Intermolecular Forces

Similar documents
Chapter 11. General Chemistry. Chapter 11/1

Physical Properties of Solutions

Properties of Solutions. Chapter 13

Colligative Properties

Physical Properties of Solutions

Solutions and Their Properties

11/4/2017. General Chemistry CHEM 101 (3+1+0) Dr. Mohamed El-Newehy. Chapter 4 Physical Properties of Solutions

Sample Problem. (b) Mass % H 2 SO 4 = kg H 2 SO 4 /1.046 kg total = 7.04%

Chapter 12.4 Colligative Properties of Solutions Objectives List and define the colligative properties of solutions. Relate the values of colligative

Molality. Molality (m) is the number of moles of solute per kilogram of solvent. mol of solute kg solvent. Molality ( m) =

Colligative properties CH102 General Chemistry, Spring 2011, Boston University

Unit 7. Solution Concentrations and Colligative Properties

x =!b ± b2! 4ac 2a moles particles solution (expt) moles solute dissolved (calculated conc ) i =

Chapter 12. Physical Properties of Solutions. Chemistry, Raymond Chang 10th edition, 2010 McGraw-Hill

Bushra Javed Valencia College CHM 1046 Chapter 12 - Solutions

Colligative Properties

Chapter 13 Properties of Solutions

solubility solubilities that increase with increasing temperature

Chapter 13. Characteristics of a Solution. Example of A Homogenous Mixtures. Solutions

AP CHEMISTRY NOTES 15-1 INTERMOLECULAR FORCES

Solutions. Chapter 14 Solutions. Ion-Ion Forces (Ionic Bonding) Attraction Between Ions and Permanent Dipoles. Covalent Bonding Forces

AP Chemistry--Chapter 11: Properties of Solutions

1) Define the following terms: a) solution, b) miscible, c) hydration, d) percent by mass (solute), e) colligative property, f) hypotonic.

COLLIGATIVE PROPERTIES

Announcements. It is critical that you are keeping up. Ask or see me if you need help. Lecture slides updated and homework solutions posted.

1) Define the following terms: a) solution, b) miscible, c) hydration, d) percent by mass (solute), e) colligative property, f) hypotonic.

CH 222 Chapter Eleven Concept Guide

Overview. Types of Solutions. Intermolecular forces in solution. Concentration terms. Colligative properties. Osmotic Pressure 2 / 46

Lecture outline: Chapter 13

Colligative Properties

1) Define the following terms: a) solution, b) miscible, c) hydration, d) percent by mass (solute), e) colligative property, f) hypotonic.

Chapter 11. Properties of Solutions Solutions

Chapter 11 Problems: 11, 15, 18, 20-23, 30, 32-35, 39, 41, 43, 45, 47, 49-51, 53, 55-57, 59-61, 63, 65, 67, 70, 71, 74, 75, 78, 81, 85, 86, 93

ALE 24. Colligative Properties (Part 2)

Chapter 13. Properties of Solutions

A) sublimation. B) liquefaction. C) evaporation. D) condensation. E) freezing. 11. Below is a phase diagram for a substance.

Chapter 11 Review Packet

70 Example: If a solution is m citric acid, what is the molar concentration (M) of the solution? The density of the solution is 1.

CHAPTER OUTLINE. I. The Structure of Water: An Introduction to Intermolecular Forces

Colligative Properties

Concentration of Solutions

Chemistry 201: General Chemistry II - Lecture

1) Define the following terms: a) solution, b) miscible, c) hydration, d) percent by mass (solute), e) colligative property, f) hypotonic.

PHYSICAL PROPERTIES OF SOLUTIONS

Properties of Solutions. Overview of factors affecting solubility Ways of expressing concentration Physical properties of solutions

StudyHub: AP Chemistry

Chem 1100 Pre-Test 3. Multiple Choice Identify the choice that best completes the statement or answers the question.

CHEM 121b Exam 1 Spring 1999

CHEMISTRY 102 FALL 2010 EXAM 1 FORM C SECTION 502 DR. KEENEY-KENNICUTT PART 1

Chapter 12. Properties of Solutions

An aqueous solution is 8.50% ammonium chloride by mass. The density of the solution is g/ml Find: molality, mole fraction, molarity.

CHEMISTRY 102 FALL 2010 EXAM 1 FORM D SECTION 502 DR. KEENEY-KENNICUTT PART 1

- Let's look at how things dissolve into water, since aqueous solutions are quite common. sucrose (table sugar)

Freezing point depression - The freezing temperature of a SOLUTION gets lower as the CONCENTRATION of a solution increases.

