1. 1 Vanessa N. Prasad-Permaul CHM 1046 Valencia Community College

2. Introduction 1. A mixture is any intimate combination of two or more pure substances 2. Can be classified as heterogeneous or homogeneous Heterogeneous -The mixing of components is visually nonuniform and have regions of different composition Homogenous -Mixing is uniform, same composition throughout -Can be classified according to the size of their particles as either solutions or colloids 2

3. Solutions Solution 1. Homogeneous mixtures 2. Contain particles with diameters in the range of 0.1–2 nm 3. Transparent but may be colored 4. Do not separate on standing Colloids 1. Milk & fog 2. Diameters 2-500 nm 3. Do not separate on standing 3

4. Types of Solutions 4

5. Solution Formation Solute  Dissolved substance, or smaller quantity substance Solvent  Liquid dissolved in, larger quantity substance Saturated solution  Contains the maximum amount of solute that will dissolve in a given solvent. 5

6. Solution Formation Unsaturated Contains less solute than a solvent has the capacity to dissolve. Supersaturated Contains more solute than would be present in a saturated solution. Crystallization The process in which dissolved solute comes out of the solution and forms crystals. 6

7. 7 EXAMPLE 12.1: A: GIVE AN EXAMPLE OF A SOLID SOLUTION PREPARED FROM A LIQUID AND A SOLID A dental filling made up of liquid mercury and solid silver is a solid solution B: GIVE AN EXAMPLE OF A LIQUID SOLUTION PREPARED BY DISSOLVING A GAS IN A LIQUID Aqueous ammonia

8. 8 EXERCISE 12.1: IDENTIFY THE SOLUTE & SOLVENT IN THE FOLLOWING SOLUTIONS. a)80g of chromium & 5g of molybdenum b)5g of MgCl2 dissolved in 1000g of H2O c)39% N2, 41% Ar, and the rest O2

9. Three Types of interactions 1. Solvent-solvent 2. Solvent-solute 3. Solute-solute “Like dissolves like” solutions will form when three types of interactions are similar in kind and magnitude 9 Energy Changes and the Solution Process

10. Example NaCl and water: Ionic solid NaCl dissolve in polar solvents like water because the strong ion-dipole attractions between Na + and Cl - ions and polar water molecules are similar in magnitude to the strong dipole-dipole attractions between water molecules and to the strong ion-ion attractions between Na + and Cl - ions Example Oil and water Oil does not dissolve in water because the two liquids have different kinds of intermolecular forces. Oil is not polar or an ionic solvent 10

11. Energy Changes and the Solution Process NaCl in H 2 O 1. Ions that are less tightly held because of their position at a corner or an edge of the crystal are exposed to water molecules 2. Water molecules will collide with the NaCl until an ion breaks free 3. More water molecules then cluster around the ion, stabilizing it by ion-dipole attractions 4. The water molecules attack the weak part of the crystal until it is dissolved 5. Ions in solution are said to be solvated they are surrounded and stabilized by an ordered shell of solvent molecules 11

12. 12 EXAMPLE 12.2: WOULD NAPTHALENE (C10H 8 ) BE MORE SOLUBLE IN ETHANOL OR IN BENZENE? EXPLAIN. Naphthalene is more soluble in benzene because nonpolar naphthalene must break the strong hydrogen bonds between ethanol molecules and replace them with weaker London forces.

13. 13 EXERCISE 12.2: WHICH OF THE FOLLOWING COMPOUNDS IS LIKELY TO BE MORE SOLUBLE IN WATER: C4H9OH or C2H9SH? EXPLAIN.

