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Solutions © 2009, Prentice-Hall, Inc. Chapter 13 Properties of Solutions John D. Bookstaver St. Charles Community College Cottleville, MO Chemistry, The.

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1 Solutions © 2009, Prentice-Hall, Inc. Chapter 13 Properties of Solutions John D. Bookstaver St. Charles Community College Cottleville, MO Chemistry, The Central Science, 11th edition Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten

2 Solutions © 2009, Prentice-Hall, Inc. Overview of Chapter 13 - Solutions Review/On your own: describe the three factors involved in solubility (solute-solute, solvent-solvent and solute-solvent interactions) explain what factors affect solubility and predict solubilities of solutes in solvents define and apply the terms miscible and immiscible, saturated, unsaturated and supersaturated solutions use Henry’s Law to calculate the solubility or pressure of a gas in liquid solution calculate solubilities (g/100 g H 2 O) interpret solubility graphs

3 Solutions © 2009, Prentice-Hall, Inc. Overview of Chapter 13 - Solutions New Material calculate concentration using mass percentage, ppm, ppb, mole fraction, molarity and molality and convert amongst them define colligative properties use Raoult’s Law to calculate vapor pressures and mole fractions of solutions calculate freezing point depression, boiling point elevation and osmotic pressure calculate the molar mass from any of the four colligative properties

4 Solutions © 2009, Prentice-Hall, Inc. Solutions Solutions are homogeneous mixtures of two or more pure substances. In a solution, the solute is dispersed uniformly throughout the solvent.

5 Solutions © 2009, Prentice-Hall, Inc. Solutions The intermolecular forces between solute and solvent particles must be strong enough to compete with those between solute particles and those between solvent particles.

6 Solutions © 2009, Prentice-Hall, Inc. How Does a Solution Form? As a solution forms, the solvent pulls solute particles apart and surrounds, or solvates, them.

7 Solutions © 2009, Prentice-Hall, Inc. How Does a Solution Form If an ionic salt is soluble in water, it is because the ion- dipole interactions are strong enough to overcome the lattice energy of the salt crystal.

8 Solutions © 2009, Prentice-Hall, Inc. Energy Changes in Solution Simply put, three processes affect the energetics of solution: –separation of solute particles, –separation of solvent particles, –new interactions between solute and solvent.

9 Solutions © 2009, Prentice-Hall, Inc. Energy Changes in Solution The enthalpy change of the overall process depends on  H for each of these steps.

10 Solutions © 2009, Prentice-Hall, Inc. Why Do Endothermic Processes Occur? Things do not tend to occur spontaneously (i.e., without outside intervention) unless the energy of the system is lowered.

11 Solutions © 2009, Prentice-Hall, Inc. Why Do Endothermic Processes Occur? Yet we know the in some processes, like the dissolution of NH 4 NO 3 in water, heat is absorbed, not released.

12 Solutions © 2009, Prentice-Hall, Inc. Enthalpy Is Only Part of the Picture The reason is that increasing the disorder or randomness (known as entropy) of a system tends to lower the energy of the system.

13 Solutions © 2009, Prentice-Hall, Inc. Enthalpy Is Only Part of the Picture So even though enthalpy may increase, the overall energy of the system can still decrease if the system becomes more disordered.

14 Solutions © 2009, Prentice-Hall, Inc. Student, Beware! Just because a substance disappears when it comes in contact with a solvent, it doesn’t mean the substance dissolved.

15 Solutions © 2009, Prentice-Hall, Inc. Student, Beware! Dissolution is a physical change — you can get back the original solute by evaporating the solvent. If you can’t, the substance didn’t dissolve, it reacted.

16 Solutions © 2009, Prentice-Hall, Inc. 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 O (g)  Na 2 SO 4 10 H 2 O (s)

17 Solutions © 2009, Prentice-Hall, Inc. Sample Exercise 13.1 (p. 534) Essentially all of the water vapor in the closed container is consumed in this reaction. If we consider our system to consist initially of Na 2 SO 4(s) and 10 H 2 O (g) a)does the system become more or less ordered in this process, and b)does the entropy of the system increase or decrease?

18 Solutions © 2009, Prentice-Hall, Inc. Practice Exercise 13.1 Does the entropy of the system increase or decrease when the stopcock is opened to allow mixing of the two gases in the apparatus?

19 Solutions © 2009, Prentice-Hall, Inc. Types of Solutions Saturated –In a saturated solution, the solvent holds as much solute as is possible at that temperature. –Dissolved solute is in dynamic equilibrium with solid solute particles.

