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Chapter 12 Solutions 2008, Prentice Hall Chemistry: A Molecular Approach, 1 st Ed. Nivaldo Tro Roy Kennedy Massachusetts Bay Community College Wellesley.

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Presentation on theme: "Chapter 12 Solutions 2008, Prentice Hall Chemistry: A Molecular Approach, 1 st Ed. Nivaldo Tro Roy Kennedy Massachusetts Bay Community College Wellesley."— Presentation transcript:

1 Chapter 12 Solutions 2008, Prentice Hall Chemistry: A Molecular Approach, 1 st Ed. Nivaldo Tro Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

2 Tro, Chemistry: A Molecular Approach2 Solution homogeneous mixtures composition may vary from one sample to another appears to be one substance, though really contains multiple materials most homogeneous materials we encounter are actually solutions e.g., air and sea water nature has a tendency toward spontaneous mixing generally, uniform mixing is more energetically favorable

3 Tro, Chemistry: A Molecular Approach3 Solutions solute is the dissolved substance seems to disappear takes on the state of the solvent solvent is the substance solute dissolves in does not appear to change state when both solute and solvent have the same state, the solvent is the component present in the highest percentage solutions in which the solvent is water are called aqueous solutions

4 Tro, Chemistry: A Molecular Approach4 Seawater drinking seawater will dehydrate you and give you diarrhea the cell wall acts as a barrier to solute moving the only way for the seawater and the cell solution to have uniform mixing is for water to flow out of the cells of your intestine and into your digestive tract

5 Tro, Chemistry: A Molecular Approach5 Common Types of Solution Solution Phase Solute Phase Solvent PhaseExample gaseous solutionsgas air (mostly N 2 & O 2 ) liquid solutions gas liquid solid liquid soda (CO 2 in H 2 O) vodka (C 2 H 5 OH in H 2 O) seawater (NaCl in H 2 O) solid solutionssolid brass (Zn in Cu) solutions that contain Hg and some other metal are called amalgams solutions that contain metal solutes and a metal solvent are called alloys

6 6 Brass TypeColor% Cu% ZnDensity g/cm 3 MP °C Tensile Strength psi Uses Gildingredish Kpre-83 pennies, munitions, plaques Commercialbronze Kdoor knobs, grillwork Jewelrybronze Kcostume jewelry Redgolden Kelectrical sockets, fasteners & eyelets Lowdeep yellow Kmusical instruments, clock dials Cartridgeyellow Kcar radiator cores Commonyellow Klamp fixtures, bead chain Muntz metalyellow Knuts & bolts, brazing rods

7 Tro, Chemistry: A Molecular Approach7 Solubility when one substance (solute) dissolves in another (solvent) it is said to be soluble salt is soluble in water bromine is soluble in methylene chloride when one substance does not dissolve in another it is said to be insoluble oil is insoluble in water the solubility of one substance in another depends on two factors – natures tendency towards mixing, and the types of intermolecular attractive forces

8 Tro, Chemistry: A Molecular Approach8 Spontaneous Mixing

9 Tro, Chemistry: A Molecular Approach9 Solubility there is usually a limit to the solubility of one substance in another gases are always soluble in each other two liquids that are mutually soluble are said to be miscible alcohol and water are miscible oil and water are immiscible the maximum amount of solute that can be dissolved in a given amount of solvent is called the solubility the solubility of one substance in another varies with temperature and pressure

10 Tro, Chemistry: A Molecular Approach10 Mixing and the Solution Process Entropy formation of a solution does not necessarily lower the potential energy of the system the difference in attractive forces between atoms of two separate ideal gases vs. two mixed ideal gases is negligible yet the gases mix spontaneously the gases mix because the energy of the system is lowered through the release of entropy entropy is the measure of energy dispersal throughout the system energy has a spontaneous drive to spread out over as large a volume as it is allowed

11 Tro, Chemistry: A Molecular Approach11 Intermolecular Forces and the Solution Process Enthalpy of Solution energy changes in the formation of most solutions also involve differences in attractive forces between particles must overcome solute-solute attractive forces endothermic must overcome some of the solvent-solvent attractive forces endothermic at least some of the energy to do this comes from making new solute-solvent attractions exothermic