14. Energy Changes and the Solution Process  G, Free energy change 1. If  G is negative the process is spontaneous, and the substance is dissolved 2. If  G is positive the process is non-spontaneous, the substance is not dissolved 3.  G =  H -T  S  H, enthalpy, heat flow in or out of the system,  H soln heat of solution  S, entropy, disorder,  S soln entropy of solution 14

15. Energy Changes and the Solution Process 15

16. Energy Changes and the Solution Process  S soln Entropy of Solution Usually a positive number because when you dissolve something you are increasing disorder  H soln Heat of Solution 1. Harder to predict because it could be exothermic (-  H soln ) or endothermic (+  H soln ) 2. The value of the heat of solution for a substance results from an interplay of the three kinds of interactions 16

17. Energy Changes and the Solution Process 1) Solvent-solvent interactions: Energy is required (endothermic) to overcome intermolecular forces between solvent molecules because the molecules must be separated and pushed apart to make room for solute particles 2) Solute-solute interactions: Energy is required (endothermic) to overcome interactions holding solute particles together in a crystal. For an ionic solid, this is the lattice energy. Substances with higher lattice energies therefore tend to be less soluble than substances with lower lattice energies. 17

18. Energy Changes and the Solution Process 3) Solvent-solute interactions: Energy is released (exothermic) when solvent molecules cluster around solute particles and solvate them. For ionic substances in water, the amount of hydration energy released is generally greater for smaller cations than for larger ones because water molecules can approach the positive nuclei of smaller ions more closely and thus bind more tightly. Hydration energy generally increases as the charge on the ion increases. 18

19. Energy Changes and the Solution Process 19 The solute–solvent interactions are stronger than solute–solute or solvent–solvent. Favorable process Exothermic rxn. Exothermic -  H soln

20. Energy Changes and the Solution Process 20 The solute–solvent interactions are weaker than solute–solute or solvent–solvent. Unfavorable process. Endothermic rxn Endothermic +  H soln

21. Energy Changes and the Solution Process  Hydration  The attraction of ions for water molecules  Hydration Energy  The energy associated with the attraction between ions and water molecules  Lattice energy  The energy holding ions together in a crystal lattice 21

22. 22 EXAMPLE 12.3: WHICH OF THE FOLLOWINGIONS WOULD BE EXPECTED TO HAVE THE GREATER ENERGY OF HYDRATION. Mg 2+ OR Al 3+ ? Al 3+ because water molecules can approach the positive nuclei of smaller ions more closely and thus bind more tightly. Hydration energy generally increases as the charge on the ion increases.

23. 23 EXERCISE 12.3: WHICH ION HAS THE LARGER HYDRATION ENERGY, Na + or K + EXPLAIN.

24. Molarity (M) M = mole of solute / Liter of solution Molality (m) m = moles of solute/mass of solvent (kg) Mole Fraction (x) X = mole of component / total moles 24 Units of Concentration

25. 25 EXAMPLE 12.6: GLUCOSE, C 6 H 12 O 6, IS A SUGAR THAT IS IN FRUITS. IT IS ALSO FOUND IN BLOOD AND IS THE BODY’S MAIN SOURCE OF ENERGY. WHAT IS THE MOLALITY OF A SOLUTION CONTAINING 5.67g OF GLUCOSE DISSOLVED IN 25.2g OF WATER? 5.67g C 6 H 12 O 6 x 1 mol C 6 H 12 O 6 = 0.0315 mol C 6 H 12 O 6 180.2g C 6 H 12 O 6 0.0315 mol C 6 H 12 O 6 = 1.25 m C 6 H 12 O 6 25.2g x 1kg 1000g

26. 26 EXERCISE 12.6: TOLUENE, C6H5CH3, IS A STARTING MATERIAL FOR TNT (TRINITROTLOUENE). FIND THE MOLALITY OF TOLUENE IN A SOLUTION THAT CONTAINS 35.6g OF TOLUENE AND 125g OF BENZENE

