Chemistry Problem Set 4 Answers Professor: C. E. Loader


 Benjamin Melton
 1 years ago
 Views:
Transcription
1 roblem Set 4 Answers rofessor: C. E. Loader. At 5.0 ºC, the vapor pressure of ice is 3.03 mm and that of liquid water is 3.63 mm. Calculate G for the change of one mole of liquid water at 5.0 ºC to solid at 5.0 ºC. The system will not be in equilibrium, so G is not zero. We have to find a series of steps between the initial and final states. For S, we raised the temperture to 0 º but won't do that now because of the difficulty of calculating G for a change in T. The clue to the solution is given by the inclusion of vapor pressures in the problem. We can calculate G for a change in pressure at constant temperature. The alternative path for the process has three steps, all at 5 º; Step : Step 2: Step 3: Liquid water at 3.63 mm vapor at 3.63 mm. Vapor at 3.63 mm vapor at 3.03 mm. Vapor at 3.03 mm solid water at 3.03 mm. Steps () and (3) are equilibrium processes ( G 0), so the total G is the G of step (2). G nrt ln 2 (8.34)(268.2) ln 08.3 J Calculate the pressure at which the conversion of graphite to diamond would become thermodynamically possible. (a) (a) At K. (b) At 000 K. The densities of diamond and graphite are 3.53 g ml and g ml respectively, kj mol and J mol K r H V r S V. All of these values may be considered to be independent of temperature C (graphite) C (diamond) At constant temperature dg Vd or d( r G) ( r V)dp equation (i) V where r G K H V T r S J/mol (298.5 K)( J/mol) J mol age of 8 printed pages
2 and r V V(diamond) V(graphite) 2.0 g mol 3.53 g ml 2.0 g mol g ml ml mol x 0 3 L mol So, from equation (i), integrating with r V taken as constant (since densities are constant) d( r G) ( r V)d ( r V) d r G 2 r G 2 r V( 2 ) equation (ii) Set bar, Then G J mol r r G 0 2 will be the pressure required for r G 2 0 (thermodynamic possibility) J/mol x 0 3 L/mol ( 2 00 ka) 2.52 x 0 6 ka or.52 x 0 4 bar (b) at 000 K, V r G 000 K r H V T r S J/mol (000 K)( J/mol K) 550 J mol Again using equation (ii), J/mol x 0 3 L/mol ( 2 00 ka) x 0 6 ka or 2.77 x 0 4 bar A higher pressure is required at higher temperature. So, why do you think high temperatures are used to make artificial diamonds? 3. Calculate the freezing point of benzene at a pressure of 2000 atm. The normal freezing point is ºC and the densities of liquid and solid benzene are g cm 3 and g cm 3, respectively. H 0.58 kj mol FUS. Using the Clapeyron equation. d dt T M V FUS T T M V FUS or. where V FUS V(l)  V(s) V FUS 78. g mol g cm g mol g cm 3 2. cm 3 mol 2. x 0 6 m 3 mol T 0.58 x 0 3 J K  mol 2. x 0 6 m 3 mol x K.79 9 x 0 7 a K Now 2000 atm x 0 8 a T.26 K T M (2000 atm) ( ) K K x 08 a.799 x 0 7 a K age 2 of 8 printed pages
3 4. Using the data provided for benzene, calculate the vapor pressure of solid benzene at 20 ºC. Benzene: Boiling point 80.0 ºC Melting point 5.53 ºC H VA kj mol 0.59 kj mol We assume that the triple point occurs at the same temperature as the normal melting point. Then we can apply the ClausiusClapeyron equation to the vaporization curve to find the vapour pressure of liquid benzene at the triple point (taken as 5.53 ºC). Remember that the vapour pressure at the normal boiling point is atm. Then ln t atm x 03 J$mol ( 8.34 J$ mol $K K  ) K Hence t 6.07 x 0 2 atm. (vapour pressure at triple point) But at the triple point, solid is in equilibrium with the vapour at this same vapour pressure. We can now apply the ClausiusClapeyron equation to the sublimation curve to determine the vapour pressure (p) at 20 ºC. For the sublimation curve we need H SUB H FUS + H VA 4.35 x 0 3 J mol atm 20 ºC TRILE OINT ressure SUBLIMATION CURVE VAORIZATION CURVE 5.53 ºC 80.0 ºC Temperature Then ln 6.07 x J$ mol 8.34 J$mol $K ( K K ) Hence p.00 x 0 2 atm (Vapour pressure of solid benzene at 20 ºC) age 3 of 8 printed pages