Born-Haber Cycle: ΔH hydration

concentration of solute (molality) Freezing point depression constant (for SOLVENT)

Colligative Properties. Vapour pressure Boiling point Freezing point Osmotic pressure

Chemistry 103 Spring 2010

Pure Liquid with solute. Pure Liquid

LESSON 11. Glossary: Solutions. Boiling-point elevation

7.02 Colligative Properties

Properties of Solutions

Chapter 13. Properties of Solutions. Lecture Presentation. John D. Bookstaver St. Charles Community College Cottleville, MO

Find molality: mass percent. molality Assume a basis of 100g solution, then find moles ammonium chloride: Find mass water: So molality is:

Properties of Solutions

Chapter 11: Properties of Solutions

- Let's look at how things dissolve into water, since aqueous solutions are quite common. sucrose (table sugar)

Chapter 11. Properties of Solutions

- Applications: In chemistry, this effect is often used to determine the molecular weight of an unknown molecule.

COLLIGATIVE PROPERTIES. Engr. Yvonne Ligaya F. Musico 1

Chapter 11: Properties of Solutions - Their Concentrations and Colligative Properties. Chapter Outline

Soluble: A solute that dissolves in a specific solvent. Insoluble: A solute that will not dissolve in a specific solvent. "Like Dissolves Like"

75 A solution of 2.500g of unknown dissolved in g of benzene has a freezing point of C. What is the molecular weight of the unknown?

CHEMISTRY XL-14A PHYSICAL EQUILIBRIUM. August 13, 2011 Robert Iafe

CHEM N-7 November 2005

SOLUTION CONCENTRATIONS

Chapter 17 - Properties of Solutions

Chapter 12. Solutions and Their Behavior. Supersaturated contains more than the saturation limit (very unstable)

Chem 1100 Pre-Test 3. Multiple Choice Identify the choice that best completes the statement or answers the question.

Chapter 11 Properties of Solutions

Chapter 10: CHM 2045 (Dr. Capps)

Properties of Solutions

Properties of Solutions

Chapter 13. Properties of Solutions

Chapter 13 Properties of Solutions

Name AP CHEM / / Chapter 11 Outline Properties of Solutions

Chapter 11. Properties of Solutions. Copyright 2017 Cengage Learning. All Rights Reserved.

Name Class Date. In the space provided, write the letter of the term or phrase that best completes each statement or best answers each question.

11) What thermodynamic pressure encourages solution formation of two nonpolar substances?

64 previous solution

Chapter 17 - Properties of Solutions

Gilbert Kirss Foster. Chapter 11. Properties of Solutions. Their Concentrations and Colligative Properties

PSI AP Chemistry: Solutions Practice Problems

Let's look at the following "reaction" Mixtures. water + salt > "salt water"

AP CHEMISTRY CHAPTER 8 PROBLEM SET #2. (Questions 1-3) Select the letter of the answer that best completes the statement or answers the question.

FORMULA SHEET (tear off)

Lecture Presentation. Chapter 12. Solutions. Sherril Soman, Grand Valley State University Pearson Education, Inc.

Chapter 13 Properties of Solutions

Mixtures. What happens to the properties (phase changes) when we make a solution? Principles of Chemistry II. Vanden Bout

Solutions. π = n RT = M RT V

Transcription:

Intermolecular Forces! When two molecules approach one another, they are attracted to some extent! Polar molecules are attracted through the electrostatic interaction of their dipole moments! Non-polar molecules are attracted through London Dispersion forces (induced dipoles) Intermolecular Forces! London Dispersion forces result when the valence electrons from one molecule are attracted to the positively charged nucleus of another molecule! The larger the molecule, the larger the London forces it experiences

Intermolecular Forces Solutes and Solutions

Solutes and Solution! The first rule of solubility is likes dissolve likes! Polar or ionic substances are soluble in polar solvents! Non-polar substances are soluble in nonpolar solvents Solutes and Solution! There must be a reason why a substance is soluble in a solvent:! either the solution process lowers the overall energy of the system! Or the solution process increases the overall entropy of the system! Entropy is a measure of the amount of disorder in a system total entropy must increase for any spontaneous change

Solutes and Solution! The forces that drive the dissolution of a solute usually involve both enthalpy and entropy terms!!h soln (energy of solution) < 0 for most species! The creation of a solution takes a more ordered system (solid phase or pure liquid phase) and makes more disordered system (solute molecules are more randomly distributed throughout the solution) Solutes and Solutions Detergents

Concentration Units Molarity: defined as moles of solute per liter of solution molessolute mol M = = litersof solution L The volume used is the total volume of the solution, not just the volume of the solvent Usually the volume of the solute is negligible compared to the solvent volume Concentration Units Molality: defined as moles of solute per mass of solvent molessolute m = = kilogramsof solvent mol kg In this case, the mass is only the mass of the solvent, not the total mass of the solution

