Study Guide for Module 11B Solutions II

Transcription

1 Chemistry 1020, Module 11B Name Study Guide for Module 11B Solutions II Reading Assignment: Sections 11.2 through 11.7 in Chemistry, 6th Edition by Zumdahl. Guide for Your Lecturer: 1. Review of Preparing Solutions 2. Solubility Rules 3. The Solution Process and Analytical Concentrations 4. Solutions Compared to Colloids and Suspensions; Review of Freezing Point, Boiling Point, Vapor Pressure; Osmosis and Osmotic Pressure 5. Qualitative Consideration of the Colligative Properties of Solutions 6. Calculation of Colligative Properties: Vapor Pressure Lowering 7. Calculation of Colligative Properties: Osmotic Pressure 8. Calculation of Colligative Properties: Freezing Point Depression 9. Calculation of Colligative Properties: Boiling Point Elevation 10. Molecular Weight of an Unknown from Freezing Point Depression or Boiling Point Elevation Homework Note: indicates problems to be stressed on drill quizzes and hour exams. Xavier University of Louisiana 43

2 Chemistry 1020, Module 11B Review of Preparing Solutions (From Module 4) Note: The procedure used to prepare the solution is the same as that taught in Chemistry 1010 and the accompanying laboratory. It is: Step 1: Step 2: Step 3: Step 4: Step 5: Measure out g of the solute (or ml of the original solution). Put approximately one-half of the water in the "total volume" into a container. (If solution is being made from solid, "total volume" = volume of solution to be made. If solution is being made from another solution, "total volume" = volume of solution to be made - volume of solution to be diluted.) Add the substance from Step 1 to the water in Step 2 and mix until all is dissolved. Add water until the total volume to be made is obtained. Mix the solution. (Note: If the solution is made in a volumetric flask, cap and invert to mix.) 1a) S. How would you prepare 105 milliliters of 0.32 M copper sulfate penta hydrate solution from pure solute? Formula: CuSO4 5H2O Type cmpd: salt Strength: strong Calculation gfw of CuSO4 5H2O = * (2* ) = g/mol M = mol solute/l soln M = (g/mw)/l M*L = g/mw g = M*L*mw L soln = 105 ml * (1 L/1000 ml) = L g = (0.32 mol/l)(0.105 L)(249.7 g/mol) = R = 8.39 g = 8.4 g CuSO4 5H2O * *Although this number has been rounded, in the lab you would weigh as accurately as the balance available allows. In General Chemistry, this would be to the nearest ±0.01 g so you would weight out a number close to 8.4 and record its weight to two places past the decimal. Steps to be used in preparing solution 1) Weigh out approximately 8.4 g CuSO4 5H2O and then record the actual mass reading on the balance to the nearest 0.01 g. 2) Add about 105/2 = 52 ml of H2O to the container. 3) Add the 8.4 g CuSO4 5H2O to the water in the container and mix until the solute dissolves. 4) Add enough additional H2O to get a total of 105 ml of solution. 5) Mix the solution. A. How would you prepare 112 milliliters of 0.23 M sodium sulfate solution from sodium sulfate hexahydrate? Calculation Steps to be used in preparing solution 1) 2) 3) 4) 5) 44 Xavier University of Louisiana

3 Chemistry 1020, Module 11B Name Review of Preparing Solutions (continued) 1a) B. How would you prepare 123 milliliters of 0.45 M cesium chlorate solution from cesium chlorate dihydrate? Calculation Steps to be used in preparing solution 1) 2) 3) 4) 5) C. How would you prepare 34 milliliters of 0.78 M glucose (gfw=180.0) from solid solute? Calculation Steps to be used in preparing solution 1) 2) 3) 4) 5) D. How would you prepare 81 milliliters of 0.13 M urea (gfw=60.0) solution from solid solute? Calculation Steps to be used in preparing solution 1) 2) 3) 4) 5) Xavier University of Louisiana 45

