions is an unsaturated solution. Chapter 12 Section 2 The Solution Process
31.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Mass of Solute Added Versus Mass of Solute Dissolved Chapter 12 Section 2 The Solution Process
32.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Solubility, continued Supersaturated Solutions • When a saturated solution is cooled, the excess solute usually comes out of solution, leaving the solution saturated at the lower temperature. • But sometimes the excess solute does not separate, and a supersaturated solution is produced, which is a solution that contains more dissolved solute than a saturated solution contains under the same conditions. Chapter 12 Section 2 The Solution Process • A supersaturated solution will form crystals of solute if disturbed or more solute is added.
33.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Solubility, continued Solubility Values • The solubility of a substance is the amount of that substance required to form a saturated solution with a specific amount of solvent at a specified temperature. • example: The solubility of sugar is 204 g per 100 g of water at 20°C. • Solubilities vary widely, and must be determined experimentally. • They can be found in chemical handbooks and are usually given as grams of solute per 100 g of solvent at a given temperature. Chapter 12 Section 2 The Solution Process
34.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Solubility of Common Compounds Chapter 12 Section 2 The Solution Process
35.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 12 Section 2 The Solution Process Solubility of Common Compounds
36.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Click below to watch the Visual Concept. Visual Concept Solubility of a Solid in a Liquid Chapter 12
37.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Solute-Solvent Interactions • Solubility varies greatly with the type of compounds involved. • “Like dissolves like” is a rough but useful rule for predicting whether one substance will dissolve in another. • What makes substances similar depends on: • type of bonding • polarity or nonpolarity of molecules • intermolecular forces between the solute and solvent Chapter 12 Section 2 The Solution Process
38.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Click below to watch the Visual Concept. Visual Concept Like Dissolves Like Chapter 12
39.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Solute-Solvent Interactions, continued Dissolving Ionic Compounds in Aqueous Solution • The polarity of water molecules plays an important role in the formation of solutions of ionic compounds in water. • The slightly charged parts of water molecules attract the ions in the ionic compounds and surround them, separating them from the crystal surface and drawing them into the solution. • This solution process with water as the solvent is referred to as hydration. The ions are said to be hydrated. Chapter 12 Section 2 The Solution Process
40.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Solute-Solvent Interactions, continued Dissolving Ionic Compounds in Aqueous Solution The hydration of the ionic solute lithium chloride is shown below. Chapter 12 Section 2 The Solution Process
41.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Solute-Solvent Interactions, continued Nonpolar Solvents • Ionic compounds are generally not soluble in nonpolar solvents such as carbon tetrachloride, CCl4, and toluene, C6H5CH3. • The nonpolar solvent molecules do not attract the ions of the crystal strongly enough to overcome the forces holding the crystal together. • Ionic and nonpolar substances differ widely in bonding type, polarity, and intermolecular forces, so their particles cannot intermingle very much. Chapter 12 Section 2 The Solution Process
42.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Solute-Solvent Interactions, continued Liquid Solutes and Solvents • Oil and water do not mix because oil is nonpolar whereas water is polar. The hydrogen bonding between water molecules squeezes out whatever oil molecules may come between them. • Two polar substances, or two nonpolar substances, on the other hand, form solutions together easily because their intermolecular forces match. • Liquids that are not soluble in each other are immiscible. Liquids that dissolve freely in one another in any proportion are miscible. Chapter 12 Section 2 The Solution Process
43.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Click below to watch the Visual Concept. Visual Concept Comparing Miscible and Immiscible Liquids Chapter 12
44.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Solute-Solvent Interactions, continued Effects of Pressure on Solubility • Changes in pressure have very little effect on the solubilities of liquids or solids in liquid solvents. However, increases in pressure increase gas solubilities in liquids. • An equilibrium is established between a gas above a liquid solvent and the gas dissolved in a liquid. • As long as this equilibrium is undisturbed, the solubility of the gas in the liquid is unchanged at a given pressure: gas + solvent →← solution Chapter 12 Section 2 The Solution Process
45.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Solute-Solvent Interactions, continued Effects of Pressure on Solubility, continued • Increasing the pressure of the solute gas above the solution causes gas particles to collide with the liquid surface more often. This causes more gas particles to dissolve in the liquid. ↑ gas + solvent →← solution ↓ gas + solvent →← solution Chapter 12 Section 2 The Solution Process • Decreasing the pressure of the solute gas above the solution allows more dissolved gas particles to escape from solution.