20 Solutions © 2009, Prentice-Hall, Inc. Types of Solutions Unsaturated –If a solution is unsaturated, less solute than can dissolve in the solvent at that temperature is dissolved in the solvent.

21 Solutions © 2009, Prentice-Hall, Inc. Types of Solutions Supersaturated –In supersaturated solutions, the solvent holds more solute than is normally possible at that temperature. –These solutions are unstable; crystallization can usually be stimulated by adding a “seed crystal” or scratching the side of the flask.

22 Solutions © 2009, Prentice-Hall, Inc. Factors Affecting Solubility Chemists use the axiom “like dissolves like." –Polar substances tend to dissolve in polar solvents. –Nonpolar substances tend to dissolve in nonpolar solvents.

23 Solutions © 2009, Prentice-Hall, Inc. Factors Affecting Solubility The more similar the intermolecular attractions, the more likely one substance is to be soluble in another.

24 Solutions © 2009, Prentice-Hall, Inc. Factors Affecting Solubility Glucose (which has hydrogen bonding) is very soluble in water, while cyclohexane (which only has dispersion forces) is not.

25 Solutions © 2009, Prentice-Hall, Inc. Factors Affecting Solubility Vitamin A is soluble in nonpolar compounds (like fats). Vitamin C is soluble in water.

26 Solutions © 2009, Prentice-Hall, Inc. Sample Exercise 13.2 (p. 539) Predict whether each of the following substances is more likely to dissolve in carbon tetrachloride (CCl 4 ) or in water: C 7 H 16, Na 2 SO 4, HCl, and I 2.

27 Solutions © 2009, Prentice-Hall, Inc. Practice Exercise 13.2 Arrange the following substances in order of increasing solubility in water:

28 Solutions © 2009, Prentice-Hall, Inc. Gases in Solution In general, the solubility of gases in water increases with increasing mass. Larger molecules have stronger dispersion forces.

29 Solutions © 2009, Prentice-Hall, Inc. Gases in Solution The solubility of liquids and solids does not change appreciably with pressure. The solubility of a gas in a liquid is directly proportional to its pressure.

30 Solutions © 2009, Prentice-Hall, Inc. Henry’s Law S g = kP g where S g is the solubility of the gas, k is the Henry’s Law constant for that gas in that solvent, and P g is the partial pressure of the gas above the liquid.

31 Solutions © 2009, Prentice-Hall, Inc. Sample Exercise 13.3 (p. 540) Calculate the concentration of CO 2 in a soft drink that is bottled with a partial pressure of CO 2 of 4.0 atm over the liquid at 25 o C. The Henry’s Law constant for CO 2 in water at this temperature is 3.1 x mol/L-atm. (0.12 M)

32 Solutions © 2009, Prentice-Hall, Inc. Practice Exercise 13.3 Calculate the concentration of CO 2 in a soft drink after the bottle is opened and equilibrates at 25 o C under a CO 2 partial pressure of 3.0 x atm. (9.3 x M)

33 Solutions © 2009, Prentice-Hall, Inc. Temperature Generally, the solubility of solid solutes in liquid solvents increases with increasing temperature.

34 Solutions © 2009, Prentice-Hall, Inc. Temperature The opposite is true of gases. –Carbonated soft drinks are more “bubbly” if stored in the refrigerator. –Warm lakes have less O 2 dissolved in them than cool lakes.

35 Solutions © 2009, Prentice-Hall, Inc. Ways of Expressing Concentrations of Solutions

36 Solutions © 2009, Prentice-Hall, Inc. Ways of Expressing Concentration All methods involve quantifying the amount of solute per amount of solvent, or solution. Dilute and concentrated are qualitative ways to describe concentration.