4 Chemistry 1020, Module 11B Review of Preparing Solutions (continued) 1a) E. How would you prepare 93 milliliters of 0.24 M sodium carbonate solution from solid solute? Calculation Steps to be used in preparing solution 1) 2) 3) 4) 5) b) S. How would you prepare 112 milliliters of 0.87 M hydrobromic acid from 3.4 M hydrobromic acid? Formula: HBr Type cmpd: acid Strength: strong Calculation M1 = 3.4 M V1 =? M2 = 0.87 M V2 = 112 ml M1V1 = M2V2 V1 = M2V2/M1 V1 = (0.87 M)(112 ml)/(3.4 M) = R V1 = ml = 29 ml of 3.4 M HBr* *Although this number has been rounded, in the lab you would measure volume as accurately as the equipment available allows. In General Chemistry, this would be to the nearest ±0.1 ml if you use a graduated cylinder or to the nearest ±0.01 ml if you use a pipette. If you don t know what kind of equipment to use, assume a graduated cylinder. Steps to be used in preparing solution 1) Measure out 28.7 ml of 3.4 M HBr 2) Put approximately ( )/2 = 42 ml of H2O in a container. 3) Add the 28.7 ml of 3.4 M HBr to the container and mix. 4) Add additional water to the container until a total of 112 ml of solution is obtained. 5) Mix the solution. A. How would you prepare 34 milliliters of 0.98 M sulfuric acid from a 5.2 M solution of sulfuric acid? Calculation Steps to be used in preparing solution 1) 2) 3) 4) 5) 46 Xavier University of Louisiana

5 Chemistry 1020, Module 11B Name Review of Preparing Solutions (continued) 1b) B. How would you prepare 123 milliliters of 1.2 M hydrochloric acid from 2.0 M hydrochloric acid? Calculation Steps to be used in preparing solution 1) 2) 3) 4) 5) C. How would you prepare 565 milliliters of 0.17 M nitric acid from 1.3 M nitric acid? Calculation Steps to be used in preparing solution 1) 2) 3) 4) 5) D. How would you prepare 81 milliliters of 1.5 M acetic acid from a 2.8 M solution of acetic acid? Calculation Steps to be used in preparing solution 1) 2) 3) 4) 5) Xavier University of Louisiana 47

6 Chemistry 1020, Module 11B Review of Preparing Solutions (continued) 1b) E. How would you prepare 78 milliliters of 0.57 M phosphoric acid from a 4.5 M solution of phosphoric acid? Calculation Steps to be used in preparing solution 1) 2) 3) 4) 5) 48 Xavier University of Louisiana

7 Chemistry 1020, Module 11B Name Solubility Rules Names and Formulas of Common Anions and Common Chemical Compounds 2a) List the following by name and formula. (NOTE: This comes directly from Chemistry 1010, Module 0.) A. Four polyatomic anions which contain S. Name Formula Name Formula B. Two polyatomic anions which contain only N and O. C. Four polyatomic anions which contain Cl. D. Four polyatomic anions which contain P. E. Four polyatomic anions which contain C. F. Two polyatomic anions which contain Cr. G. List four other important polyatomic anions. H. Four Group VIIA monoatomic anions. I. Two Group VIA monoatomic anions. Xavier University of Louisiana 49

8 Chemistry 1020, Module 11B 2b. Cations whose salts are generally soluble: Na +, K +, NH4 +, and the other Group IA metal ions Anions whose salts are all soluble: NO3 -, CH3COO -, and ClO4 - Anions whose salts are generally soluble: Cl -, except for AgCl, Hg2Cl2*, and PbCl2 Br -, except for AgBr, Hg2Br2*, PbBr2, and HgBr2 I -, except for AgI, Hg2I2*, PbI2, and HgI2 SO4 2- except for CaSO4, SrSO4, BaSO4, PbSO4, Hg2SO4*, &Ag2SO4 Anions whose salts are generally insoluble: S 2-, except for those of IA and IIA metals and (NH4)2S. CO3 2-, except for those of IA metals and (NH4)2CO3 SO3 2-, except for those of IA metals and (NH4)2SO3 PO4 3-, except for those of IA metals and (NH4)3PO4 OH -, except for those of IA metals, Ba(OH)2, Sr(OH)2, and Ca(OH)2 *Hg2 2+ is mercury(i) ion (or mercurous ion). It is the only diatomic metallic cation you need to know. A. List the names and formulas of seven cations almost all of whose salts are water soluble. Then list exceptions. Formula Name Exceptions B. List the names and formulas of three anions almost all of whose salts are water soluble. Then list exceptions. C. List the names and formulas of four anions most of whose salts are water soluble. Then list exceptions. D. List the names and formulas of five anions most of whose salts are not soluble in water. Then list exceptions. 50 Xavier University of Louisiana