4 5. The normal melting point of naphthalene (molar mass 28.2 g mol ) is 80.2 ºC and it rises by 2.32 ºC when the pressure increases by 00 atmospheres. It boils at 27.9 ºC and it's molar enthalpy of vaporization is 40.5 kj mol. The density of solid naphthalene is.5 g ml and that of the liquid is 0% less. Calculate. (a) The molar enthalpy of fusion of naphthalene g$mol V(s).5 g$ ml V(s).5 ml mol 28.2 g$mol V(l).035 g$ml V(l) 23.9 ml mol V FUS V FUS 2.4 ml mol L mol dt d T mp V FUS 2.32 K 00 x 0.3 ka ( K)(.024 L/mol) H 9.3 kj mol FUS (b) The vapor pressure of liquid naphthalene at the triple point (you may assume that this is the same temperature as the normal melting point). Apply the ClausiusClapeyron equation to the vaporization curve, assuming that the triple point temperature is the same as the melting point, ln( ) H VA atm R T mpt T bpt ln() J$ mol 8.34 J$ mol $ K ( K K ) ln() x 0 2 atm 2.2 ka (vapour pressure at the triple point) (c) The vapor pressure of solid naphthalene at 75.0 ºC. H SUB + H VA 9.3 kj mol kj mol age 4 of 8 printed pages
5 H kj mol SUB The vapor pressure of solid naphthalene and liquid naphthalene are the same at the triple point. Apply the ClausiusClapeyron equation to the sublimitation curve 2.2 ka ln 2.2 ka H SUB R ( K J$ mol 8.34 J$ mol $ K ) K ln ( ) 2.2 ka ln ka ka (vapor pressure of solid at 75 ºC) (d) The free energy change when.00 mol of supercooled liquid naphthalene solidifies at 75.0 ºC. We already have the vapour pressure of solid naphthalene at 75 ºC. Use the extrapolated liquid vaporization curve to find the vapour pressure of supercooled liquid naphthalene at 75 ºC: ln 2.2 ka H VA R ( K K ) 2.2 ka ln ka ka (vapor pressure of (l) at 75 ºC) At 75 ºC (l) G? (s) G O G 3 O (g),.73 ka G 2 (g),.56 ka G G 2 nrt ln 2 x 8.34 x x ln(.56 G 293 J mol (spontaneous of course).73 ) age 5 of 8 printed pages
6 NOTE: That (l) naphthalene is in equilibrium with vapour at.73 ka at 75 ºC (see above), so G zero. Likewise, solid naphthalene is equilibrium with vapour at.56 ka at 75 ºC (part (c) above) and G 3 zero. This part of the problem is the same as problem # in this problem set g of hydrocarbon of the type; H(CH 2 ) n H, is dissolved in 90 g of ethylene bromide. The freezing point of the solution is 0.53 K lower than the pure solvent (K F 2.4 K m ). Calculate the value of n in the formula of the hydrocarbon. What assumption is it necessary to make about the hydrocarbon? Assume that the hydrocarbon is nonvolatile. Then T F K F m T m F 0.53 K molal K F 2.4 K (molal) g ethylene bromide contains mol hydrocarbon 90 x g ethylene bromide contains mol hydrocarbon mol hydrocarbon But, from question, this hydrocarbon has a mass of 0.80 g 0.80 g Molar mass of the hydrocarbon mol 00 g mol For H(CH 2 ) n H, the molar mass 4n + 2 4n n n 7 4 The formula is H(CH 2 ) n H. 7. Bovine serum albumin (BSA) has a molar mass of g mol. Calculate the osmotic pressure of a solution of 2.0 g BSA in 0.00 L of water at 25.0 ºC. Assume V SOLN 0.00 L π π 2 g M 2 V g mol 0.00 L C x 0 4 mol L C 2 RT where C 2 in mol L, R in L atm, π in atm x 0 4 mol L x L atm K mol x 298 K 7.5 x 0 3 atm 8. The normal boiling point of nitrobenzene is 20.9 ºC. If the vapor pressure at 80 ºC is mm Hg, what is the vapor pressure at 200 ºC? age 6 of 8 printed pages
7 This problem has two steps. Since H, isn't given, you must calculate it. Remember that the normal boiling point is at atm. You have two pressures with corresponding temperatures, and, using the ClausiusClapeyron equation H kj mol V (ln 760 ) 484.(453.2) In order to obtain the vapor pressure at a given temperature, use the ClausiusClapeyron equation again and solve for the desired quantity. ln 2 ln + H V R T 2 T T 2 T Choosing mm as, x 0 ln 2 ln(353.2) ( 453.2) mm Hg 9. When 4.07 g of a new water soluble polymer was dissolved in 82.4 ml of water the measured osmotic pressure was 6.23 ka at 22.0 ºC. (a) Calculate the molar mass of the polymer a π C 2 RT so C x mol L J K mol x K 4.07 g molar mass.95 x 0 4 g mol x 0 3 mol L x 82.4 x 0 3 L (b) Calculate the freezing point of the solution. Assuming the density of the water is.0 g ml , the molality is 4.07 g m.95 x 0 4 g mol 82.4 % 0 3 kg m T K f m T.86 K m % m K. So the freezing point of the solution is still 22.0 ºC to 3 significant digits  the change in fp is barely measurable. 0. henol (C 6 H 5 OH) is partially dimerized in the solvent bromoform. When 2.58 g of phenol is dissolved in 00 g of bromoform, the bromoform freezing point is lowered by 2.37 ºC. ure bromoform has a molal freezingpointdepression age 7 of 8 printed pages