37 Solutions © 2009, Prentice-Hall, Inc. Quantitative Expressions of Concentration Require specific information regarding such quantities as Masses Moles Liters of the solute, solvent, or solution

38 Solutions © 2009, Prentice-Hall, Inc. Quantitative Expressions of Concentration Mass percentage, which includes ppm, ppb Mole fraction Molarity Molality

39 Solutions © 2009, Prentice-Hall, Inc. Mass Percentage Mass % of A = mass of A in solution total mass of solution  100

40 Solutions © 2009, Prentice-Hall, Inc. Parts per Million and Parts per Billion ppm = mass of A in solution total mass of solution  10 6 Parts per Million (ppm) = mg solute/kg solution Parts per Billion (ppb) =  g solute/kg solution ppb = mass of A in solution total mass of solution  10 9

41 Solutions © 2009, Prentice-Hall, Inc. Sample Exercise 13.4 (p. 542) a)A solution is made by dissolving 13.5 g glucose (C 6 H 12 O 6 ) in kg of water. What is the mass percentage of solute in this solution? (11.9%) b)A 2.5-g sample of groundwater was found to contain 5.4  g of Zn 2+. What is the concentration of Zn 2+ in parts per million? (2.2 ppm)

42 Solutions © 2009, Prentice-Hall, Inc. Practice Exercise 13.4 a)Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water. (2.91%) b)A commercial bleaching solution contains 3.62 mass % sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of bleaching solution? (90.5 g NaOCl)

43 Solutions © 2009, Prentice-Hall, Inc. Mole Fraction, Molarity and Molality Common expressions of concentration based on the number of moles of one or more components Remember: mass can be converted to moles using MM

44 Solutions © 2009, Prentice-Hall, Inc. moles of A total moles in solution X A = Mole Fraction (X) In some applications, one needs the mole fraction of solvent, not solute — make sure you find the quantity you need! Note: mole fraction has no units. Note: mole fractions range from 0 to 1 Particularly useful for gases

45 Solutions © 2009, Prentice-Hall, Inc. mol of solute L of solution M = Molarity (M) You will recall this concentration measure from Chapter 4. Since volume is temperature- dependent, molarity can change with temperature.

46 Solutions © 2009, Prentice-Hall, Inc. mol of solute kg of solvent m = Molality (m) Since both moles and mass do not change with temperature, molality (unlike molarity) is not temperature- dependent.

47 Solutions © 2009, Prentice-Hall, Inc. Sample Exercise 13.5 (p. 544) A solution is made by dissolving 4.35 g glucose (C 6 H 12 O 6 ) in 25.0 mL of water. Calculate the molality of glucose in the solution. (Assume D H2O = 1.00 g/mL) (0.964 m)

48 Solutions © 2009, Prentice-Hall, Inc. Practice Exercise 13.5 What is the molality of a solution made by dissolving 36.5 g of naphthalene (C 10 H 8 ) in 425 g of toluene (C 7 H 8 )? (0.671 m)

49 Solutions © 2009, Prentice-Hall, Inc. Where we are going next Problems that involve conversions between different types of units and other applications that require preparatory steps

50 Solutions © 2009, Prentice-Hall, Inc. Changing Molarity to Molality If we know the density of the solution, we can calculate the molality from the molarity and vice versa.

51 Solutions © 2009, Prentice-Hall, Inc. Sample Exercise 13.6 (p. 544) A solution of hydrochloric acid contains 36% HCl by mass. a) Calculate the mole fraction of HCl in the solution. (0.22) b) Calculate the molality of HCl in the solution. (15 m)

52 Solutions © 2009, Prentice-Hall, Inc. Practice Exercise 13.6 A commercial bleach solution contains 3.62 mass % NaOCl in water. Calculate a)the mole fraction of NaOCl in the solution and(9.00 x ) b)the molality of NaOCl in the solution (0.505 m)

53 Solutions © 2009, Prentice-Hall, Inc. Sample Exercise 13.7 (p. 546) A solution with a density of g/mL contains 5.0 g of toluene (C 7 H 8 ) and 225 g of benzene. Calculate the molarity of the solution. (0.21 M)

54 Solutions © 2009, Prentice-Hall, Inc. Practice Exercise 13.7 A solution containing equal masses of glycerol (C 3 H 8 O 3 ) and water has a density of 1.10 g/mL. Calculate a) the molality of glycerol; (10.9 m) b) the mole fraction of glycerol; (0.163) c) the molarity of glycerol in the solution. (5.97 M)

55 Solutions © 2009, Prentice-Hall, Inc. Warmup Caffeine (C 8 H 10 N 4 O 2 ) is a stimulant found in coffee and tea. If a solution of caffeine in chloroform (CHCl 3 ) as a solvent has a concentration of m, calculate a) the percent caffeine by mass (1.44%) b) the mole fraction of caffeine ( ) (p. 567, Q# 13.54)

56 Solutions © 2009, Prentice-Hall, Inc. Overview Solutions Colligative properties, especially to find MM, including Friday’s lab