9 Chemistry 1020, Module 11B Name The Solution Process and Analytical Concentrations 3a) Draw a sketch showing solid sodium chloride dissolving in water to form solvated sodium ion and chloride ions. Write the balanced chemical equation for the solution of NaCl in water. b) List three steps which together might be used to explain how the solution process occurs. (See pp. 516) Step 1: Step 2: Step 3: c) Write a balanced chemical equation showing the given substance dissolving in water. S. sodium acetate Formula: NaCH3COO Type cmpd: salt Strength: strong NaCH3COO(s) Na + (aq) + CH3COO - (aq) A. ammonium sulfide B. copper(ii) nitrate C. zinc(ii) sulfate D. calcium chloride E. rubidium sulfite Xavier University of Louisiana 51

10 Chemistry 1020, Module 11B The Solution Process and Analytical Concentrations (continued) 3d) Use an INITIAL/CHANGE/FINAL chart to determine the real concentrations of the species in the solution. (pp ) S M Na2SO4 Name: sodium sulfate Type cmpd: salt Strength: strong Na2SO4(s) 2 Na + (aq) + SO4 2- (aq) initial _ final _ 0.30 Therefore, real concentrations of species in solution are: [Na + ] = 0.60 M, [SO4 2- ] = 0.30 M A M CaCl2 Name Type cmpd: Strength: B M HNO3 Name: Type cmpd: Strength: C M KOH Name: Type cmpd: Strength: D M Na3PO4 Name: Type cmpd: Strength: E M K2SO4 Name: Type cmpd: Strength: 52 Xavier University of Louisiana

11 Chemistry 1020, Module 11B Name Solutions Compared to Colloids and Suspensions; Selected Properties of Solutions 4a) Distinguish among solutions, colloids, and suspensions. (All three are mixtures. They differ in terms of the size of "solute" particles. In a solution, the solute particles are small and the mixture is homogeneous all the way down to the molecular level. In a suspension the solute particles are still small enough to stay suspended for a while but are large enough to settle out {precipitate} if the mixture is left to stand for a while. In a colloid, the solute particles are intermediate in size of the solute particles in a solution and the size in a suspension. As a result, they are small enough to stay suspended indefinitely {i.e. don't settle out as in the suspension} but are too large to consider the solution as truly homogeneous.) b) Outline a procedure you could use to determine if a given mixture were a solution, colloid, or a suspension. (Let the mixture sit for a while. If a precipitate settles out, it is a suspension. If not, shine light through it. If the light beam passes through without being reflected, it is a solution. If the light beam is reflected to the side {i.e. can be seen passing through the mixture in the way a light beam can be seen passing through a cloud of dust), the mixture is a colloid. The solute particles in a solution are not large enough to reflect the light, those in the colloid are.) c) Describe what happens (overall) to solute and solvent as two solutions of differing concentrations diffuse together. d) Sketch the overall movement of urea and water if a 0.5 M solution of urea is placed in contact with a 0.2 M solution without mixing. e) Define osmosis. (p. 535, A37) f) Briefly compare what happens to solute and solvent during diffusion (question 4c ) with what happens during osmosis (question 4e). Xavier University of Louisiana 53

12 Chemistry 1020, Module 11B Solutions Compared to Colloids and Suspensions; Selected Properties of Solutions (continued) 4g) S. Answer the following questions concerning the diagram below which represents a cell in which two solutions are separated by a semipermeable membrane. (pp ) 0.5 m 0.03 m urea urea 1-In which direction does the water flow as the system goes to equilibrium? From the more dilute 0.03 m solution to the more concentrated 0.5 m solution. 2-In which direction does the solute flow as the system goes to equilibrium? Solute does not flow. The membrane is semipermeable which means that only H2O can flow. 3-What happens to the concentration of the 0.03 m side of the system as it goes to equilibrium? It increases as water flows out. 4-What happens to the concentration of the 0.5 m side of the system as it goes to equilibrium? It decreases as water flows in. 5-What happens to the volume of solution in the 0.03 m side of the system as it goes to equilibrium? It decreases as water flows out. 6-What happens to the volume of solution in the 0.5 m side of the system as it goes to equilibrium? It increases as water flows in. A. Answer the following questions concerning the diagram below which represents a balloon made from a semipermeable membrane filled with a 0.5 m glucose solution and immersed in a 0.2 m glucose solution in a beaker. 1-In which direction does the water flow as the system goes to equilibrium? 2-In which direction does the solute flow as the system goes to equilibrium? 3-What happens to the concentration of the solution inside the balloon as the system goes to equilibrium? 4-What happens to the concentration of the solution in the beaker as the system goes to equilibrium? 5-What happens to the volume of solution in the balloon as the system goes to equilibrium? 6-What happens to the volume of solution outside the balloon as the system goes to equilibrium? 54 Xavier University of Louisiana