12 Tro, Chemistry: A Molecular Approach12 Intermolecular Attractions

13 Tro, Chemistry: A Molecular Approach13 Relative Interactions and Solution Formation when the solute-to-solvent attractions are weaker than the sum of the solute-to-solute and solvent-to-solvent attractions, the solution will only form if the energy difference is small enough to be overcome by the entropy Solute-to-Solvent> Solute-to-Solute + Solvent-to-Solvent Solution Forms Solute-to-Solvent= Solute-to-Solute + Solvent-to-Solvent Solution Forms Solute-to-Solvent< Solute-to-Solute + Solvent-to-Solvent Solution May or May Not Form

14 Tro, Chemistry: A Molecular Approach14 Solution Interactions

15 Tro, Chemistry: A Molecular Approach15 Will It Dissolve? Chemists Rule of Thumb – Like Dissolves Like a chemical will dissolve in a solvent if it has a similar structure to the solvent when the solvent and solute structures are similar, the solvent molecules will attract the solute particles at least as well as the solute particles to each other

16 Tro, Chemistry: A Molecular Approach16 Classifying Solvents SolventClass Structural Feature Water, H 2 OpolarO-H Methyl Alcohol, CH 3 OHpolarO-H Ethyl Alcohol, C 2 H 5 OHpolarO-H Acetone, C 3 H 6 OpolarC=O Toluene, C 7 H 8 nonpolarC-C & C-H Hexane, C 6 H 14 nonpolarC-C & C-H Diethyl Ether, C 4 H 10 OnonpolarC-C, C-H & C-O, (nonpolar > polar) Carbon TetrachloridenonpolarC-Cl, but symmetrical

17 Tro, Chemistry: A Molecular Approach17 Example 12.1a predict whether the following vitamin is soluble in fat or water Vitamin C The 4 OH groups make the molecule highly polar and it will also H-bond to water. Vitamin C is water soluble

18 Tro, Chemistry: A Molecular Approach18 Example 12.1b predict whether the following vitamin is soluble in fat or water Vitamin K 3 The 2 C=O groups are polar, but their geometric symmetry suggests their pulls will cancel and the molecule will be nonpolar. Vitamin K 3 is fat soluble

19 Tro, Chemistry: A Molecular Approach19 Energetics of Solution Formation overcome attractions between the solute particles – endothermic overcome some attractions between solvent molecules – endothermic for new attractions between solute particles and solvent molecules – exothermic the overall H depends on the relative sizes of the H for these 3 processes H soln = H solute + H solvent + H mix

20 Tro, Chemistry: A Molecular Approach20 Solution Process 1. add energy in to overcome solute-solute attractions 2. add energy in to overcome some solvent-solvent attractions 3. form new solute-solvent attractions, releasing energy

21 Tro, Chemistry: A Molecular Approach21 Energetics of Solution Formation if the total energy cost for breaking attractions between particles in the pure solute and pure solvent is less than the energy released in making the new attractions between the solute and solvent, the overall process will be exothermic if the total energy cost for breaking attractions between particles in the pure solute and pure solvent is greater than the energy released in making the new attractions between the solute and solvent, the overall process will be endothermic

22 Tro, Chemistry: A Molecular Approach22 Heats of Hydration for aqueous ionic solutions, the energy added to overcome the attractions between water molecules and the energy released in forming attractions between the water molecules and ions is combined into a term called the heat of hydration attractive forces in water = H-bonds attractive forces between ion and water = ion-dipole H hydration = heat released when 1 mole of gaseous ions dissolves in water

23 Tro, Chemistry: A Molecular Approach23 Heat of Hydration

24 Tro, Chemistry: A Molecular Approach24 Ion-Dipole Interactions when ions dissolve in water they become hydrated each ion is surrounded by water molecules

25 Tro, Chemistry: A Molecular Approach25 Solution Equilibrium the dissolution of a solute in a solvent is an equilibrium process initially, when there is no dissolved solute, the only process possible is dissolution shortly, solute particles can start to recombine to reform solute molecules – but the rate of dissolution >> rate of deposition and the solute continues to dissolve eventually, the rate of dissolution = the rate of deposition – the solution is saturated with solute and no more solute will dissolve