57 Solutions © 2009, Prentice-Hall, Inc. Colligative Properties Changes in colligative properties depend only on the number of solute particles present, not on the identity of the solute particles. Physical properties “depending on the collection”, i.e. collective effects of the solute particles

58 Solutions © 2009, Prentice-Hall, Inc. Colligative Properties Application of molality Examples: C 12 H 22 O 11  C 12 H 22 O 11 one particle NaCl  Na + + Cl - two particles CaCl 2  Ca Cl - three particles This will have the greatest effect (of these examples)

59 Solutions © 2009, Prentice-Hall, Inc. Colligative Properties Among colligative properties are –Vapor pressure lowering –Boiling point elevation* –Melting point depression* - Friday’s lab –Osmotic pressure* *quantitative

60 Solutions © 2009, Prentice-Hall, Inc. Vapor Pressure Because of solute- solvent intermolecular attraction, higher concentrations of nonvolatile solutes make it harder for solvent to escape to the vapor phase.

61 Solutions © 2009, Prentice-Hall, Inc. Vapor Pressure Focus on the solvent e.g. NaCl dissolved in H 2 O – lower water VP than pure H 2 O Lower VP because the Na + and Cl - particles (nonvolatile solutes) get surrounded by H 2 O  fewer “free” H 2 O particles left to vaporize at surface Stronger IMFs  harder for solvent molecules to leave

62 Solutions © 2009, Prentice-Hall, Inc. Vapor Pressure The vapor pressure of a solution is lower than that of the pure solvent.

63 Solutions © 2009, Prentice-Hall, Inc. Raoult’s Law For an ideal solution: P A = X A P  A where –X A is the mole fraction of compound A, and –P  A is the normal vapor pressure of A at that temperature. NOTE: This is one of those times when you want to make sure you have the vapor pressure of the solvent.

64 Solutions © 2009, Prentice-Hall, Inc. Raoult’s Law Real solutions show approximately ideal behavior when: -The solute concentration is low -The solute and solvent have similarly sized molecules -The solute and solvent have similar types of IMFs -Raoult’s Law breaks down when the solvent- solvent and solute-solute IMFs are >>> or <<< solute-solvent IMFs.

65 Solutions © 2009, Prentice-Hall, Inc. Sample Exercise 13.8 (p. 547) Glycerin (C 3 H 8 O 3 ) is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25 o C. Calculate the vapor pressure at 25 o C of a solution made by adding 50.0 mL of glycerin to mL of water. The vapor pressure of pure water at 25 o C is 23.8 torr. (23.2 torr)

66 Solutions © 2009, Prentice-Hall, Inc. Practice Exercise 13.8 The vapor pressure of pure water at 110 o C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.00 atm at 110 o C. Assuming that Raoult’s law is obeyed, what is the mole fraction of ethylene glycol in the solution? (rearrange equation to find mole fraction) (0.290)

67 Solutions © 2009, Prentice-Hall, Inc. Boiling Point Elevation and Freezing Point Depression Nonvolatile solute- solvent interactions also cause solutions to have higher boiling points and lower freezing points than the pure solvent.

68 Solutions © 2009, Prentice-Hall, Inc. Boiling Point Elevation  particles of nonvolatile solute   VP   boiling point -Since bp = VP at atm P, if the VP is lower, more energy must be used to overcome atm P = higher T at boiling -It takes more energy to break apart solvent/solute bonds to allow the solvent to vaporize. -For H 2 O: 1 mole particles = o C (molal bp elevation constant) 1 kg H 2 O Note: 1 m solution of NaCl is 2 m in total solute particles

69 Solutions © 2009, Prentice-Hall, Inc. Boiling Point Elevation The change in boiling point is proportional to the molality of the solution:  T b = K b  m where K b is the molal boiling point elevation constant, a property of the solvent. m = molality  T b is added to the normal boiling point of the solvent.

70 Solutions © 2009, Prentice-Hall, Inc. Boiling Point Elevation The change in freezing point can be found similarly:  T f = K f  m Here K f is the molal freezing point depression constant of the solvent.  T f is subtracted from the normal boiling point of the solvent.