13 Chemistry 1020, Module 11B Name Solutions Compared to Colloids and Suspensions; Selected Properties of Solutions (continued) 4g) B. Answer the following questions concerning the diagram below which represents a cell in which two solutions are separated by a semipermeable membrane. 1-In which direction does the water flow as the system goes to equilibrium? The solution inside the upside funnel above is 0.03 m in urea. That outside is 0.1 m in urea. There is a semipermable membrane across the top of the funnel (on the bottom of the diagram) separating the two. 2-In which direction does the solute flow as the system goes to equilibrium? 3-What happens to the concentration outside the funnel as the system goes to equilibrium? 4-What happens to the concentration inside the funnel as the system goes to equilibrium? 5-What happens to the volume of solution outside the funnel as the system goes to equilibrium? 6-What happens to the volume of solution inside the funnel as the system goes to equilibrium? C. Answer the following questions concerning the diagram below which represents a U-tube with a 0.4 m glucose solution on the left (below) and a 0.1 m glucose solution on the right. There is a semipermeable membrane separating the two halves of the tube.. 1-In which direction does the water flow as the system goes to equilibrium? 2-In which direction does the solute flow as the system goes to equilibrium? 3-What happens to the concentration of the solution on the left as the system goes to equilibrium? 0.4 m 0.1 m glucose glucose 4-What happens to the concentration of the solution on the right as the system goes to equilibrium? 5-What happens to the volume of solution on the left as the system goes to equilibrium? 6-What happens to the volume of solution on the right as the system goes to equilibrium? Xavier University of Louisiana 55

14 Chemistry 1020, Module 11B Qualitative Consideration of the Colligative Properties of Solutions 5a) Define colligative property (p. 531,A31) b) Define vapor pressure. (p. 484, A41) c) If liquid water is placed in a flask and the flask is closed, some of the water evaporates and gives rise to vapor pressure. Draw a stoppered flask showing molecules of water in both the liquid and vapor states. Then explain what causes the vapor pressure. (p. 484) d) Define melting point. (p. 491) e) Define boiling point. (p. 491) f) Define osmotic pressure. (p. 535) g) What happens to each of the following when the concentration of a solution increases? -vapor pressure of water above the solution (pp ) -osmotic pressure of the solution (p ) -freezing point of the solution (p. 533) -boiling point of the solution (p. 531) h) List four colligative properties of solutions AND indicate where you will study them in the remainder of this module. Colligative Property Learning Goal Where Covered 1) 2) 3) 4) i) Define "particle concentration". (particle concentration = concentration of solute * number of particles each solute molecule breaks into when in solution) j) What are the two factors which determine the particle concentration of a solution? (1) concentration of solute, 2) the extent of dissociation of the solute) 56 Xavier University of Louisiana

15 Chemistry 1020, Module 11B Name Qualitative Consideration of the Colligative Properties of Solutions (continued) 5k) For each of the given pairs of solutions, indicate the strength of the solutes and the particle concentrations. Then answer the questions in the 4th and 5th columns. INCLUDE EXPLANATIONS. Solutions Strength Particle Which solution has the Which solution conducts Concentrations indicated property? S m urea nonelectrolyte 0.16 m Higher fp? The two have the same parti m glucose nonelectrolyte 0.16 m cle concentrations so they have the same fp. S m NaCl strong 0.46 m Lower vapor pressure? electrolyte 0.23 m NaCl since 0.13 m NaCl strong 0.26 m its particle conc. is electrolyte higher. A m urea Higher osmotic press.? 0.16 m glucose B m glucose Lower boiling point? 0.29 m glucose electrity better? Neither conduct since no ions are in either m NaCl since it has higher concentration of ions. C m Na2SO m glucose Higher vapor pressure? Solutions Strength Particle Concentrations Which solution has the indicated property 1? D m urea Higher bp? Which solution conducts electrity better? 0.16 m NaNO3 E m NaCl Higher bp? 0.44 m glucose F m urea Lower vapor press.? 0.23 m glucose G m NaBr Higher freezing point? 0.21 m NaBr H m urea Lower vapor pressure? 0.38 m urea I m NaNO3 Lower freezing point? 0.17 m KBr Xavier University of Louisiana 57