26 Tro, Chemistry: A Molecular Approach26 Solution Equilibrium

27 Tro, Chemistry: A Molecular Approach27 Solubility Limit a solution that has the maximum amount of solute dissolved in it is said to be saturated depends on the amount of solvent depends on the temperature and pressure of gases a solution that has less solute than saturation is said to be unsaturated a solution that has more solute than saturation is said to be supersaturated

28 Tro, Chemistry: A Molecular Approach28 How Can You Make a Solvent Hold More Solute Than It Is Able To? solutions can be made saturated at non-room conditions – then allowed to come to room conditions slowly for some solutes, instead of coming out of solution when the conditions change, they get stuck in-between the solvent molecules and the solution becomes supersaturated supersaturated solutions are unstable and lose all the solute above saturation when disturbed e.g., shaking a carbonated beverage

29 Tro, Chemistry: A Molecular Approach29 Adding Solute to a Supersaturated Solution of NaC 2 H 3 O 2

30 Tro, Chemistry: A Molecular Approach30 Temperature Dependence of Solubility of Solids in Water solubility is generally given in grams of solute that will dissolve in 100 g of water for most solids, the solubility of the solid increases as the temperature increases when H solution is endothermic solubility curves can be used to predict whether a solution with a particular amount of solute dissolved in water is saturated (on the line), unsaturated (below the line), or supersaturated (above the line)

31 Tro, Chemistry: A Molecular Approach31 Solubility Curves

32 Tro, Chemistry: A Molecular Approach32

33 Tro, Chemistry: A Molecular Approach33

34 Tro, Chemistry: A Molecular Approach34 Temperature Dependence of Solubility of Gases in Water solubility is generally given in moles of solute that will dissolve in 1 Liter of solution generally lower solubility than ionic or polar covalent solids because most are nonpolar molecules for all gases, the solubility of the gas decreases as the temperature increases the H solution is exothermic because you do not need to overcome solute-solute attractions

35 Tro, Chemistry: A Molecular Approach35

36 Tro, Chemistry: A Molecular Approach36

37 Tro, Chemistry: A Molecular Approach37

38 38 Pressure Dependence of Solubility of Gases in Water the larger the partial pressure of a gas in contact with a liquid, the more soluble the gas is in the liquid

39 Tro, Chemistry: A Molecular Approach39 Henrys Law the solubility of a gas (S gas ) is directly proportional to its partial pressure, (P gas ) S gas = k H P gas k H is called Henrys Law Constant

40 Tro, Chemistry: A Molecular Approach40 Relationship between Partial Pressure and Solubility of a Gas

41 Tro, Chemistry: A Molecular Approach41

42 Tro, Chemistry: A Molecular Approach42 persrst

43 Tro, Chemistry: A Molecular Approach43 Ex 12.2 – What pressure of CO 2 is required to keep the [CO 2 ] = 0.12 M at 25°C? the unit is correct, the pressure higher than 1 atm meets our expectation from general experience Check: Solve: S = k H P, k H = 3.4 x M/atm Concept Plan: Relationships: S = [CO 2 ] = 0.12 M, P of CO 2, atm Given: Find: [CO 2 ]P

44 Tro, Chemistry: A Molecular Approach44 Concentrations solutions have variable composition to describe a solution, need to describe components and relative amounts the terms dilute and concentrated can be used as qualitative descriptions of the amount of solute in solution concentration = amount of solute in a given amount of solution occasionally amount of solvent

45 Tro, Chemistry: A Molecular Approach45 Solution Concentration Molarity moles of solute per 1 liter of solution used because it describes how many molecules of solute in each liter of solution if a sugar solution concentration is 2.0 M, 1 liter of solution contains 2.0 moles of sugar, 2 liters = 4.0 moles sugar, 0.5 liters = 1.0 mole sugar molarity = moles of solute liters of solution

46 Tro, Chemistry: A Molecular Approach46 Molarity and Dissociation the molarity of the ionic compound allows you to determine the molarity of the dissolved ions CaCl 2 (aq) = Ca +2 (aq) + 2 Cl -1 (aq) A 1.0 M CaCl 2 (aq) solution contains 1.0 moles of CaCl 2 in each liter of solution 1 L = 1.0 moles CaCl 2, 2 L = 2.0 moles CaCl 2 Because each CaCl 2 dissociates to give one Ca +2 = 1.0 M Ca +2 1 L = 1.0 moles Ca +2, 2 L = 2.0 moles Ca +2 Because each CaCl 2 dissociates to give 2 Cl -1 = 2.0 M Cl -1 1 L = 2.0 moles Cl -1, 2 L = 4.0 moles Cl -1