71 Solutions © 2009, Prentice-Hall, Inc. Freezing Point Depression  particles of nonvolatile solute   freezing point -When a solution freezes, crystals of almost pure solvent form first -Solute molecules are usually not soluble in the solid phase of the solvent, so become more concentrated in the liquid phase   VP  triple point occurs at a lower T because of the lower VP for the solution. -The melting-point (freezing-point) curve is a vertical line from the triple point -  the solution freezes at a lower temperature than the pure solvent

72 Solutions © 2009, Prentice-Hall, Inc. Boiling Point Elevation and Freezing Point Depression Note that in both equations,  T does not depend on what the solute is, but only on how many particles are dissolved.  T b = K b  m  T f = K f  m

73 Solutions © 2009, Prentice-Hall, Inc. Boiling Point Elevation and Freezing Point Depression e.g. What is the boiling point of a solution that contains 1.25 mol CaCl 2 in g of water? mol CaCl 2 = 1.25 mol CaCl 2 x g = m x 3 = 2.68 m g H 2 O g H 2 O 1 kg = # particles when CaCl 2 dissociates in H 2 O  T b = K b  m = o C x 2.68 m = 1.37 o C o C = o C

74 Solutions © 2009, Prentice-Hall, Inc. Sample Exercise 13.9 (p. 550) Automotive antifreeze consists of ethylene glycol (C 2 H 6 O 2 ), a nonvolatile nonelectrolyte. Calculate the boiling point and freezing point of a 25.0 mass % solution of ethylene glycol in water. (102.7 o C; o C)

75 Solutions © 2009, Prentice-Hall, Inc. Practice Exercise 13.9 Calculate the freezing point of a solution containing kg of chloroform (CHCl 3 ) and 42.0 g of eucalyptol (C 10 H 18 O), a fragrant substance found in the leaves of eucalyptus trees. (See Table 13.4) (-65.6 o C)

76 Solutions © 2009, Prentice-Hall, Inc. Sample Exercise (p. 551) List the following aqueous solutions in order of their expected freezing points: m CaCl 2 ; 0.15 m HCl; m HC 2 H 3 O 2 ; 0.10 m C 12 H 22 O 11.

77 Solutions © 2009, Prentice-Hall, Inc. Practice Exercise Which of the following solutes will produce the largest increase in boiling point upon addition of 1 kg of water: 1 mol Co(NO 3 ) 2, 2 mol of KCl, 3 mol of ethylene glycol (C 2 H 6 O 2 )? (2 mol KCl)

78 Solutions © 2009, Prentice-Hall, Inc. Check in End of last semester colligative properties, esp. freezing point depression and boiling point elevation Calculations: predict new bp or fp Next: Colligative props of electrolytes Osmotic pressure How to determine molar mass using colligative properties - freezing point depression lab

79 Solutions © 2009, Prentice-Hall, Inc. Colligative Properties of Electrolytes Since these properties depend on the number of particles dissolved, solutions of electrolytes (which dissociate in solution) should show greater changes than those of nonelectrolytes.

80 Solutions © 2009, Prentice-Hall, Inc. Colligative Properties of Electrolytes However, a 1M solution of NaCl does not show twice the change in freezing point that a 1M solution of methanol does.

81 Solutions © 2009, Prentice-Hall, Inc. van’t Hoff Factor One mole of NaCl in water does not really give rise to two moles of ions.

82 Solutions © 2009, Prentice-Hall, Inc. van’t Hoff Factor Some Na + and Cl - reassociate for a short time, so the true concentration of particles is somewhat less than two times the concentration of NaCl.

83 Solutions © 2009, Prentice-Hall, Inc. van’t Hoff Factor Reassociation is more likely at higher concentration. Therefore, the number of particles present is concentration- dependent.

84 Solutions © 2009, Prentice-Hall, Inc. van’t Hoff Factor We modify the previous equations by multiplying by the van’t Hoff factor, i.  T f = K f  m  i (is is used incorrectly in your handout on p. 18)

85 Solutions © 2009, Prentice-Hall, Inc. Osmosis Some substances form semipermeable membranes, allowing some smaller particles to pass through, but blocking other larger particles. In biological systems, most semipermeable membranes allow water to pass through, but solutes are not free to do so.

86 Solutions © 2009, Prentice-Hall, Inc. Osmosis In osmosis, there is net movement of solvent from the area of higher solvent concentration (lower solute concentration) to the are of lower solvent concentration (higher solute concentration).

87 Solutions © 2009, Prentice-Hall, Inc. Osmotic Pressure The pressure required to stop osmosis, known as osmotic pressure, , is nVnV  = ( ) RT = MRT where M is the molarity of the solution. If the osmotic pressure is the same on both sides of a membrane (i.e., the concentrations are the same), the solutions are isotonic.