16 Chemistry 1020, Module 11B Calculation of Colligative Properties: Vapor Pressure Lowering 6a) In general how does vapor pressure of a solution change as its concentration changes? (p ) b) In words, what is the quantitative relationship between vapor pressure of a solution a nonelectrolyte and the concentration of that solution? (The vapor pressure of the solution is directly proportional to the mole fraction of water in the solution. ) c) Write the equation for the statement in b). (Psoln = Xwater in soln*ppure water) d) In real life, chemists are generally more interested in the how much the vapor pressure of a solution differs from the vapor pressure of pure water than in the vapor pressure of the solution itself. In addition, they usually know the mole fraction of solute in the solution rather than mole fraction of water. Use Xsolute + Xpure water = 1 to obtain an equation which relates change in vapor pressure as the solution forms (i.e. the difference between the vapor pressure of the solution from that of pure water) to the mole fraction of solute. Note: This is the equation usually used when considering colligative properties of water. ( VP = Ppure water - Psoln = Xsolute*Ppure water ) e) When can the equation in d above be used? (i.e. What are its limitations?) f) S. A certain solution is prepared at 20 o C by adding 42.0 grams of glucose (gfw = 180.0) to 251 ml of water. How does the vapor pressure of a solution differ from that of pure water at the same temperature? (pp ) Formula: C6H12O6 Type cmpd: not a, b, or s Strength: nonelectrolyte Psoln = Pwater*Xwater Pwater = torr (from data sheet) Xwater = (mol water)/(total moles) mol water = 251 ml H2O * * 1 g H 2O 1 ml H2O * 1 mol H2O g H2O = = mol H2O mol glucose = 42.0 g glucose * 1 mol glucose * g glucose = mol glucose Xwater = mol H2O total moles = = Psoln = Pwater*Xwater = (17.54 torr)(0.9838) = torr * = 17.3 torr P = torr = 0.24 torr difference which rounds to 0.2 torr Therefore, it is 0.2 torr lower. *Reminder: Carry one more place in intermediate steps which involve multiplication or division. Then round round to the number you want when you finish multiplication/division before you do addition or subtraction. A. How much does the vapor pressure of pure water at 60 o C change if the water is used to prepare a solution by adding 21.3 grams of urea (CH4N2O) to 302 ml of H20? 58 Xavier University of Louisiana

17 Chemistry 1020, Module 11B Name Calculation of Colligative Properties: Vapor Pressure Lowering (continued) 6f) B. What is the vapor pressure of a solution prepared by adding 45.3 grams of glucose (C6H12O6) to 153 ml of water at 50 o C? C. How much is the vapor pressure of a solution lowered by adding 25.9 grams of nonelectrolyte with a molecular weight of to 151 ml of water at 90 o C? D. How much is the vapor pressure lowered by adding 55.7 grams of glucose (C6H12O6) to 125 ml of water at 60 o C? E. What is the vapor pressure at 80 o C of a solution prepared by adding 42.4 grams of glucose (gfw = 180.0) to 125 ml of water? Xavier University of Louisiana 59

18 Chemistry 1020, Module 11B Calculation of Colligative Properties: Vapor Pressure Lowering (continued) 6f) F. How many grams of urea (gfw = 60.0) must be added to 212 ml of water to make a solution whose vapor pressure at 70 o C is 213 torr? Calculation of Colligative Properties: Osmotic Pressure 7a) In general how does osmotic pressure of a solution change as its concentration changes? (p ) b) State a quantitative relationship between osmotic pressure and concentration of a solution in words. (p. 537) c) S. A solution of urea is found to have an osmotic pressure of 4.2 atm. What would be the osmotic pressure of a solution with twice the concentration? Since osmotic pressure is directly proportional to concentration, doubling the concentration of the solution would cause the osmotic pressure to double. Therefore, the osmotic pressure of the solution twice the concentration would be 2*4.2 atm = 8.4 atm. A. A solution of urea is found to have an osmotic pressure of 8.4 atm. What would be the osmotic pressure of a solution with three times the concentration? B A solution of glucose is found to have an osmotic pressure of 5.6 atm. What would be the osmotic pressure of a solution with one-half the concentration? C. A solution of urea is found to have an osmotic pressure of 2.4 atm. What would be the osmotic pressure of a solution with one-fourth the concentration? D. A solution of glucose is found to have an osmotic pressure of 9.6 atm. What would be the osmotic pressure of a solution with one-third the concentration? 60 Xavier University of Louisiana