47 Tro, Chemistry: A Molecular Approach47 Solution Concentration Molality, m moles of solute per 1 kilogram of solvent defined in terms of amount of solvent, not solution like the others does not vary with temperature because based on masses, not volumes

48 Tro, Chemistry: A Molecular Approach48 Percent parts of solute in every 100 parts solution mass percent = mass of solute in 100 parts solution by mass if a solution is 0.9% by mass, then there are 0.9 grams of solute in every 100 grams of solution or 0.9 kg solute in every 100 kg solution

49 Tro, Chemistry: A Molecular Approach49 Percent Concentration

50 Tro, Chemistry: A Molecular Approach50 Using Concentrations as Conversion Factors concentrations show the relationship between the amount of solute and the amount of solvent 12%(m/m) sugar(aq) means 12 g sugar 100 g solution or 12 kg sugar 100 kg solution; or 12 lbs. 100 lbs. solution 5.5%(m/v) Ag in Hg means 5.5 g Ag 100 mL solution 22%(v/v) alcohol(aq) means 22 mL EtOH 100 mL solution The concentration can then be used to convert the amount of solute into the amount of solution, or vice versa

51 Tro, Chemistry: A Molecular Approach51 Preparing a Solution need to know amount of solution and concentration of solution calculate the mass of solute needed start with amount of solution use concentration as a conversion factor 5% by mass 5 g solute 100 g solution Dissolve the grams of solute in enough solvent to total the total amount of solution.

52 Tro, Chemistry: A Molecular Approach52 Example - How would you prepare g of 5.00% by mass glucose solution (normal glucose)? Given:250.0 g solution Find:g Glucose Equivalence:5.00 g Glucose 100 g solution Solution Map: g solution g Glucose Apply Solution Map: Answer: Dissolve 12.5 g of glucose in enough water to total g

53 53 Solution Concentration PPM grams of solute per 1,000,000 g of solution mg of solute per 1 kg of solution 1 liter of water = 1 kg of water for water solutions we often approximate the kg of the solution as the kg or L of water grams solute grams solution x 10 6 mg solute kg solution mg solute L solution

54 Tro, Chemistry: A Molecular Approach54 Ex 12.3 – What volume of 10.5% by mass soda contains 78.5 g of sugar? the unit is correct, the magnitude seems reasonable as the mass of sugar 10% the volume of solution Check: Solve: 100 g soln = 10.5 g sugar, 1 mL soln = 1.04 g Concept Plan: Relationships: 78.5 g sugar volume, mL Given: Find: g soluteg solnmL soln

55 Tro, Chemistry: A Molecular Approach55 Solution Concentrations Mole Fraction, X A the mole fraction is the fraction of the moles of one component in the total moles of all the components of the solution total of all the mole fractions in a solution = 1 unitless the mole percentage is the percentage of the moles of one component in the total moles of all the components of the solution = mole fraction x 100% mole fraction of A = X A = moles of components A total moles in the solution

56 Tro, Chemistry: A Molecular Approach56 Ex 12.4a – What is the molarity of a solution prepared by mixing 17.2 g of C 2 H 6 O 2 with kg of H 2 O to make 515 mL of solution? the unit is correct, the magnitude is reasonableCheck: Solve: M = mol/L, 1 mol C 2 H 6 O 2 = g, 1 mL = L Concept Plan: Relationships: 17.2 g C 2 H 6 O 2, kg H 2 O, 515 mL soln M Given: Find: g C 2 H 6 O 2 mol C 2 H 6 O 2 mL solnL soln M

57 Tro, Chemistry: A Molecular Approach57 Ex 12.4b – What is the molality of a solution prepared by mixing 17.2 g of C 2 H 6 O 2 with kg of H 2 O to make 515 mL of solution? the unit is correct, the magnitude is reasonableCheck: Solve: m = mol/kg, 1 mol C 2 H 6 O 2 = g Concept Plan: Relationships: 17.2 g C 2 H 6 O 2, kg H 2 O, 515 mL soln m Given: Find: g C 2 H 6 O 2 mol C 2 H 6 O 2 kg H 2 O m