8 constant of 4. ºC kg mol and freezes at 8.3 ºC. Calculate the equilibrium constant, K m. for the dimerization reaction of phenol in bromoform at 6 ºC, assuming an ideally dilute solution. 2 C 6 H 5 OH (C 6 H 5 OH) 2 T K f m a m a m a 0.68 mol kg (i) where m a is the apparent molality of the phenol. (i.e. based on actual # mole present at equilibrium) Also m p 2.58 g 94. g mol /0.00 kg m p mol kg (ii) where m p is the molality of the phenol if it it were all present as monomer (i.e. no dimerization). So, in molalities, I (all monomer) C E 2 C 6 H 5 OH (C 6 H 5 OH) x +x ( x) x (from (ii)) At equilibrium, actual # mole of solute per kilogram of solvent But from (i) above, this # of mole also x 0.68 x mol Finally, (m 2 ) eq x K m (m p ) 2 eq ( x) 2 ( x) + x x ( (0.060)) age 8 of 8 printed pages
vap H = RT 1T 2 = 30.850 kj mol 1 100 kpa = 341 K
Thermodynamics: Examples for chapter 6. 1. The boiling point of hexane at 1 atm is 68.7 C. What is the boiling point at 1 bar? The vapor pressure of hexane at 49.6 C is 53.32 kpa. Assume that the vapor
More informationCOLLIGATIVE PROPERTIES:
COLLIGATIVE PROPERTIES: A colligative property is a property that depends only on the number of solute particles present, not their identity. The properties we will look at are: lowering of vapor pressure;
More informationChapter 13. Properties of Solutions
13.4 Ways of Expressing Concentration All methods involve quantifying the amount of solute per amount of solvent (or solution). Concentration may be expressed qualitatively or quantitatively. The terms
More informationPHASE CHEMISTRY AND COLLIGATIVE PROPERTIES
PHASE CHEMISTRY AND COLLIGATIVE PROPERTIES Phase Diagrams Solutions Solution Concentrations Colligative Properties Brown et al., Chapter 10, 385 394, Chapter 11, 423437 CHEM120 Lecture Series Two : 2011/01
More informationColligative Properties. Vapour pressure Boiling point Freezing point Osmotic pressure
Colligative Properties Vapour pressure Boiling point Freezing point Osmotic pressure Learning objectives Describe meaning of colligative property Use Raoult s law to determine vapor pressure of solutions
More informationThermodynamics. Chapter 13 Phase Diagrams. NC State University
Thermodynamics Chapter 13 Phase Diagrams NC State University Pressure (atm) Definition of a phase diagram A phase diagram is a representation of the states of matter, solid, liquid, or gas as a function
More informationChapter 13. Properties of Solutions
Sample Exercise 13.1 (p. 534) By the process illustrated below, water vapor reacts with excess solid sodium sulfate to form the hydrated form of the salt. The chemical reaction is Na 2 SO 4(s) + 10 H 2
More informationSample Exercise 13.1 Predicting Solubility Patterns
Sample Exercise 13.1 Predicting Solubility Patterns Predict whether each of the following substances is more likely to dissolve in the nonpolar solvent carbon tetrachloride (CCl 4 ) or in water: C 7 H
More informationChem. 1A Final Exam Review Problems From ch. 11, 12 & 13
Chem. A Final Exam Review Problems From ch., 2 & 3 f Multiple Choice Identify the choice that best completes the statement or answers the question.. Place the following cations in order from lowest to
More information48 Practice Problems for Ch. 17  Chem 1C  Joseph
48 Practice Problems for Ch. 17  Chem 1C  Joseph 1. Which of the following concentration measures will change in value as the temperature of a solution changes? A) mass percent B) mole fraction C) molality
More information( )( L L)
Chemistry 360 Dr. Jean M. Standard Problem Set 5 Solutions 1. Determine the amount of pressurevolume work performed by 1 mole of water freezing to ice at 0 C and 1 atm pressure. The density of liquid
More informationCHAPTER 14 Solutions
CHAPTER 14 Solutions The Dissolution Process 1. Effect of Temperature on Solubility 2. Molality and Mole Fraction Colligative Properties of Solutions 3. Lowering of Vapor Pressure and Raoult s Law 4. Fractional
More informationColligative Properties: Freezing Point Depression and Molecular Weight
Purpose: Colligative Properties: Freezing Point Depression and Molecular Weight The first purpose of this lab is to experimentally determine the van't Hoff (i) factor for two different substances, sucrose
More information1. Define the term colligative property and list those physical properties of a solution that can be classified as colligative properties.