46.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Click below to watch the Visual Concept. Visual Concept Pressure, Temperature, and Solubility of Gases Chapter 12
47.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Solute-Solvent Interactions, continued Henry’s Law • Henry’s law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of that gas on the surface of the liquid. • In carbonated beverages, the solubility of carbon dioxide is increased by increasing the pressure. The sealed containers contain CO2 at high pressure, which keeps the CO2 dissolved in the beverage, above the liquid. • When the beverage container is opened, the pressure above the solution is reduced, and CO2 begins to escape from the solution. Chapter 12 Section 2 The Solution Process • The rapid escape of a gas from a liquid in which it is dissolved is known as effervescence.
48.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Click below to watch the Visual Concept. Visual Concept Henry’s Law Chapter 12
49.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Click below to watch the Visual Concept. Visual Concept Effervescence Chapter 12
50.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Solute-Solvent Interactions, continued Effects of Temperature on Solubility • Increasing the temperature usually decreases gas solubility. • As temperature increases, the average kinetic energy of molecules increases. • A greater number of solute molecules are therefore able to escape from the attraction of solvent molecules and return to the gas phase. • At higher temperatures, therefore, equilibrium is reached with fewer gas molecules in solution, and gases are generally less soluble. Chapter 12 Section 2 The Solution Process
51.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Solute-Solvent Interactions, continued Effects of Temperature on Solubility • Increasing the temperature usually increases solubility of solids in liquids, as mentioned previously. • The effect of temperature on solubility for a given solute is difficult to predict. • The solubilities of some solutes vary greatly over different temperatures, and those for other solutes hardly change at all. • A few solid solutes are actually less soluble at higher temperatures. Chapter 12 Section 2 The Solution Process
52.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Solubility vs. Temperature Chapter 12 Section 2 The Solution Process
53.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Enthalpies of Solution • The formation of a solution is accompanied by an energy change. • If you dissolve some potassium iodide, KI, in water, you will find that the outside of the container feels cold to the touch. • But if you dissolve some sodium hydroxide, NaOH, in the same way, the outside of the container feels hot. • The formation of a solid-liquid solution can either absorb energy (KI in water) or release energy as heat (NaOH in water) Chapter 12 Section 2 The Solution Process
54.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Enthalpies of Solution, continued • Before dissolving begins, solute particles are held together by intermolecular forces. Solvent particles are also held together by intermolecular forces. • Energy changes occur during solution formation because energy is required to separate solute molecules and solvent molecules from their neighbors. • A solute particle that is surrounded by solvent molecules is said to be solvated. Chapter 12 Section 2 The Solution Process
55.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Enthalpies of Solution, continued The diagram above shows the enthalpy changes that occur during the formation of a solution. Chapter 12 Section 2 The Solution Process
56.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Enthalpies of Solution, continued • The net amount of energy absorbed as heat by the solution when a specific amount of solute dissolves in a solvent is the enthalpy of solution. • The enthalpy of solution is negative (energy is released) when the sum of attractions from Steps 1 and 2 is less than Step 3, from the diagram on the previous slide. • The enthalpy of solution is positive (energy is absorbed) when the sum of attractions from Steps 1 and 2 is greater than Step 3. Chapter 12 Section 2 The Solution Process
57.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Objectives • Given the mass of solute and volume of solvent, calculate the concentration of solution. • Given the concentration of a solution, determine the amount of solute in a given amount of solution. • Given the concentration of a solution, determine the amount of solution that contains a given amount of solute. Chapter 12 Section 3 Concentration of Solutions
58.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Concentration • The concentration of a solution is a measure of the amount of solute in a given amount of solvent or solution. • Concentration is a ratio: any amount of a given solution has the same concentration. • The opposite of concentrated is dilute. • These terms are unrelated to the degree to which a solution is saturated: a saturated solution of a solute that is not very soluble might be very dilute. Chapter 12 Section 3 Concentration of Solutions
59.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Concentration Units Section 3 Concentration of SolutionsChapter 12
60.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Click below to watch the Visual Concept. Visual Concept Concentration Chapter 12
61.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molarity • Molarity is the number of moles of solute in one liter of solution. • For example, a “one molar” solution of sodium hydroxide contains one mole of NaOH in every liter of solution. • The symbol for molarity is M. The concentration of a one molar NaOH solution is written 1 M NaOH. Chapter 12 Section 3 Concentration of Solutions
62.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molarity, continued • To calculate molarity, you must know the amount of solute in moles and the volume of solution in liters. • When weighing out the solute, this means you will need to know the molar mass of the solute in order to convert mass to moles. • example: One mole of NaOH has a mass of 40.0 g. If this quantity of NaOH is dissolved in enough water to make 1.00 L of solution, it is a 1.00 M solution. Chapter 12 Section 3 Concentration of Solutions
63.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molarity, continued • The molarity of any solution can be calculated by dividing the number of moles of solute by the number of liters of solution: molarity (M) = amount of solute (mol) volume of solution (L) Chapter 12 Section 3 Concentration of Solutions • Note that a 1 M solution is not made by adding 1 mol of solute to 1 L of solvent. In such a case, the final total volume of the solution might not be 1 L. • Solvent must be added carefully while dissolving to ensure a final volume of 1 L.