19 Chemistry 1020, Module 11B Name Calculation of Colligative Properties: Osmotic Pressure (continued) 7d) Write the equation showing how osmotic pressure of solution is related to concentration. (p. 537) e) What is the constant in the equation in d above called? f) When can the equation in d above be used? (i.e. What are its limitations?) g) S. Calculate the osmotic pressure at room temperature (25 o C) of a solution prepared by adding 12 grams of urea (gfw=60.0) to 93 grams of water. The density of the solution is 1.01 g/ml. (Section 11.6 in text) Formula: CO(NH2)2 Type cmpd: not a, b, or s Strength: nonelectrolyte Osmotic pressure = = MRT and M = (mol urea)/(l soln) 1 mol urea mol urea = 12 g urea * = mol urea 60.0 g urea 1.00 ml L soln = (12 g urea + 93 g H2O) * 1.01 g * 1 L 1000 ml = L soln mol urea = MRT = L * L*atm mol*k * ( K) = R = 46.9 atm = 47 atm A. Calculate the osmotic pressure at 50 o C of a solution prepared by adding 23 g of urea (gfw = 60.0) to 112 g of H2O. The density of the solution is 1.08 g/ml. B. Calculate the osmotic pressure at 75 o C of a solution prepared by adding 43 grams of glucose (gfw = 180.0) to 234 grams of water. The density of the solution is 1.07 g/ml. Xavier University of Louisiana 61

20 Chemistry 1020, Module 11B Calculation of Colligative Properties: Osmotic Pressure (continued) 7g) C. How many grams of glucose (gfw = 180.0) must be added to enough water to make 183 ml of a solution with an osmotic pressure of 1.3 atm at 45 o C? D. How many grams of urea (gfw = 60.0) must be added to enough water to obtain 115 ml of a solution with an osmotic pressure of 2.7 atm at 35 o C? E. Calculate the osmotic pressure at 75 o C of a solution prepared by adding 52 g of a substance with gfw = to 512 g of water. The density of the solution is 1.03 g/ml. The solution does not conduct electricity when a low voltage is applied. F. How many grams of urea (gfw = 60.0) must be added to enough water to make 255 ml of a solution with an osmotic pressure of 3.2 atm at 37 o C? 62 Xavier University of Louisiana

21 Chemistry 1020, Module 11B Name Calculation of Colligative Properties: Freezing Point Depression 8a) In general how does freezing point of a solution change as its concentration changes? (p. 533) b) State a quantitative relationship between change in freezing point and concentration of a solution in words. (p. 534) c) S. A solution of urea is found to freeze at o C. At what temperature would a solution with twice the concentration freeze? Since change in freezing point is directly proportional to concentration, doubling the concentration of the solution would cause the freezing point to decrease twice as much. That is, T for next solution is 2*0.23 o C = 0.46 o C. Therefore, the solution freezes at o C. A. A solution of urea is found to freeze at o C. At what temperature would a solution with three times the concentration freeze? B A solution of glucose is found to freeze at o C. At what temperature would a solution with one-fourth the concentration freeze? C. A solution of urea is found to freeze at o C. At what temperature would a solution with one-half the concentration freeze? D. A solution of urea is found to freeze at o C. At what temperature would a solution with four times the concentration freeze? d) Write the equation showing how change in freezing point of solution is related to concentration. (p. 534) e) What is the constant in the equation in d above called? f) When can the equation in d above be used? (i.e. What are its limitations?) Xavier University of Louisiana 63

22 Chemistry 1020, Module 11B Calculation of Colligative Properties: Freezing Point Depression (continued) 8g) S. Calculate the freezing point of a solution prepared by dissolving 22 grams of glucose, gfw = 180.0, in 61 grams of water. (pp. 534) Formula: C6H12O6 Type cmpd: not a, b, or s Strength: nonelectrolyte Tf = Kf*m m = moles solute/kg solvent = moles glucose/kg water 1 mol glucose moles solute = 22 g glucose * 180 g glucose = mol glucose 1 kg H2O kg solvent = 61 g H2O * 1000 g H2O = kg H 2O Tf = 1.86 o C m * mol glucose kg H2O = 3.72o C which rounds to 3.7 o C* Tf = freezing point - Tf = 0.00 o C o C = -3.7 o C* *Note: The rule for rounding after multiplying or dividing is different from that after adding or subtrracting. Therefore, use significant figures to round after multiplying, then add or subtract. Finally, round so the result has the same number of decimal places as the least precise measurement used in the calculation. Thus, when 3.7 o C is subtracted from 0.00 o C, you get -3.7 o C. A. Calculate the freezing point of a solution prepared by dissolving 1.8 grams of urea, gfw = 60.0 in 301 g H2O. B. How many grams of glucose (gfw=180.0) must be added to 152 grams of water in order to obtain a solution with a freezing point of o C? 64 Xavier University of Louisiana