27. 27 EXAMPLE 12.7: WHAT ARE THE MOLE FRACTIONS OF GLUCOSE AND WATER IN A SOLUTION CONTAINING 5.67g OF GLUCOSE DISSOLVED IN 25.2g OF WATER? 5.67g C6H12O6 x 1 mol C6H12O6 = 0.0315 mol C6H12O6 180.2g C6H12O6 25.2g H2O x 1 mol H2O = 1.40 mol H2O 18.0g H2O 1.40mol + 0.0315mol = 1.432mol MOLE FRACTION OF GLUCOSE 0.0315mol = 0.0220 x 100% = 2.20% 1.432mol MOLE FRACTION OF WATER 1.40mol = 0.978 x 100% = 97.8% 1.432mol

28. 28 EXERCISE 12.7: CALCULATE THE MOLE FRACTIONS OF TOLUENE AND BENZENE IN THE SOLUTION FROM EXERCISE 12.6

29. Units of Concentration Mass Percent (mass %) Mass % = (mass of component / total mass of sol’n) x 100% Parts per million, ppm = (mass of component / total mass of solution) x 10 6 Parts per billion, ppb = (mass of component / total mass of solution) x 10 9 29

30. 30 EXAMPLE 12.5: HOW WOULD YOU PREPARE 425g OF AN AQUEOUS SOLUTION CONTAINING 2.40% BY MASS OF SODIUM ACETATE, NaC2H3O2? MASS OF SOLUTE: 0.0240 x 425g = 10.2g MASS OF WATER: 425g – 10.2g = 414.8 = 415g MASS OF SOLUTION: 425g  YOU WOULD PREPARE THIS SOLUTION BY DISSOLVING 10.2g OF NaC2H3O2 IN 415g OF WATER FOR A TOTAL MASS OF 425g.

31. 31 EXERCISE 12.5: AN EXPERIMENT CALLS FOR 35.0g OF HYDROCHLORIC ACID THAT IS 20.2% HCl BY MASS. HOW MANY GRAMS OF HCl IS THIS? HOW MANY GRAMS OF WATER IS THIS?

32. Some Factors Affecting Solubility Solubility The amount of solute per unit of solvent needed to form a saturated solution Miscible Mutually soluble in all proportions Effect of Temperature on Solubility 1. Most solid substances become more soluble as temperature rises 2. Most gases become less soluble as temperature rises 32

33. Some Factors Affecting Solubility Effect of Pressure on Solubility 1. No effect on liquids or solids 2. The solubility of a gas in a liquid at a given temperature is directly proportional to the partial pressure of the gas over the solution, @ 25°C Henry’s Law solubility = k x P k = constant characteristic of specific gas, mol/L  atm P = partial pressure of the gas over the sol’n 33

34. Some Factors Affecting Solubility a) Equal numbers of gas molecules escaping liquid and returning to liquid b) Increase pressure, increase # of gas molecules returning to liquid, solubility increases c) A new equilibrium is reached, where the #’s of escaping = # of returning 34

35. 35 EXAMPLE 12.4: 27g OF ACETYLENE (C2H2) DISSOLVES IN 1L OF ACETONE AT 1.0atm. IF THE PARTIAL PRESSURE OF ACETYLENE IS INCREASED TO 12atm, WHAT IS THE SOLUBILITY OF ACETONE? S2 = P2 S1 P1 S2 = 12atm 27g C2H2/L acetone 1.0atm S2 = 27g C2H2 x 12atm = 3.2x10 2 g C2H2/L acetone L acetone 1.0atm  320g of acetylene will dissolve in 1L of acetone @ 12atm

36. 36 EXERCISE 12.4: A LITER OF WATER @ 25 0 C DISSOLVES 0.0404g OF O2 WHEN THE PARTIAL PRESSURE OF THE OXYGEN IS 1.00atm. WHAT IS THE SOLUBILITY OF OXYGEN FROM AIR, IN WHICH THE PARTIAL PRESSURE OF O2 IS 159mmHg?