Solutions Colligative Properties DCI Name Section 1. Define the term colligative property and list those physical properties of a solution that can be classified as colligative properties. Colligative
More informationR = J/mol K R = L atm/mol K
version: master Exam 1  VDB/LaB/Spk This MC portion of the exam should have 19 questions. The point values are given with each question. Bubble in your answer choices on the bubblehseet provided. Your
More informationSimple Mixtures. Atkins 7th: Sections ; Atkins 8th: The Properties of Solutions. Liquid Mixtures
The Properties of Solutions Simple Mixtures Atkins 7th: Sections 7.47.5; Atkins 8th: 5.45.5 Liquid Mixtures Colligative Properties Boiling point elevation Freezing point depression Solubility Osmosis
More informationSample Test 1 SAMPLE TEST 1. CHAPTER 12
13 Sample Test 1 SAMPLE TEST 1. CHAPTER 12 1. The molality of a solution is defined as a. moles of solute per liter of solution. b. grams of solute per liter of solution. c. moles of solute per kilogram
More information1 Exercise 5.5b pg 201
In this solution set, an underline is used to show the last significant digit of numbers. For instance in x 2.51693 the 2,5,1, and 6 are all significant. Digits to the right of the underlined digit, the
More informationColligative Properties
Colligative Properties Vapor pressures have been defined as the pressure over a liquid in dynamic equilibrium between the liquid and gas phase in a closed system. The vapor pressure of a solution is different
More informationExample: orange juice from frozen concentrate.
Dilution: a process in which the concentration (molarity) of a solution is lowered. The amount of solute (atoms, moles, grams, etc.) remains the same, but the volume is increased by adding more solvent.
More information12.3 Colligative Properties
12.3 Colligative Properties Changes in solvent properties due to impurities Colloidal suspensions or dispersions scatter light, a phenomenon known as the Tyndall effect. (a) Dust in the air scatters the
More informationAssigned questions for Lecture 14 are listed below. The questions occur in the following editions of Physical Chemistry by P.W.
Assigned questions for Lecture 14 are listed below. The questions occur in the following editions of Physical Chemistry by P.W. Atkins: 10th edition 9th edition 8th edition Note: The letter P in front
More informationCHEM 36 General Chemistry EXAM #1 February 13, 2002
CHEM 36 General Chemistry EXAM #1 February 13, 2002 Name: Serkey, Anne INSTRUCTIONS: Read through the entire exam before you begin. Answer all of the questions. For questions involving calculations, show
More informationColligative Properties of Nonvolatile Solutes 01. Colligative Properties of Nonvolatile Solutes 02. Colligative Properties of Nonvolatile Solutes 04
Colligative Properties of Nonvolatile Solutes 01 Colligative Properties of Nonvolatile Solutes 02 Colligative Properties: Depend on the amount not on the identity There are four main colligative properties:
More informationColligative properties. Chemistry 433. Freezing point depression. Freezing point depression. Freezing point depression 10/28/2008
Chemistry 433 Lecture 20 Colligative Properties Freezing Point Depression Boiling Point Elevation Osmosis NC State University Colligative properties There are a number of properties of a dilute solution
More informationChapter 13: Physical Properties of Solutions
Chapter 13: Physical Properties of Solutions Key topics: Molecular Picture (interactions, enthalpy, entropy) Concentration Units Colligative Properties terminology: Solution: a homogeneous mixture Solute:
More informationExperiment 5: Molecular Weight Determination From Freezing Point Depression
Experiment 5: Molecular Weight Determination From Freezing Point Depression PURPOSE To become familiar with colligative properties and to use them to determine the molar mass of a substance APPARATUS AND
More informationChapter 13 Properties of Solutions. Classification of Matter
Chapter 13 Properties of Solutions Learning goals and key skills: Describe how enthalpy and entropy changes affect solution formation Describe the relationship between intermolecular forces and solubility,
More informationEXERCISES. 16. What is the ionic strength in a solution containing NaCl in c=0.14 mol/dm 3 concentration and Na 3 PO 4 in 0.21 mol/dm 3 concentration?
EXERISES 1. The standard enthalpy of reaction is 512 kj/mol and the standard entropy of reaction is 1.60 kj/(k mol) for the denaturalization of a certain protein. Determine the temperature range where
More informationChemical Process calculation III
Chapter 7 Ideal and Real Gases Gas, Liquid, and Solid Chemical Process calculation III Gas: a substance in a form like air, relatively low in density and viscosity Liquid: a substance that flows freely
More information1) What is the overall order of the following reaction, given the rate law?