64.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Click below to watch the Visual Concept. Visual Concept Preparation of a Solution Based on Molarity Chapter 12
65.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molarity, continued Sample Problem A You have 3.50 L of solution that contains 90.0 g of sodium chloride, NaCl. What is the molarity of that solution? Chapter 12 Section 3 Concentration of Solutions
66.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molarity, continued Sample Problem A Solution Given: solute mass = 90.0 g NaCl solution volume = 3.50 L Unknown: molarity of NaCl solution Solution: 90.0 g NaCl × 1 mol NaCl 58.44 g NaCl = 1.54 mol NaCl 1.54 mol NaCl 3.50 L of solution = 0.440 M NaCl Chapter 12 Section 3 Concentration of Solutions
67.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molarity, continued Sample Problem B You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain? Chapter 12 Section 3 Concentration of Solutions
68.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molarity, continued Sample Problem B Solution Given: volume of solution = 0.8 L concentration of solution = 0.5 M HCl Unknown: moles of HCl in a given volume Solution: 0.5 mol HCl 1.0 L of solution × 0.8 L of solution = 0.4 mol HCl Chapter 12 Section 3 Concentration of Solutions
69.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molarity, continued Sample Problem C To produce 40.0 g of silver chromate, you will need at least 23.4 g of potassium chromate in solution as a reactant. All you have on hand is 5 L of a 6.0 M K2CrO4 solution. What volume of the solution is needed to give you the 23.4 g K2CrO4 needed for the reaction? Chapter 12 Section 3 Concentration of Solutions
70.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Sample Problem C Solution Given: volume of solution = 5 L concentration of solution = 6.0 M K2CrO4 mass of solute = 23.4 K2CrO4 mass of product = 40.0 g Ag2CrO4 Unknown: volume of K2CrO4 solution in L Molarity, continued Chapter 12 Section 3 Concentration of Solutions
71.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 6.0 M K2 CrO4 = 0.120 mol K2 CrO4 x L K2 CrO4 solution Sample Problem C Solution, continued Solution: Molarity, continued 1 mol K2 CrO4 = 194.2 g K2 CrO4 23.4 g K2 CrO4 × 1 mol K2 CrO4 194.2 g K2 CrO4 = 0.120 mol K2 CrO4 x = 0.020 L K2 CrO4 solution Chapter 12 Section 3 Concentration of Solutions
72.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molality • Molality is the concentration of a solution expressed in moles of solute per kilogram of solvent. • A solution that contains 1 mol of solute dissolved in 1 kg of solvent is a “one molal” solution. • The symbol for molality is m, and the concentration of this solution is written as 1 m NaOH. Chapter 12 Section 3 Concentration of Solutions
73.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu • The molality of any solution can be calculated by dividing the number of moles of solute by the number of kilograms of solvent: molality (m) = amount of solute (mol) mass of solvent (kg) Chapter 12 Section 3 Concentration of Solutions Molality, continued • Unlike molarity, which is a ratio of which the denominator is liters of solution, molality is per kilograms of solvent. • Molality is used when studying properties of solutions related to vapor pressure and temperature changes, because molality does not change with temperature.