23 Chemistry 1020, Module 11B Name Calculation of Colligative Properties: Freezing Point Depression (continued) 8g) C. Calculate the freezing point of a solution prepared by dissolving 32 grams of glucose, gfw = 180.0, in 112 grams of water. D. Calculate the freezing point of a solution prepared by dissolving 15 grams of urea, gfw = 60.0, in 251 grams of water. E. Calculate the freezing point of a solution prepared by dissolving 25 grams of glucose (gfw = 180.0) in 123 g of water. F. How many grams of a substance with gfw = 75.0 must be added to 412 grams of water to obtain a solution with a freezing point of o C? The solution does not conduct electricity. Xavier University of Louisiana 65

24 Chemistry 1020, Module 11B Calculation of Colligative Properties: Boiling Point Elevation 9a) In general how does boiling point of a solution change as its concentration changes? (p ) b) State a quantitative relationship between change in boiling point and concentration of a solution in words. (p. 532) c) S. A solution of urea is found to boil at o C. At what temperature would a solution with twice the concentration boil if pressure is kept constant? Since change in boiling point is directly proportional to concentration, doubling the concentration of the solution would cause the boiling point to increase twice as much. That is, T for new solution is 2*1.12 o C = 2.24 o C. Therefore, the solution boils at o C. (Note: You have to add the change to the boiling point of water to get the boiling point of the solution.) A. A solution of urea is found to boil at o C. At what temperature would a solution with three times the concentration boil if pressure is kept constant? B A solution of glucose is found to boil at o C. At what temperature would a solution with one-fourth the concentration boil if pressure is kept constant? C. A solution of urea is found to boil at o C. At what temperature would a solution with one-half the concentration boil if pressure is kept constant? D. A solution of urea is found to boil at o C. At what temperature would a solution with four times the concentration boil if pressure is kept constant? d) Write the equation showing how change in boiling point of solution is related to concentration. (p. 532) e) What is the constant in the equation in d above called? f) When can the equation in d above be used? (i.e. What are its limitations?) 66 Xavier University of Louisiana

25 Chemistry 1020, Module 11B Name Calculation of Colligative Properties: Boiling Point Elevation (continued) 9g) S. Calculate the boiling point of a solution prepared by dissolving 15 grams of urea (gfw=60.0) in 105 grams of water. Formula: CO(NH2)2 Type cmpd: not a, b, or s Strength: nonelectrolyte Tb = Kb*m m = moles solute/kg solvent = moles urea/kg water 1 mol urea moles solute = 15 g urea * 60.0 g urea = mol urea 1 kg H2O kg solvent = 105 g H2O * 1000 g H2O = kg H 2O Tb = 0.51 o C m * mol urea kg H2O = 1.21o C which rounds to 1.2 o C* Tb = boiling point + Tb = o C o C = o C* *Note: The rule for rounding after multiplying or dividing is different from that after adding or subtrracting. Therefore, use significant figures to round after multiplying, then add or subtract. Finally, round so the result has the same number of decimal places as the least precise measurement used in the calculation. Thus, when 1.2 o C is added to o C, you get o C o C rounds to o C. A. What is the boiling point of a solution prepared by adding 57 grams of glucose (gfw=180.0) to 125 grams of water? B. How many grams of glucose (gfw=180.0) must be added to 205 grams of water in order to obtain a solution which has a boiling point of o C? Xavier University of Louisiana 67

26 Chemistry 1020, Module 11B Calculation of Colligative Properties: Boiling Point Elevation (continued) 9g) C. What is the boiling point of solution prepared by adding 63 grams of a substance with gfw = to 142 grams of water? The solution does not conduct electricity when low voltage is applied. D. How many grams of water must be added to 12 grams of urea (gfw=60.0) in order to obtain a solution with a boiling point of o C? E. What is the boiling point of a solution prepared by adding 32 grams of a urea (gfw = 60.0) to 152 ml of water? F. How many grams of urea (gfw = 60.0) must be added to 455 ml of water to obtain a solution with a boiling point of o C? 68 Xavier University of Louisiana