37. Physical Behavior of Solutions: Colligative Properties Colligative properties Properties that depend on the amount of a dissolved solute but not its chemical identity There are four main colligative properties: 1. Vapor pressure lowering 2. Freezing point depression 3. Boiling point elevation 4. Osmotic pressure 37

38. Physical Behavior of Solutions: Colligative Properties In comparing the properties of a pure solvent with those of a solution… 1. Vapor pressure of sol’n is lower 2. Boiling point of sol’n is higher 3. Freezing point of sol’n is lower 4. Osmosis, the migration of solvent molecules through a semipermeable membrane, occurs when solvent and solution are separated by the membrane 38

39. Vapor-pressure Lowering of Solutions: Raoult’s Law 1. A liquid in a closed container is in equilibrium with its vapor and that the amount of pressure exerted by the vapor is called the vapor pressure. 2. When you compare the vapor pressure of a pure solvent with that of a solution at the same temperature the two values are different. 3. If the solute is nonvolatile and has no appreciable vapor pressure of its own (solid dissolved) the vapor pressure of the solution is always lower that that of the pure solvent. 4. If the solute is volatile and has a significant vapor pressure (2 liquids) the vapor pressure of the mixture is intermediate between the vapor pressures of the two pure liquids. 39

40. Solutions with a Nonvolatile Solute When solute molecules displace solvent molecules at the surface, the vapor pressure drops since fewer gas molecules are needed to equalize the escape rate and capture rates at the liquid surface. 40

41. Solutions with a Nonvolatile Solute!!! Raoult’s Law P soln = P solv · X solv P soln = vapor pressure of the solution P solv = vapor pressure of the pure solvent X solv = mole fraction of the solvent in the solution 41

42. Raoult’s Law applies to only Ideal solutions 1. Law works best when solute concentrations are low and when solute and solvent particles have similar intermolecular forces. 2. If intermolecular forces between solute particles and solvent molecules are weaker than solvent molecules alone, solvent molecules are less tightly held, vapor pressure is higher than Raoult predicts 3. If intermolecular forces between solute and solvent are stronger than solvent alone, solvent molecules are more tightly held and the vapor pressure is lower than predicted 4. No Van’t Hoff factor!!! is a measure of the effect of a solute upon colligative propertiescolligative properties 42

43. Solutions with a Nonvolatile Solute Close-up view of part of the vapor pressure curve for a pure solvent and a solution of a nonvolatile solute. Which curve represents the pure solvent, and which the solution? Why? 43

44.  The lower vapor pressure of a sol’n relative to that of a pure solvent is due to the difference in their entropies of vaporization,  S vap. Because the entropy of the solvent in a sol’n is higher to begin with,  S vap is smaller for the sol’n than for the pure solvent. As a result vaporization of the solvent from the sol’n is less favored (less negative  G vap ), and the vapor pressure of the solution is lower. 44

45. Solutions with a Volatile Solute!! P total = P A + P B P total = (P° A · X A ) + (P° B · X B ) P° A = vapor pressure of pure A X A = mole fraction of A P° B = vapor pressure of pure B X B = mole fraction of B P total should be intermediate to A & B 45

46. Close-up view of part of the vapor pressure curves for two pure liquids and a mixture of the two. Which curves represent the mixture? 46 1)Red 2)Green 3)Blue

47. 47 EXAMPLE 12.12: CALCULATE THE VAPOR-PRESSURE WHEN 5.67g OF GLUCOSE IS DISSOLVED 25.2g OF WATER @ 25 o C. THE VAPOR PRESSURE OF WATER @ 25 o C IS 23.8mmHg. WHAT IS THE VAPOR PRESSURE OF THE SOLUTION?  P = P A O X B = 23.8 mmHg x 0.0220 = 0.524 mmHg THE VAPOR PRESSURE OF THE SOLUTION IS: P A = P A O -  P = (23.8mmHg – 0.524mmHg) = 23.3mmHg

48. 48 EXERCISE 12.12: NAPTHALENE IS USED TO MAKE MOTHBALLS. A SOLUTION IS MADE BY DISSOLVING 0.515g OF NAPTHALENE IN 60.8g OF CHLOROFORM. CALCULATE THE VAPOR PRESSURE LOWERING OF CHLOROFORM @ 20 o C FROM NAPTHALENE. THE VAPOR PRESSURE OF CHLOROFORM @ 20 o C IS 156mmHg. NAPTHALENE CAN BE ASSUMED TO BE NON-VOLATILE COMPARED WITH CHLOROFORM. WHAT IS THE VAPOR PRESSURE OF THE SOLUTION?