PRACTICE PROBLEMS FOR TEST 2 (March 11, 2009) 1) What is the overall order of the following reaction, given the rate law? A) 1st order B) 2nd order C) 3rd order D) 4th order E) 0th order 2NO(g) + H 2(g)
More informationPhysical pharmacy. dr basam al zayady
Physical pharmacy Lec 7 dr basam al zayady Ideal Solutions and Raoult's Law In an ideal solution of two volatile liquids, the partial vapor pressure of each volatile constituent is equal to the vapor pressure
More informationCHEMISTRY 443, Fall, 2011 (11F) Circle Section Number: Final Exam, December 14, 2011
NAME: CHEMISTRY 443, Fall, 2011 (11F) Circle Section Number: 10 11 80 81 Final Exam, December 14, 2011 Answer each question in the space provided; use back of page if extra space is needed. Answer questions
More informationVaporLiquid Equilibria
31 Introduction to Chemical Engineering Calculations Lecture 7. VaporLiquid Equilibria Vapor and Gas Vapor A substance that is below its critical temperature. Gas A substance that is above its critical
More informationMolecular Mass by Freezing Point Depression
Molecular Mass by Freezing Point Depression Kyle Miller November 28, 2006 1 Purpose The purpose of this experiment is to determine the molecular mass of organic compounds which are dissolved in a solvent
More informationFinal Exam CHM 3410, Dr. Mebel, Fall 2005
Final Exam CHM 3410, Dr. Mebel, Fall 2005 1. At 31.2 C, pure propane and nbutane have vapor pressures of 1200 and 200 Torr, respectively. (a) Calculate the mole fraction of propane in the liquid mixture
More information12A. A Molar Mass from FreezingPoint Depression
12A. A Molar Mass from FreezingPoint Depression Time: 2 hours Required chemicals and solutions: Reagent Requirement/5 Pairs Preparation of 1 L p C 6 H 4 Cl 2 2.0 g Cyclohexane 100 ml Other required materials:
More informationChem 420/523 Chemical Thermodynamics Homework Assignment # 6
Chem 420/523 Chemical hermodynamics Homework Assignment # 6 1. * Solid monoclinic sulfur (S α ) spontaneously converts to solid rhombic sulfur (S β ) at 298.15 K and 0.101 MPa pressure. For the conversion
More informationProblem # 2 Determine the kinds of intermolecular forces present in each element or compound:
Chapter 11 Homework solutions Problem # 2 Determine the kinds of intermolecular forces present in each element or compound: A. Kr B. NCl 3 C. SiH 4 D. HF SOLUTION: Kr is a single atom, hence it can have
More informationColligative properties CH102 General Chemistry, Spring 2014, Boston University
Colligative properties CH102 General Chemistry, Spring 2014, Boston University here are four colligative properties. vaporpressure lowering boilingpoint elevation freezingpoint depression osmotic pressure
More informationUnit 13 Practice Test
Name: Class: Date: Unit 13 Practice Test Multiple Choice Identify the choice that best completes the statement or answers the question. 1) The dissolution of water in octane (C 8 H 18 ) is prevented by.
More informationa) Consider mixing two liquids where mixing is exothermic ( Hsoln < 0). Would you expect a solution to form (yes/maybe/no)? Justify your answer.
Problems  Chapter 13 (with solutions) 1) The following question concerns mixing of liquids. a) Consider mixing two liquids where mixing is exothermic (Hsoln < 0). Would you expect a solution to form (yes/maybe/no)?
More informationEntropy Changes & Processes
Entropy Changes & Processes Chapter 4 of Atkins: he Second Law: he Concepts Section 4.3, 7th edition; 3.3, 8th edition Entropy of Phase ransition at the ransition emperature Expansion of the Perfect Gas
More informationLab 9. Colligative Properties an Online Lab Activity
Prelab Assignment Before coming to lab: Lab 9. Colligative Properties an Online Lab Activity Chemistry 162  K. Marr Revised Winter 2014 This lab exercise does not require a report in your lab notebook.
More informationChem 338 Homework Set #5 solutions October 10, 2001 From Atkins: 5.2, 5.9, 5.12, 5.13, 5.15, 5.17, 5.21
Chem 8 Homework Set #5 solutions October 10, 2001 From Atkins: 5.2, 5.9, 5.12, 5.1, 5.15, 5.17, 5.21 5.2) The density of rhombic sulfur is 2.070 g cm  and that of monoclinic sulfur is 1.957 g cm . Can
More informationName Class Date. In the space provided, write the letter of the term or phrase that best completes each statement or best answers each question.
Assessment Chapter Test A Chapter: States of Matter In the space provided, write the letter of the term or phrase that best completes each statement or best answers each question. 1. The kineticmolecular
More informationa) Use the following equation from the lecture notes: = ( 8.314 J K 1 mol 1) ( ) 10 L
hermodynamics: Examples for chapter 4. 1. One mole of nitrogen gas is allowed to expand from 0.5 to 10 L reversible and isothermal process at 300 K. Calculate the change in molar entropy using a the ideal
More informationTo calculate the value of the boiling point constant for water. To use colligative properties to determine the molecular weight of a substance.
Colligative Properties of Solutions: A Study of Boiling Point Elevation Amina ElAshmawy, Collin County Community College (With contributions by Timm Pschigoda, St. Joseph High School, St. Joseph, MI)
More informationThe first law: transformation of energy into heat and work. Chemical reactions can be used to provide heat and for doing work.