27 Chemistry 1020, Module 11B Name Molecular Weight of an Unknown from Freezing Point Depression or Boiling Point Elevation 10. S. 1.5 grams of an unknown nonelectrolyte is dissolved in 52 grams of water to produce a solution which has a freezing point of o C. What is the molecular weight of the unknown. (pp ) Tf = 0 o C - Tf Tf = 0 o C - (-0.53 o C) Tf = 0.53 o C Tf = Kf*m = (g solute/mw solute) = Kf * kg solvent Tf * (kg solvent) = Kf * g solute mw solute Tf * (kg solvent) * (mw solute) = Kf * (g solute) Kf * (g solute) mw solute = (kg solvent) * ( Tf) = mw solu= (1.86 o C/m) * (1.5 g) (0.052 kg) * (0.53 o C) R = g/mol = 1.0*10 2 g/mol A grams of an unknown nonelectrolyte is dissolved in 124 grams of water to produce a solution which has a freezing point of o C. What is the molecular weight of the unknown? B grams of an unknown nonelectrolyte is dissolved in 35.7 grams of water to produce a solution which has a freezing point of o C. What is the molecular weight of the unknown? C grams of an unknown nonelectrolyte is dissolved in 25.4 grams of water to produce a solution with a boiling point of o C. What is the molecular weight of the unknown. D grams of an unknown nonelectrolyte is dissolved in 46.6 grams of water to produce a solution with a boiling point of o C. What is the molecular weight of the unknown? Xavier University of Louisiana 69

28 Chemistry 1020, Module 11B Molecular Weight of an Unknown from Freezing Point Depression or Boiling Point Elevation (continued) 10. E grams of an unknown nonelectrolyte is dissolved in 35.5 grams of water to produce a solution which has a freezing point of o C. What is the molecular weight of the unknown? F grams of an unknown nonelectrolyte is dissolved in 75.2 grams of water to produce a solution which as a boiling point of o C. What is the molecular weight of the unknown? Bonding/Model Activity to Improve Ability to Visualize in 3-D A. a) Draw the Lewis structure of CH3NO2 (nitromethane) B. a) Draw the Lewis structure of CH3CH2Cl ( ethylchloride) C. a) Draw the Lewis structure of CCl4 (carbon tetrachloride) Xavier University of Louisiana

29 Chemistry 1020, Module 11B Name Bonding/Model Activity to Improve Ability to Visualize in 3-D (continued) D. a) Draw the Lewis structure of C2H4 (ethylene) A-D b) Then use your model set to assemble a model of the species. Challenge Questions A. The vapor pressure of water at 60 o C is 152 torr and the solubility of NaCl at that temperature is 51 grams/100 grams of water. Use this information to sketch a curve illustrating how the vapor pressure of a solution of NaCl varies as grams of NaCl is added to ml of water 1 gram at a time. (i.e. Sketch vp versus g solute added.) B. 45 grams of NaCl is added to grams of water and the water is boiled off. Sketch the boiling point of the solution versus time if the solubility of NaCl in water at 100 o C is 53 g/100ml. C. What is the percent dissociation of a weak electrolyte if the freezing point of a m solution of the substance is o C. The electrolyte has the formula AB and partly dissociates into A + and B - when it dissolves. D. A m solution of acetic acid, CH 3 COOH, has a freezing point of o C. What is the percent dissociation of acetic acid in the solution? E. The boiling point of a solution prepared by adding 2.31 grams of an organic compound (93.8% C and 6.2% H) to 86.7 grams of carbon tetrachloride was K. The boiling point of pure carbon tetrachloride is K and the Kb of carbon tetrachloride is 4.48 K/m. What is the molecular formula of the organic compound? F. What is the boiling point of a solution of 10.0 grams of P4 in 45.2 ml of carbon disulfide if the boiling point elevation constant for CS2 is 2.35 o C/m and the boiling point is o C? Note: All parts of this module as well as the following Skills Module on Rounding Rules must be filled out when you arrive at drill. You will have quizzes on both. Xavier University of Louisiana 71

30 Chemistry 1020, Module 11B Revised by JW Carmichael 5/9/2000; SJB 5/8/2001; SJB 2002 & Xavier University of Louisiana