49. Boiling Point Elevation and Freezing Point Depression of Solutions 49

50. Boiling Point Elevation and Freezing Point Depression of Solutions 1. Red line is pure solvent 2. Green line solution of nonvolatile solute 3. Vapor pressure of sol’n is lower 4. Temp at which vapor pressure = 1 atm for sol’n is higher 5. Boiling point of sol’n is higher by  T b 6. Liquid/vapor phase transition line is lower for sol’n 7. Triple point temp is lower for sol’n 8. Solid/liquid phase transition has shifted to a lower temp. 9. The freezing point of the sol’n is lower by  T f 50

51. Boiling Point Elevation and Freezing Point Depression of Solutions  T b = K b · m  T f = K f · m K b = molal boiling-point elevation constant K f = molal freezing-point depression constant m = molality NO Van’t Hoff Factor!! 51

52. Boiling Point Elevation and Freezing Point Depression of Solutions The higher boiling point of a solution relative to that of a pure solvent is due to a difference in their entropies of vaporization,  S vap. Because the solvent in a solution has a higher entropy to begin with,  S vap is smaller for the solution than for the pure solvent. As a result, the boiling point of the solution T b is higher than that of the pure solvent. 52

53. Boiling Point Elevation and Freezing Point Depression of Solutions The lower freezing point of a solution relative to that of a pure solvent is due to a difference in their entropies of fusion,  S fusion. Because the solvent in a solution has a higher entropy level to begin with,  S fusion is larger for the solution than for the pure solvent. As a result the freezing point of the solution T f is lower than that of the pure solvent. 53

54. 54 EXAMPLE 12.13: AN AQUEOUS SOLUTION IS 0.0222 m GLUCOSE. WHAT ARE THE BOILING POINT AND THE FREEZING POINT OF THIS SOLUTION?  T b = K b c m = 0.512 o C/m x 0.0222 m = 0.0114 o C  T f = K f c m = 1.86 o C/m x 0.0222 m = 0.0413 o C THE BOILING POINT OF THE SOLUTION IS 100.000 o C + 0.0114 o C = 100.011 o C THE FREEZING POINT OF THE SOLUTION IS 0.000 o C – 0.0413 o C = -0.041 o C

55. 55 EXERCISE 12.13: HOW MANY GRAMS OF ETHYLENE GLYCOL MUST BE ADDED TO 37.8g OF WATER TO GIVE A FREEZING POINT OF -0.150 o C?

56. Osmosis and Osmotic Pressure 1. Semipermeable membranes allow water or other small molecules to pass through, but they block the passage of large solute molecules or ions. 2. When a solution and a pure solvent are separated by the right kind of semipermeable membrane, solvent molecules pass through the membrane in a process known as osmosis. 3. Passage of solvent through the membrane takes place in both directions 56

57. Osmosis and Osmotic Pressure 4. Passage from the pure solvent side to the solution side is more favored and faster. 5. The amount of liquid on the pure solvent side decreases 6. The amount of liquid on the solution side increases 7. The concentration of the solution decreases 57

58. Osmosis and Osmotic Pressure 58

59. Osmosis and Osmotic Pressure Osmotic Pressure 1. The amount of pressure necessary to achieve equilibrium 2.  = MRT  = osmotic pressure M = molarity R = gas constant,.08206 L atm/K mol T = temperature in kelvins 59