88 Solutions © 2009, Prentice-Hall, Inc. Osmosis in Blood Cells If the solute concentration outside the cell is greater than that inside the cell, the solution is hypertonic. Water will flow out of the cell, and crenation results.

89 Solutions © 2009, Prentice-Hall, Inc. Osmosis in Cells If the solute concentration outside the cell is less than that inside the cell, the solution is hypotonic. Water will flow into the cell, and hemolysis results.

90 Solutions © 2009, Prentice-Hall, Inc. Sample Exercise (p. 553) (on your own) The average osmotic pressure of blood is 7.7 atm at 25 o C. What concentration of glucose (C 6 H 12 O 6 ) will be isotonic with blood? (0.31 M)

91 Solutions © 2009, Prentice-Hall, Inc. Practice Exercise (on your own) What is the osmotic pressure at 20 o C of a M sucrose (C 12 H 22 O 11 ) solution? (0.048 atm or 37 torr)

92 Solutions © 2009, Prentice-Hall, Inc. Using Colligative Properties to Determine Molar Mass Any of the four colligative properties may be used to determine molar mass Thursday’s lab

93 Solutions © 2009, Prentice-Hall, Inc. Sample Exercise (p. 555) A solution of an unknown nonvolatile nonelectrolyte was prepared by dissolving g of the substance in 40.0 g of CCl 4. The boiling point of the resultant solution was o C higher than that of the pure solvent. Calculate the molar mass of the solute. (88.0 g/mol)

94 Solutions © 2009, Prentice-Hall, Inc. Practice Exercise Camphor (C 10 H 16 O) melts at o C, and it has a particularly large freezing-point- depression constant, K f = 40.0 o C/m. When g of an organic substance of unknown molar mass is dissolved in g of liquid camphor, the freezing point of the mixture is found to be o C. What is the molar mass of the solute? (110 g/mol)

95 Solutions © 2009, Prentice-Hall, Inc. Sample Exercise (p. 556) The osmotic pressure of an aqueous solution of a certain protein was measured in order to determine its molar mass. The solution contained 3.50 mg of protein dissolved in sufficient water to form 5.00 mL of solution. The osmotic pressure of the solution at 25 o C was found to be 1.54 torr. Calculate the molar mass of the protein. (8.45 x 10 3 g/mol)

96 Solutions © 2009, Prentice-Hall, Inc. Practice Exercise A sample of 2.05 g of the plastic polystyrene was dissolved in enough toluene to form L of solution. The osmotic pressure of this solution was found to be 1.21 kPa at 25 o C. Calculate the molar mass of the polystyrene. (4.20 x 10 4 g/mol)

97 Solutions © 2009, Prentice-Hall, Inc.

98 Solutions © 2009, Prentice-Hall, Inc. Colloids (review on your own) Suspensions of particles larger than individual ions or molecules, but too small to be settled out by gravity are called colloids.

99 Solutions © 2009, Prentice-Hall, Inc. Tyndall Effect Colloidal suspensions can scatter rays of light. This phenomenon is known as the Tyndall effect.

100 Solutions © 2009, Prentice-Hall, Inc. Colloids in Biological Systems Some molecules have a polar, hydrophilic (water-loving) end and a non-polar, hydrophobic (water- hating) end.

101 Solutions © 2009, Prentice-Hall, Inc. Colloids in Biological Systems Sodium stearate is one example of such a molecule.

102 Solutions © 2009, Prentice-Hall, Inc. Colloids in Biological Systems These molecules can aid in the emulsification of fats and oils in aqueous solutions.

103 Solutions © 2009, Prentice-Hall, Inc. Sample Integrative Exercise 13 (p. 560) A L solution is made by dissolving g of CaCl 2(s) in water. a) Calculate the osmotic pressure of this solution at 27 o C, assuming that it is completely dissociated into its component ions. (2.93 atm) b)The measured osmotic pressure of this solution is 2.56 atm at 27 o C. Explain why it is less than the value calculated in (a), and calculate the van’t Hoff factor, i, for the solute in this solution. (2.62)

104 Solutions © 2009, Prentice-Hall, Inc. Sample Integrative Exercise 13 (p. 560) c) The enthalpy of solution for CaCl 2 is  H = kJ/mol. If the final temperature of the solution was 27.0 o C, what was its initial temperature? (Assume that the density of the solution is 1.00 g/mL, that its specific heat is 4.18 J/g-K, and that the solution loses no heat to its surroundings.) (26.2 o C)


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