74.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Click below to watch the Visual Concept. Visual Concept Comparing Molarity and Molality Chapter 12
75.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Making a Molal Solution Chapter 12 Section 3 Concentration of Solutions
76.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molality, continued Sample Problem D A solution was prepared by dissolving 17.1 g of sucrose (table sugar, C12H22O11) in 125 g of water. Find the molal concentration of this solution. Chapter 12 Section 3 Concentration of Solutions
77.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molality, continued Sample Problem D Solution Given: solute mass = 17.1 C12H22O11 solvent mass = 125 g H2O Unknown: molal concentration Solution: First, convert grams of solute to moles and grams of solvent to kilograms. 17.1 g C12 H22 O11 × 1 mol C12 H22 O11 342.34 g C12 H22 O11 = 0.0500 mol C12 H22 O11 125 g H2 O 1000 g/ kg = 0.125 kg H2 O Chapter 12 Section 3 Concentration of Solutions
78.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molality, continued Sample Problem D Solution, continued Then, divide moles of solute by kilograms of solvent. 0.0500 mol C12 H22 O11 0.125 kg H2 O = 0.400 m C12 H22 O11 Chapter 12 Section 3 Concentration of Solutions
79.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molality, continued Sample Problem E A solution of iodine, I2, in carbon tetrachloride, CCl4, is used when iodine is needed for certain chemical tests. How much iodine must be added to prepare a 0.480 m solution of iodine in CCl4 if 100.0 g of CCl4 is used? Chapter 12 Section 3 Concentration of Solutions
80.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molality, continued Sample Problem E Solution Given: molality of solution = 0.480 m I2 mass of solvent = 100.0 g CCl4 Unknown: mass of solute Solution: First, convert grams of solvent to kilograms. 100.0 g CCl4 1000 g/ kg = 0.100 kg CCl4 Chapter 12 Section 3 Concentration of Solutions
81.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 0.480 mol I2 × 253.8 g I2 mol I2 = 12.2 g I2 Molality, continued Sample Problem E Solution, continued Solution, continued: Then, use the equation for molality to solve for moles of solute. 0.480 m = x mol I2 0.1 kg H2 O x = 0.0480 mol I2 Chapter 12 Section 3 Concentration of Solutions Finally, convert moles of solute to grams of solute.
82.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu End of Chapter 12 Show
83.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice 1. Water is an excellent solvent because A. it is a covalent compound. B. it is a nonconductor of electricity. C. its molecules are quite polar. D. it is a clear, colorless liquid. Chapter 12 Standardized Test Preparation
84.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice 1. Water is an excellent solvent because A. it is a covalent compound. B. it is a nonconductor of electricity. C. its molecules are quite polar. D. it is a clear, colorless liquid. Chapter 12 Standardized Test Preparation
85.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice 2. Two liquids are likely to be immiscible if A. both have polar molecules. B. both have nonpolar molecules. C. one is polar and the other is nonpolar. D. one is water and the other is methyl alcohol, CH3OH. Chapter 12 Standardized Test Preparation
86.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice 2. Two liquids are likely to be immiscible if A. both have polar molecules. B. both have nonpolar molecules. C. one is polar and the other is nonpolar. D. one is water and the other is methyl alcohol, CH3OH. Chapter 12 Standardized Test Preparation
87.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice 3. The solubility of a gas in a liquid would be increased by an A. addition of an electrolyte. B. addition of an emulsifier. C. agitation of the solution. D. increase in its partial pressure. Chapter 12 Standardized Test Preparation
88.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice 3. The solubility of a gas in a liquid would be increased by an A. addition of an electrolyte. B. addition of an emulsifier. C. agitation of the solution. D. increase in its partial pressure. Chapter 12 Standardized Test Preparation
89.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice 4. Which of the following types of compounds is most likely to be a strong electrolyte? A. a polar compound B. a nonpolar compound C. a covalent compound D. an ionic compound Chapter 12 Standardized Test Preparation
90.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice 4. Which of the following types of compounds is most likely to be a strong electrolyte? A. a polar compound B. a nonpolar compound C. a covalent compound D. an ionic compound Chapter 12 Standardized Test Preparation
91.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice 5. A saturated solution can become supersaturated under which of the following conditions? A. It contains electrolytes. B. The solution is heated and then allowed to cool. C. More solvent is added. D. More solute is added. Chapter 12 Standardized Test Preparation
92.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice 5. A saturated solution can become supersaturated under which of the following conditions? A. It contains electrolytes. B. The solution is heated and then allowed to cool. C. More solvent is added. D. More solute is added. Chapter 12 Standardized Test Preparation
93.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice Chapter 12 Standardized Test Preparation 6. Molarity is expressed in units of A. moles of solute per liter of solution. B. liters of solution per mole of solute. C. moles of solute per liter of solvent. D. liters of solvent per mole of solute.