58 Tro, Chemistry: A Molecular Approach58 Ex 12.4c – What is the percent by mass of a solution prepared by mixing 17.2 g of C 2 H 6 O 2 with kg of H 2 O to make 515 mL of solution? the unit is correct, the magnitude is reasonableCheck: Solve: 1 kg = 1000 g Concept Plan: Relationships: 17.2 g C 2 H 6 O 2, kg H 2 O, 515 mL soln %(m/m) Given: Find: g C 2 H 6 O 2 g solventg soln %

59 Tro, Chemistry: A Molecular Approach59 the unit is correct, the magnitude is reasonable Ex 12.4d – What is the mole fraction of a solution prepared by mixing 17.2 g of C 2 H 6 O 2 with kg of H 2 O to make 515 mL of solution? Check: Solve: = mol A /mol tot, 1 mol C 2 H 6 O 2 = g, 1 mol H 2 O = g Concept Plan: Relationships: 17.2 g C 2 H 6 O 2, kg H 2 O, 515 mL soln Given: Find: g C 2 H 6 O 2 mol C 2 H 6 O 2 g H 2 Omol H 2 O

60 Tro, Chemistry: A Molecular Approach60 the unit is correct, the magnitude is reasonable Ex 12.4d – What is the mole percent of a solution prepared by mixing 17.2 g of C 2 H 6 O 2 with kg of H 2 O to make 515 mL of solution? Check: Solve: = mol A /mol tot, 1 mol C 2 H 6 O 2 = g, 1 mol H 2 O = g Concept Plan: Relationships: 17.2 g C 2 H 6 O 2, kg H 2 O, 515 mL soln Given: Find: g C 2 H 6 O 2 mol C 2 H 6 O 2 g H 2 Omol H 2 O

61 Tro, Chemistry: A Molecular Approach61 Converting Concentration Units assume a convenient amount of solution given %(m/m), assume 100 g solution given %(m/v), assume 100 mL solution given ppm, assume 1,000,000 g solution given M, assume 1 liter of solution given m, assume 1 kg of solvent given X, assume you have a total of 1 mole of solutes in the solution determine amount of solution in non-given unit(s) if assume amount of solution in grams, use density to convert to mL and then to L if assume amount of solution in L or mL, use density to convert to grams determine the amount of solute in this amount of solution, in grams and moles determine the amount of solvent in this amount of solution, in grams and moles use definitions to calculate other units

62 62

63 Tro, Chemistry: A Molecular Approach63 Vapor Pressure of Solutions the vapor pressure of a solvent above a solution is lower than the vapor pressure of the pure solvent the solute particles replace some of the solvent molecules at the surface The pure solvent establishes an liquid vapor equilibrium Addition of a nonvolatile solute reduces the rate of vaporization, decreasing the amount of vapor Eventually, equilibrium is re- established, but a smaller number of vapor molecules – therefore the vapor pressure will be lower

64 Tro, Chemistry: A Molecular Approach64 Thirsty Solutions a concentrated solution will draw solvent molecules toward it due to the natural drive for materials in nature to mix similarly, a concentrated solution will draw pure solvent vapor into it due to this tendency to mix the result is reduction in vapor pressure

65 Tro, Chemistry: A Molecular Approach65 Thirsty Solutions Beakers with equal liquid levels of pure solvent and a solution are place in a bell jar. Solvent molecules evaporate from each one and fill the bell jar, establishing an equilibrium with the liquids in the beakers. When equilibrium is established, the liquid level in the solution beaker is higher than the solution level in the pure solvent beaker – the thirsty solution grabs and holds solvent vapor more effectively

66 Tro, Chemistry: A Molecular Approach66 Raoults Law the vapor pressure of a volatile solvent above a solution is equal to its mole fraction of its normal vapor pressure, P° P solvent in solution = solvent P° since the mole fraction is always less than 1, the vapor pressure of the solvent in solution will always be less than the vapor pressure of the pure solvent