The first law: transformation of energy into heat and work Chemical reactions can be used to provide heat and for doing work. Compare fuel value of different compounds. What drives these reactions to proceed
More informationMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
A.P. Chemistry Practice Test: Ch. 11, Solutions Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) Formation of solutions where the process is
More informationThermodynamics and Kinetics. Lecture 14 Properties of Mixtures Raoult s Law Henry s Law Activity NC State University
Thermodynamics and Kinetics Lecture 14 Properties of Mixtures Raoult s Law Henry s Law Activity NC State University Measures of concentration There are three measures of concentration: molar concentration
More informationThe Solution Process CHEMISTRY. Properties of Solutions. The Central Science. Prof. Demi Levendis Room GH807 Gate House
CHEMISTRY The Central Science Properties of Solutions The Solution Process Solutions: Air; brass; body fluids; sea water When a solution forms some questions we can ask are: What happens on a molecular
More informationCHEM 10123/10125, Exam 1 February 8, 2012 (50 minutes)
CHEM 10123/10125, Exam 1 February 8, 2012 (50 minutes) Name (please print) 1. SHOW ALL WORK. A saturated, aqueous NaCl solution is 5.40 M NaCl (58.44 g/mol) and is 26.0% NaCl by weight. a.) (10 points)
More informationName Date Class. SECTION 16.1 PROPERTIES OF SOLUTIONS (pages 471 477)
16 SOLUTIONS SECTION 16.1 PROPERTIES OF SOLUTIONS (pages 471 477) This section identifies the factors that affect the solubility of a substance and determine the rate at which a solute dissolves. Solution
More informationExperiment 1: Colligative Properties
Experiment 1: Colligative Properties Determination of the Molar Mass of a Compound by Freezing Point Depression. Objective: The objective of this experiment is to determine the molar mass of an unknown
More informationGibbs Free Energy and Chemical Potential. NC State University
Chemistry 433 Lecture 14 Gibbs Free Energy and Chemical Potential NC State University The internal energy expressed in terms of its natural variables We can use the combination of the first and second
More informationVapor Pressure Lowering
Colligative Properties A colligative property is a property of a solution that depends on the concentration of solute particles, but not on their chemical identity. We will study 4 colligative properties
More informationCHEMISTRY The Molecular Nature of Matter and Change
CHEMISTRY The Molecular Nature of Matter and Change Third Edition Chapter 13 The Properties of Mixtures: Solutions and Colloids Copyright The McGrawHill Companies, Inc. Permission required for reproduction
More information2. Why does the solubility of alcohols decrease with increased carbon chain length?
Colligative properties 1 1. What does the phrase like dissolves like mean. 2. Why does the solubility of alcohols decrease with increased carbon chain length? Alcohol in water (mol/100g water) Methanol
More informationDetermination of Molar Mass by Boiling Point Elevation of Urea Solution
Determination of Molar Mass by Boiling Point Elevation of Urea Solution CHRISTIAN E. MADU, PhD AND BASSAM ATTILI, PhD COLLIN COLLEGE CHEMISTRY DEPARTMENT Purpose of the Experiment Determine the boiling
More informationFrom the book (10, 12, 16, 18, 22, 24 52, 54, 56, 58, 62, 64, 66, 68, 74, 76, 78, 80, 82, 86, 88, 90, 92, 106 and 116)
Chem 112 Solutions From the book (10, 12, 16, 18, 22, 24 52, 54, 56, 58, 62, 64, 66, 68, 74, 76, 78, 80, 82, 86, 88, 90, 92, 106 and 116) 1. Which of the following compounds are nonelectrolytes? A. NaF
More informationEXPERIMENT 15 FREEZING POINT: A COLLIGATIVE PROPERTY OF SOLUTIONS
EXPERIMENT 15 FREEZING POINT: A COLLIGATIVE PROPERTY OF SOLUTIONS INTRODUCTION Scientists seldom work alone today. Instead they work in teams on projects, sharing the labor of carrying out the experiments
More informationChapter Thirteen. Physical Properties Of Solutions
Chapter Thirteen Physical Properties Of Solutions 1 Solvent: Solute: Solution: Solubility: Types of Solutions Larger portion of a solution Smaller portion of a solution A homogeneous mixture of 2 or more
More information0.279 M Change g to mol: g/mol = mol Molarity = mol L = mol 0.325L = M
118 ChemQuest 39 Name: Date: Hour: Information: Molarity Concentration is a term that describes the amount of solute that is dissolved in a solution. Concentrated solutions contain a lot of dissolved solute,
More information1. solid, vapor, critical point correct. 2. solid, liquid, critical point. 3. liquid, vapor, critical point. 4. solid, liquid, triple point
mcdonald (pam78654) HW 7B: Equilibria laude (89560) 1 This printout should have 18 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0
More informationChapter 14 Solutions
Chapter 14 Solutions 1 14.1 General properties of solutions solution a system in which one or more substances are homogeneously mixed or dissolved in another substance two components in a solution: solute
More informationCHAPTER 14 THE CLAUSIUSCLAPEYRON EQUATION
CHAPTER 4 THE CAUIUCAPEYRON EQUATION Before starting this chapter, it would probably be a good idea to reread ections 9. and 9.3 of Chapter 9. The ClausiusClapeyron equation relates the latent heat
More informationThermodynamics Answers to Tutorial # 1
Thermodynamics Answers to Tutorial # 1 1. (I) Work done in free expansion is Zero as P ex = 0 (II) Irreversible expansion against constant external pressure w = P ex (V 2 V 1 ) V 2 = nrt P 2 V 1 = nrt
More informationESSAY. Write your answer in the space provided or on a separate sheet of paper.