94.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice 6. Molarity is expressed in units of A. moles of solute per liter of solution. B. liters of solution per mole of solute. C. moles of solute per liter of solvent. D. liters of solvent per mole of solute. Chapter 12 Standardized Test Preparation
95.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice 7. What mass of NaOH is contained in 2.5 L of a 0.010 M solution? A. 0.010 g B. 1.0 g C. 2.5 g D. 0.40 g Chapter 12 Standardized Test Preparation
96.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice 7. What mass of NaOH is contained in 2.5 L of a 0.010 M solution? A. 0.010 g B. 1.0 g C. 2.5 g D. 0.40 g Chapter 12 Standardized Test Preparation
97.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice 8. Which one of the following statements is false? A. Gases are generally more soluble in water under high pressures than under low pressures. B. As temperature increases, the solubilities of some solids in water increase and the solubilities of other solids in water decrease. C. Water dissolves many ionic solutes because of its ability to hydrate ions in solution. D. Many solids dissolve more quickly in a cold solvent than in a warm solvent. Chapter 12 Standardized Test Preparation
98.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice 8. Which one of the following statements is false? A. Gases are generally more soluble in water under high pressures than under low pressures. B. As temperature increases, the solubilities of some solids in water increase and the solubilities of other solids in water decrease. C. Water dissolves many ionic solutes because of its ability to hydrate ions in solution. D. Many solids dissolve more quickly in a cold solvent than in a warm solvent. Chapter 12 Standardized Test Preparation
99.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Short Answer 9. Several experiments are carried out to determine the solubility of cadmium iodide, CdI2, in water. In each experiment, a measured mass of CdI2 is added to 100 g of water at 25°C and the mixture is stirred. Any undissolved CdI2 is then filtered off and dried, and its mass is determined. Results for several such experiments are shown in the table on the next slide. What is the solubility of CdI2 in water at this temperature? Chapter 12 Standardized Test Preparation
100.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Short Answer 9. continued Mass of CdI2 added, g Mass of undissolved CdI2 recovered, g 17.9 0.0 38.2 0.0 53.6 0.0 79.3 0.0 93.6 7.4 104.3 18.1 Chapter 12 Standardized Test Preparation
101.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Short Answer 9. Answer: 86.2 g Mass of CdI2 added, g Mass of undissolved CdI2 recovered, g 17.9 0.0 38.2 0.0 53.6 0.0 79.3 0.0 93.6 7.4 104.3 18.1 Chapter 12 Standardized Test Preparation
102.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Extended Response 10. Explain why oil and water do not mix. Chapter 12 Standardized Test Preparation
103.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Extended Response 10. Explain why oil and water do not mix. Answer: Water is polar, and oil is nonpolar. When the two are combined, the strong hydrogen bonding between water molecules squeezes out the oil droplets, forming separate layers. Chapter 12 Standardized Test Preparation
104.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Extended Response 11. Write a set of instructions on how to prepare a solution that is 0.100 M KBr, using solid KBr (molar mass 119 g/mol) as the solute. Your instructions should include a list of all materials and equipment needed. Chapter 12 Standardized Test Preparation
105.
Copyright © by
Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Extended Response 11. Write a set of instructions on how to prepare a solution that is 0.100 M KBr, using solid KBr (molar mass 119 g/mol) as the solute. Your instructions should include a list of all materials and equipment needed. Answer: Your answer should summarize the steps using 11.9 g of KBr in a 1.00 L volumetric flask (or values proportional to these). Chapter 12 Standardized Test Preparation
Download now