67 Tro, Chemistry: A Molecular Approach67 Ex 12.5 – Calculate the vapor pressure of water in a solution prepared by mixing 99.5 g of C 12 H 22 O 11 with mL of H 2 O Solve: P° H2O = 23.8 torr, 1 mol C 12 H 22 O 11 = g, 1 mol H 2 O = g Concept Plan: Relationships: 99.5 g C 12 H 22 O 11, mL H 2 O P H2O Given: Find: g C 12 H 22 O 11 mol C 12 H 22 O 11 g H 2 Omol H 2 O mL H 2 O P H2O

68 Tro, Chemistry: A Molecular Approach68 Ionic Solutes and Vapor Pressure according to Raoults Law, the effect of solute on the vapor pressure simply depends on the number of solute particles when ionic compounds dissolve in water, they dissociate – so the number of solute particles is a multiple of the number of moles of formula units the effect of ionic compounds on the vapor pressure of water is magnified by the dissociation since NaCl dissociates into 2 ions, Na + and Cl, one mole of NaCl lowers the vapor pressure of water twice as much as 1 mole of C 12 H 22 O 11 molecules would

69 Tro, Chemistry: A Molecular Approach69 Effect of Dissociation

70 Tro, Chemistry: A Molecular Approach70 Ex 12.6 – What is the vapor pressure of H 2 O when mol Ca(NO 3 ) 2 is mixed with mol H 2 55°C? the unit is correct, the pressure lower than the normal vapor pressure makes sense Check: Solve: P = P° Concept Plan: Relationships: mol Ca(NO 3 ) 2, mol H 2 O, P° = torr P of H 2 O, torr Given: Find: H2O, P° H2O P H2O Ca(NO 3 ) 2 Ca NO 3 3(0.102 mol solute)

71 Tro, Chemistry: A Molecular Approach71 Raoults Law for Volatile Solute when both the solvent and the solute can evaporate, both molecules will be found in the vapor phase the total vapor pressure above the solution will be the sum of the vapor pressures of the solute and solvent for an ideal solution P total = P solute + P solvent the solvent decreases the solute vapor pressure in the same way the solute decreased the solvents P solute = solute P° solute and P solvent = solvent P° solvent

72 Tro, Chemistry: A Molecular Approach72 Ideal vs. Nonideal Solution in ideal solutions, the made solute-solvent interactions are equal to the sum of the broken solute-solute and solvent-solvent interactions ideal solutions follow Raoults Law effectively, the solute is diluting the solvent if the solute-solvent interactions are stronger or weaker than the broken interactions the solution is nonideal

73 Tro, Chemistry: A Molecular Approach73 Vapor Pressure of a Nonideal Solution when the solute-solvent interactions are stronger than the solute-solute + solvent-solvent, the total vapor pressure of the solution will be less than predicted by Raoults Law because the vapor pressures of the solute and solvent are lower than ideal when the solute-solvent interactions are weaker than the solute-solute + solvent-solvent, the total vapor pressure of the solution will be larger than predicted by Raoults Law

74 Tro, Chemistry: A Molecular Approach74 Deviations from Raoults Law

75 Tro, Chemistry: A Molecular Approach75

76 Tro, Chemistry: A Molecular Approach76 Freezing Point Depression the freezing point of a solution is lower than the freezing point of the pure solvent for a nonvolatile solute therefore the melting point of the solid solution is lower the difference between the freezing point of the solution and freezing point of the pure solvent is directly proportional to the molal concentration of solute particles FP solvent – FP solution ) = T f = m K f the proportionality constant is called the Freezing Point Depression Constant, K f the value of K f depends on the solvent the units of K f are °C/m

77 Tro, Chemistry: A Molecular Approach77 KfKf

78 Tro, Chemistry: A Molecular Approach78 Ex 12.8 – What is the freezing point of a 1.7 m aqueous ethylene glycol solution, C 2 H 6 O 2 ? the unit is correct, the freezing point lower than the normal freezing point makes sense Check: Solve: T f = m K f, K f for H 2 O = 1.86 °C/m, FP H2O = 0.00°C Concept Plan: Relationships: 1.7 m C 2 H 6 O 2 (aq) T f, °C Given: Find: m T f