Concentration Units Mole fraction: mole fraction is defined as the number of moles solute divided by the total number of moles of all species in the solution molessolute X A = = totalmolesin solution Mole fraction is a unitless number n n A tot! Colligative properties are a set properties that depend only on the amount of solute in a solution, and not on the chemical identity of the solute! Colligative properties include:! Vapor pressure lowering! Freezing point depression! Boiling point elevation! Osmotic pressure

Vapor pressure lowering! When solute is added to a pure solvent, solvent molecules are tied up in keeping the solute molecules in solution! Because solvent molecules are more strongly attracted to the solute than to themselves, it requires more energy to remove them from the solution compared to the pure solvent Vapor pressure lowering! As a consequence, the vapor pressure of the solution is lowered! Raoult s Law states: P 1 = X 1 P 1 o P 1 = vapor pressure of the solution X 1 = mole fraction of solvent P 1 o = vapor pressure of the pure solvent

Vapor pressure lowering Example: What is the vapor pressure of a saturated NaCl solution at 25 o C? P o = 23.76 Torr " H2O = 0.99707 g/ml 35.7 g NaCl per 100 ml H 2 O Step 1 Determine mole fraction of solution 35.7 g NaCl/58.443 g/mol =0.611 mol NaCl (100 ml)(.9971 g/ml)/(15.0152 g.mol) =5.53 mol H 2 O Vapor pressure lowering Example: What is the vapor pressure of a saturated NaCl solution at 25 o C? Step 1 Determine mole fraction of solution 5.53mol X = =.819 2(.611mol) + 5.53mol

Vapor pressure lowering Example: What is the vapor pressure of a saturated NaCl solution at 25 o C? Step 2 Determine vapor pressure of solution P = X P o = = 19.5Torr (.819)(23. 76Torr) The vapor pressure over a saturated NaCl solution is nearly 20% lower than that of pure water Boiling Point Elevation! Because the vapor pressure of solution is lower than the vapor pressure of the pure solvent, the solution s boiling point will be elevated! Remember that the boiling point is that temperature where the vapor pressure of the solution is equal to the pressure over the solution

Boiling Point Elevation!T b = T b (soln) T b (solvent) = K b m K b = boiling point elevation constant m = molality of solute Boiling Point Elevation Example: What is boiling point of a NaCl solution that is saturated at 25 o C? K b (H 2 O) = 0.512 K kg/mol m = (1.22 mol solute)/(.09771 kg H 2 O) = 12.5 m!t b = (12.5 mol/kg)(0.512 K kg/mol) = 6.40 K T b = 106.4 o C

Freezing Point Depression! The freezing point of a solution will be lower than that of the pure solvent because the solute molecules interrupt the crystal structure of the solid solvent!t f = K f m K f = freezing pt depression constant m = molality of solute Freezing Point Depression Example: Determine the freezing point of a solution that is 40% by volume ethylene glycol in water K f (H 2 O) = 1.86 K kg/mol "(C 2 H 6 O 2 ) =1.109 g/ml

Freezing Point Depression Example: Determine the freezing point of a solution that is 40% by volume ethylene glycol in water Step 1 Assume we have 1.00 L of the solution; determine molality of ethylene glycol (400 ml C 2 H 6 O 2 )(1.109 g/ml)/(62.069 g/mol) = 7.15 mol C 2 H 6 O 2 m = (7.15 mol)/(.600 kg H 2 O) =11.9 m Freezing Point Depression Example: Determine the freezing point of a solution that is 40% by volume ethylene glycol in water Step 2 Determine freezing point depression!t f = (11.9 m)(1.86 K kg/mol) = 22.1 K T f = -22.1 o C

Osmotic Pressure! A solution is initially separated from a reservoir of pure solvent by a semipermeable membrane! A semi-permeable membrane allows the solvent to move through in either direction, but does not allow the solute species to pass through pure solvent solution semi-permeable membrane Osmotic Pressure! The solvent wants to have equal concentrations on each side of the system! The solvent will flow from the pure solvent side to the solution side of the system to minimize the concentration difference between each side of the system pure solvent solution flow of solvent semi-permeable membrane

Osmotic Pressure! The solution side is now higher than the solvent side which creates a pressure difference! The increased pressure on the solution side tends to push solvent back to the pure solvent side of the system pure solvent solution semi-permeable membrane Osmotic Pressure! The osmotic pressure of the solution is the pressure exerted by the solution when the system reaches equilibrium balance between force exerted by pure solvent to equalize concentration and force exerted by increased height of solution pure solvent solution semi-permeable membrane

Osmotic Pressure! The osmotic pressure is given by: # = c R T i where c = molar concentration R = gas constant T = temperature (in K) i = solute particles per unit formula pure solvent solution semi-permeable membrane Osmotic Pressure Example: Determine the osmotic pressure of a 1.00 M solution of NaCl c = 1.00 mol/l i = 2 (Na + and Cl - ) # = 2(1.00 mol/l)(.08206 L atm/mol K)(298 K) = 48.9 atm