Test 1 General Chemistry CH116 Summer, 2012 University of Massachusetts, Boston Name ESSAY. Write your answer in the space provided or on a separate sheet of paper. 1) Sodium hydride reacts with excess
More informationSolutions. Chapter 13. Properties of Solutions. Lecture Presentation
Lecture Presentation Chapter 13 Properties of Yonsei University homogeneous mixtures of two or more pure substances: may be gases, liquids, or solids In a solution, the solute is dispersed uniformly throughout
More informationChemistry Notes for class 12 Chapter 2 Solutions
1 P a g e Chemistry Notes for class 12 Chapter 2 Solutions Solution is a homogeneous mixture of two or more substances in same or different physical phases. The substances forming the solution are called
More information13.3 Factors Affecting Solubility SoluteSolvent Interactions Pressure Effects Temperature Effects
Week 3 Sections 13.313.5 13.3 Factors Affecting Solubility SoluteSolvent Interactions Pressure Effects Temperature Effects 13.4 Ways of Expressing Concentration Mass Percentage, ppm, and ppb Mole Fraction,
More informationObserve and measure the effect of a solute on the freezing point of a solvent. Determine the molar mass of a nonvolatile, nonelectrolyte solute
Chapter 10 Experiment: Molar Mass of a Solid OBJECTIVES: Observe and measure the effect of a solute on the freezing point of a solvent. Determine the molar mass of a nonvolatile, nonelectrolyte solute
More informationChapter 6 Multiphase Systems
Chapter 6 Multiphase Systems SingleComponent Systems Phase Diagram: a plot that shows conditions under which a pure substance exists in a particular phase e.g. a liquid, a solid, or a gas. Often, the
More informationColligative Properties
Colligative Properties Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) To access a customizable version of this book, as well as other interactive content, visit www.ck12.org
More informationDetermination of the Molar Mass of an Unknown Solid by Freezing Point Depression
Determination of the Molar Mass of an Unknown Solid by Freezing Point Depression GOAL AND OVERVIEW In the first part of the lab, a series of solutions will be made in order to determine the freezing point
More informationChapter 7 : Simple Mixtures
Chapter 7 : Simple Mixtures Using the concept of chemical potential to describe the physical properties of a mixture. Outline 1)Partial Molar Quantities 2)Thermodynamics of Mixing 3)Chemical Potentials
More informationThe Freezing Point Depression of tbutanol
CHEM 122L General Chemistry Laboratory Revision 2.0 The Freezing Point Depression of tbutanol To learn about the Freezing Point Depression of Solvents. To learn about Colligative Properties. To learn
More informationChapter 13 Properties of Solutions
Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 13 Properties of are homogeneous mixtures of two or more pure substances. In a solution,
More informationBomb Calorimetry. Example 4. Energy and Enthalpy
Bomb Calorimetry constant volume often used for combustion reactions heat released by reaction is absorbed by calorimeter contents need heat capacity of calorimeter q cal = q rxn = q bomb + q water Example
More informationStates of Matter CHAPTER 10 REVIEW SECTION 1. Name Date Class. Answer the following questions in the space provided.
CHAPTER 10 REVIEW States of Matter SECTION 1 SHORT ANSWER Answer the following questions in the space provided. 1. Identify whether the descriptions below describe an ideal gas or a real gas. ideal gas
More informationColligative Properties Discussion Chem. 1A
Colligative Properties Discussion Chem. 1A The material covered today is found in sections Chapter 12.5 12.7 This material will not be covered in lecture, you will have homework assigned. Chem. 1A Colligative
More informationPreLab Notebook Content: Your notebook should include the title, date, purpose, procedure; data tables.