79 Tro, Chemistry: A Molecular Approach79 Boiling Point Elevation the boiling point of a solution is higher than the boiling point of the pure solvent for a nonvolatile solute the difference between the boiling point of the solution and boiling point of the pure solvent is directly proportional to the molal concentration of solute particles BP solution – BP solvent ) = T b = m K b the proportionality constant is called the Boiling Point Elevation Constant, K b the value of K b depends on the solvent the units of K b are °C/m

80 Tro, Chemistry: A Molecular Approach80 Ex 12.9 – How many g of ethylene glycol, C 2 H 6 O 2, must be added to 1.0 kg H 2 O to give a solution that boils at 105°C? Solve: T b = m K b, K b H 2 O = °C/m, BP H2O = 100.0°C MM C2H6O2 = g/mol, 1 kg = 1000 g Concept Plan: Relationships: 1.0 kg H 2 O, T b = 105°C mass C 2 H 6 O 2, g Given: Find: T b mkg H 2 Omol C 2 H 6 O 2 g C 2 H 6 O 2

81 Tro, Chemistry: A Molecular Approach81 Osmosis osmosis is the flow of solvent through a semi- permeable membrane from solution of low concentration to solution of high concentration the amount of pressure needed to keep osmotic flow from taking place is called the osmotic pressure the osmotic pressure,, is directly proportional to the molarity of the solute particles R = (atm L)/(mol K) = MRT

82 Tro, Chemistry: A Molecular Approach82

83 Tro, Chemistry: A Molecular Approach83 Ex – What is the molar mass of a protein if 5.87 mg per 10 mL gives an osmotic pressure of 2.45 torr at 25°C? Solve: MRT, T(K)=T(°C) , R= atmL/molK M = mol/L, 1 mL = L, MM = g/mol, 1 atm = 760 torr Concept Plan: Relationships: 5.87 mg/10 mL, P = 2.45 torr, T = 25°C molar mass, g/mol Given: Find: T MmLLmol protein

84 Tro, Chemistry: A Molecular Approach84 Colligative Properties colligative properties are properties whose value depends only on the number of solute particles, and not on what they are Vapor Pressure Depression, Freezing Point Depression, Boiling Point Elevation, Osmotic Pressure the vant Hoff factor, i, is the ratio of moles of solute particles to moles of formula units dissolved measured vant Hoff factors are often lower than you might expect due to ion pairing in solution

85 Tro, Chemistry: A Molecular Approach85

86 Tro, Chemistry: A Molecular Approach86 An isosmotic solution has the same osmotic pressure as the solution inside the cell – as a result there is no net the flow of water into or out of the cell. A hyperosmotic solution has a higher osmotic pressure than the solution inside the cell – as a result there is a net flow of water out of the cell, causing it to shrivel A hyposmotic solution has a lower osmotic pressure than the solution inside the cell – as a result there is a net flow of water into the cell, causing it to swell

87 Tro, Chemistry: A Molecular Approach87 Colloids a colloidal suspension is a heterogeneous mixture in which one substance is dispersed through another most colloids are made of finely divided particles suspended in a medium the difference between colloids and regular suspensions is generally particle size – colloidal particles are from 1 to 100 nm in size

88 Tro, Chemistry: A Molecular Approach88 Properties of Colloids the particles in a colloid exhibit Brownian motion colloids exhibit the Tyndall Effect scattering of light as it passes through a suspension colloids scatter short wavelength (blue) light more effectively than long wavelength (red) light

89 Tro, Chemistry: A Molecular Approach89

90 Tro, Chemistry: A Molecular Approach90

91 Tro, Chemistry: A Molecular Approach91 Soap and Micelles soap molecules have both a hydrophilic (polar/ionic) head and a hydrophobic (nonpolar) tail part of the molecule is attracted to water, but the majority of it isnt the nonpolar tail tends to coagulate together to form a spherical structure called a micelle

92 Tro, Chemistry: A Molecular Approach92

93 Tro, Chemistry: A Molecular Approach93 Soap Action most dirt and grease is made of nonpolar molecules – making it hard for water to remove it from the surface soap molecules form micelles around the small oil particles with the polar/ionic heads pointing out this allows the micelle to be attracted to water and stay suspended

94 Tro, Chemistry: A Molecular Approach94 The polar heads of the micelles attract them to the water, and simultaneously repel other micelles so they will not coalesce and settle out.

95 Tro, Chemistry: A Molecular Approach95


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