60. Some uses of Colligative Properties 1. The most important use of colligative properties in the laboratory is for determining the molecular mass of an unknown substance. 2. Any of the four colligative properties can be used but using osmotic pressure gives the most accurate results 60

61. If you have to calculate the molar mass of a compound which of the following will give you the most accurate results? 1) Osmotic pressure 2) Vapor pressure lowering 3) Boiling point elevation 4) Freezing point depression 61

62. 62 EXAMPLE 12.8: AN AQUEOUS SOLUTION IS 0.120 m GLUCOSE. WHAT ARE THE MOLE FRACTIONS OF EACH COMPONENT IN THE SOLUTION? 0.120 m = 0.120mol IN 1.00kg OF WATER 1.00kg H2O x 1000g x 1 mol = 55.6mol H2O 1kg 18.0 g H2O MOLE FRACTION OF GLUCOSE : 0.120 mol = 0.00215 (0.120mol + 55.6mol) MOLE FRACTION OF WATER: 55.6mol = 0.998 (0.120mol + 55.6mol)

63. 63 EXERCISE 12.8: A SOLUTION IS 0.120 m METHANOL DISSOLVED IN ETHANOL. CALCULATE THE MOLE FRACTIONS OF CH3OH AND CH3CH2OH IN THE SOLUTION.

64. 64 EXAMPLE 12.9: A SOLUTION IS 0.150 mol FRACTION GLUCOSE, AND 0.850 mol FRACTION WATER. WHAT IS THE MOLALITY OF GLUCOSE IN THE SOLUTION? 0.850 mol x 18.0 g H2O = 15.3 g H2O (0.0153kg H2O) 1 mol H2O 0.150 mol C6H12O6 = 9.80 m C6H12O6 0.00153kg SOLVENT

65. 65 EXERCISE 12.9: A SOLUTION IS 0.250 mol FRACTION METHANOL AND 0.750 mol FRACTION ETHANOL. WHAT IS THE MOLALITY OF METHANOL IN THE SOLUTION?

66. 66 EXAMPLE 12.10: AN AQUEOUS SOLUTION IS 0.273 m KCl. WHAT IS THE MOLAR CONCENTRATION OF KCl? THE DENSITY OF THE SOLUTION IS 1.011 x 10 3 g/L. 0.273 mol KCl x 74.6g KCl = 20.4g KCl 1 mol KCl 1.000 x 10 3 g + 20.4g = 1.020 x 10 3 g 1.020 x 10 3 g x 1L = 1.009L 1.011x10 3 g 0.273mol KCl = 0.271 M KCl 1.009L solution

67. 67 EXERCISE 12.10: UREA (NH2)2CO IS USED AS FERTILIZER. WHAT IS THE MOLAR CONCENTRATION OF AN AQUEOUS SOLUTION THAT IS 3.42 m UREA? THE DENSITY OF THE SOLUTION IS 1.045 g/mL.

68. 68 EXAMPLE 12.11: AN AQUEOUS SOLUTION IS 0.907 M Pb(NO3)2. WHAT IS THE MOLALITY? THE DENSITY OF THE SOLUTION IS 1.252g/mL. 1.000 L x 1000mL x 1.252 g = 1.252 x 10 3 g 1L 1mL 0.907 mol Pb(NO3)2 x 331.2 g Pb(NO3)2 = 3.00 x 102 Pb(NO3)2 1 mol 1.252 x 103 g – 3.00 x 12 g = 9.52 x 102g (0.952kg) 0.907 mol Pb(NO3)2 = 0.953 m Pb(NO3)2 0.952kg solvent

69. 69 EXERCISE 12.11: AN AQUEOUS SOLUTION IS 2.00 M UREA. THE DENSITY IS 1.029 g/mL. WHAT IS THE MOLAL CONCENTRATION OF UREA IN THE SOLUTION?