Determination of Molar Mass by Freezing Point Depression M. Burkart & M. Kim Experimental Notes: Students work in pairs. Safety: Goggles and closed shoes must be worn. Dispose of all chemical in the plastic
More informationChapter 14. Mixtures
Chapter 14 Mixtures Warm Up What is the difference between a heterogeneous and homogeneous mixture? Give 1 example of a heterogeneous mixture and 1 example of a homogeneous mixture. Today s Agenda QOTD:
More informationTypes of Solutions. Chapter 17 Properties of Solutions. Types of Solutions. Types of Solutions. Types of Solutions. Types of Solutions
Big Idea: Liquids will mix together if both liquids are polar or both are nonpolar. The presence of a solute changes the physical properties of the system. For nonvolatile solutes the vapor pressure, boiling
More informationChapter 5 Thermochemistry
Chapter 5 Thermochemistry I. Nature of Energy Energy units SI unit is joule, J From E = 1/2 mv 2, 1J = 1kg. m 2 /s 2 Traditionally, we use the calorie as a unit of energy. 1 cal = 4.184J (exactly) The
More informationName Date Class THERMOCHEMISTRY. SECTION 17.1 THE FLOW OF ENERGY HEAT AND WORK (pages 505 510)
17 THERMOCHEMISTRY SECTION 17.1 THE FLOW OF ENERGY HEAT AND WORK (pages 505 510) This section explains the relationship between energy and heat, and distinguishes between heat capacity and specific heat.
More informationCOPYRIGHT FOUNTAINHEAD PRESS
Colligative Properties of Solutions Freezing Point Depression Objectives: To investigate the colligative property of freezing point depression; to examine the relationship between freezing point depression
More informationBoyle s law  For calculating changes in pressure or volume: P 1 V 1 = P 2 V 2. Charles law  For calculating temperature or volume changes: V 1 T 1
Common Equations Used in Chemistry Equation for density: d= m v Converting F to C: C = ( F  32) x 5 9 Converting C to F: F = C x 9 5 + 32 Converting C to K: K = ( C + 273.15) n x molar mass of element
More information10.7 Kinetic Molecular Theory. 10.7 Kinetic Molecular Theory. Kinetic Molecular Theory. Kinetic Molecular Theory. Kinetic Molecular Theory
Week lecturestentative 0.7 KineticMolecular Theory 40 Application to the Gas Laws 0.8 Molecular Effusion and Diffusion 43 Graham's Law of Effusion Diffusion and Mean Free Path 0.9 Real Gases: Deviations
More informationChapter 12. Solutions. Lecture Presentation
12.1 Thirsty Solutions: Why You Shouldn t Drink Seawater 544 12.2 Types of Solutions and Solubility 546 12.3 Energetics of Solution Formation 551 12.4 Solution Equilibrium and Factors Affecting Solubility
More informationCHM1045 Practice Test 3 v.1  Answers Name Fall 2013 & 2011 (Ch. 5, 6, 7, & part 11) Revised April 10, 2014
CHM1045 Practice Test 3 v.1  Answers Name Fall 013 & 011 (Ch. 5, 6, 7, & part 11) Revised April 10, 014 Given: Speed of light in a vacuum = 3.00 x 10 8 m/s Planck s constant = 6.66 x 10 34 J s E (.18x10
More informationCHAPTER 13: ANSWERS TO ASSIGNED PROBLEMS Hauser General Chemistry I revised 8/03/08
CHAPTER 13: ANSWERS TO ASSIGNED PROBLEMS Hauser General Chemistry I revised 8/03/08 13.21 The solubility of Cr(NO 3 ) 3 9 H 2 O in water is 208 g per 100 g of water at 15 C. A solution of Cr(NO 3 ) 3
More informationThermodynamics of Mixing
Thermodynamics of Mixing Dependence of Gibbs energy on mixture composition is G = n A µ A + n B µ B and at constant T and p, systems tend towards a lower Gibbs energy The simplest example of mixing: What
More informationSolutions. Occur in all phases. Ways of Measuring. Ways of Measuring. Energy of Making Solutions. 1. Break apart Solvent. Page 1
s Occur in all phases The solvent does the dissolving. The solute is dissolved. There are examples of all types of solvents dissolving all types of solvent. We will focus on aqueous solutions. Ways of
More informationTwo Ways to Form Solutions. Role of Disorder in Solutions 2/27/2012. Types of Reactions
Role of Disorder in Solutions Disorder (Entropy) is a factor Solutions mix to form maximum disorder Two Ways to Form Solutions 1. Physical Dissolving (Solvation) NaCl(s) Na + (aq) + Cl  (aq) C 12 H 22
More informationConcentration of Solutions and Molarity
Concentration of Solutions and Molarity The concentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solvent. A dilute solution is one that contains a small
More informationFreezing Point Depression, the van t Hoff Factor, and Molar Mass
, the van t Hoff Factor, and Molar Mass Objectives To understand colligative properties. To find the freezing point depression of a solution. To determine the van't Hoff factor for acetic acid dissolved
More informationH 2 (g) + ½ O 2 (g) H 2 O(l) H o f [NO(g)] = 90.2 kj/mol; H o f [H 2 O(g)] = kj/mol H o f [NH 3 (g)] = kj/mol; H o f [O 2 (g)] =?
Chapter 16 Thermodynamics GCC CHM152 Thermodynamics You are responsible for Thermo concepts from CHM 151. You may want to review Chapter 8, specifically sections 2, 5, 6, 7, 9, and 10 (except work ). Thermodynamics:
More information