70. Example 1 Arrange the following in order of their expected increasing solubility in water: Br 2, KBr, C 7 H 8 70

71. Example 2 1. Na+ 2. Cs+ 3. Li+ 4. Rb+ 1. Mg 2+ 2. Na + 3. Li + 71 Which would you expect to have the larger (more negative) hydration energy?

72. Example 3 What is the mass % concentration of a saline sol’n prepared by dissolving 1.00 mol of NaCl in 1.00 L of water? Density H2O =1.00 g/mL MM NaCl = 58.443 g/mol 72

73. Example 4 Assuming that seawater is an aqueous solution of NaCl what is its molarity? The density of seawater is 1.025 g/mL at 20  C and the NaCl concentration is 3.50 mass %  Assume 1 L to make easier, 1000 mL 73

74. Example 5 What is the molality of a solution prepared by dissolving 0.385 g of cholesterol, C 27 H 46 O in 40.0 g of chloroform, CHCl 3 ? What is the mole fraction of cholesterol in the solution? 74

75. Example 6 What mass in grams of a 0.500 m solution of sodium acetate, CH 3 CO 2 Na, in water would you use to obtain 0.150 mol of sodium acetate? 75

76. Example 7 The density at 20°C of a 0.258 m solution of glucose in water is 1.0173 g/mL and the molar mass of glucose is 180.2 g/mol. What is the molarity of the solution?  Assume 1 kg 76

77. Example 8 The density at 20°C of a 0.500 M solution of acetic acid in water is 1.0042 g/mL. What is the concentration of this solution in molality? The molar mass of acetic acid, CH 3 CO 2 H, is 60.05 g/mol.  Assume 1 L 77

78. Example 9 Which of the following will become less soluble in water as the temperature is increased? 1) NaOH (s) 2) CO 2(g) 78

79. Example 10 The solubility of CO 2 in water is 3.2 x 10 -2 M @ 25°C and 1 atm pressure. What is the Henry’s-Law constant for CO 2 in mol/L atm? solubility = k x P 79

80. Example 11 What is the vapor pressure (in mm Hg) of a solution prepared by dissolving 5.00 g of benzoic acid (C 7 H 6 O 2 ) in 100.00 g of ethyl alcohol (C 2 H 6 O) at 35°C? The vapor pressure of the pure ethyl alcohol at 35°C is 100.5 mm Hg 80

81. Example 12 What is the vapor pressure ( in mm Hg) of a sol’n prepared by dissolving 25.0 g of ethyl alcohol (C 2 H 5 OH) in 100.0 g of water at 25°C? The vapor pressure of pure water is 23.8 mm Hg and the vapor pressure of ethyl alcohol is 61.2 mm Hg at 25°C 81

82. Example 13 a) Is the boiling point of the second pure liquid higher or lower than that of the first liquid? b) On the diagram where is the approximate position of the second pure liquid? 82 The following phase diagram shows part of the vapor pressure curves for a pure liquid (green curve) and a solution (red curve) of the first liquid with a second volatile liquid (not shown)

83. Example 14 What is the normal boiling point in °C of a solution prepared by dissolving 1.50 g of aspirin (C 9 H 8 O 4 ) in 75.00 g of chloroform (CHCl 3 )? The normal boiling point of chloroform is 61.7 °C and K b of chloroform is 3.63 °C kg/mol 83

84. Example 15 What osmotic pressure in atm would you expect for a solution of 0.125 M C 6 H 12 O 6 that is separated from pure water by a semipermeable membrane at 310 K?  = MRT 84

85. Example 16 A solution of unknown substance in water at 300 K gives rise to an osmotic pressure of 3.85 atm. What is the molarity of the solution?  = MRT M =  /RT 85

86. Example 17  What is the molar mass of sucrose if a solution prepared by dissolving 0.822 g of sucrose in water and diluting to a volume of 300.0 mL has an osmotic pressure of 149 mm Hg at 298 K? 86