1 SOLUTIONS AND THEIR BEHAVIOR CHAPTER 14 CHAPTER 14

11

SOLUTIONS AND THEIR SOLUTIONS AND THEIR BEHAVIORBEHAVIOR

CHAPTER 14CHAPTER 14

22

CHAPTER OVERVIEWCHAPTER OVERVIEW

• This chapter examines homogeneous This chapter examines homogeneous mixtures called solutions, which are mixtures called solutions, which are made up of a solute and a solvent. made up of a solute and a solvent.

• Concentrations of solutions can be Concentrations of solutions can be expressed in a variety of units. expressed in a variety of units.

• Properties of solutions that depend Properties of solutions that depend only on the number of solute only on the number of solute particles and not their type are called particles and not their type are called colligative properties. colligative properties.

33

14.114.1 UNITS OF CONCENTRATION UNITS OF CONCENTRATION

• Molarity, M, moles solute per liter Molarity, M, moles solute per liter solution solution

• Molality, m, moles of solute per Molality, m, moles of solute per kilogram solvent kilogram solvent

• Mole fraction, XMole fraction, XAA, moles A divided , moles A divided by moles totalby moles total

• Weight percent (mass percent), Weight percent (mass percent), wt.% A, wt.% A,

(mass A divided by mass total) x 100%(mass A divided by mass total) x 100%

44

Molarity -vs- MolalityMolarity -vs- Molality

Each flask contains 19.4 g of K2CrO4

Water to the 1.00 L mark

Exactly 1.00 kg of water added

55

UNITS OF CONCENTRATIONUNITS OF CONCENTRATION

• parts per million, ppm, is calculated parts per million, ppm, is calculated like percent, but multiply by 10like percent, but multiply by 1066

• Remember that the mass of the Remember that the mass of the solution equals the mass of the solution equals the mass of the solute solute plusplus solvent. solvent.

• Conversions between molarity and Conversions between molarity and the other concentration units the other concentration units requires the density of the solution.requires the density of the solution.

SolutionsSolutionsSolutionsSolutions

Why does a raw egg swell or shrink when Why does a raw egg swell or shrink when placed in different solutions?placed in different solutions?

6

Some DefinitionsSome DefinitionsSome DefinitionsSome Definitions

A solution is a A solution is a HOMOGENEOUS HOMOGENEOUS mixture of 2 or mixture of 2 or more substances more substances in a single phase. in a single phase.

One constituent is One constituent is usually regarded usually regarded as the SOLVENT as the SOLVENT and the others as and the others as SOLUTES.SOLUTES.

7

88

14.2 THE SOLUTION PROCESS14.2 THE SOLUTION PROCESS

• The key to understanding the solution The key to understanding the solution process is intermolecular forces: solvent - process is intermolecular forces: solvent - solvent; solute - solute; solute - solvent. solvent; solute - solute; solute - solvent.

99

THE SOLUTION PROCESSTHE SOLUTION PROCESS

• What prevents solubility is an energy What prevents solubility is an energy barrier when the latter interaction is barrier when the latter interaction is significantly weaker than the former significantly weaker than the former interactions.interactions.

• Like dissolves like is a general rule, but Like dissolves like is a general rule, but not and explanation of the solubility not and explanation of the solubility phenomenon.phenomenon.

1010

• Liquids Dissolving in LiquidsLiquids Dissolving in Liquids

–MiscibleMiscible liquids are soluble in all liquids are soluble in all proportions. proportions.

– ImmiscibleImmiscible liquids do not mix, but form liquids do not mix, but form separate layers. separate layers.

• Isopropanol is miscible with water but gasoline is not. Explain why.Isopropanol is miscible with water but gasoline is not. Explain why.

1111

SOLUTIONSSOLUTIONS

• A saturated solution is one which has A saturated solution is one which has reached its equilibrium solubility at reached its equilibrium solubility at that temperature. that temperature.

• An unsaturated solution is one that has An unsaturated solution is one that has not reached its equilibrium solubility. not reached its equilibrium solubility.

• A supersaturated solution is one in A supersaturated solution is one in which the equilibrium solubility has which the equilibrium solubility has been temporarily exceeded.been temporarily exceeded.

SUPERSATURATED SUPERSATURATED SOLUTIONSSOLUTIONS contain contain more than is possible more than is possible and are unstable.and are unstable.

DefinitionsDefinitions

A saturated solution A saturated solution contains the contains the maximum quantity of maximum quantity of solute that dissolves solute that dissolves at that temperature.at that temperature.

Solutions can be classified asSolutions can be classified as

unsaturated unsaturated or or

saturatedsaturated..

12

Energetics of the Solution Energetics of the Solution ProcessProcess

If the enthalpy of If the enthalpy of formation of the formation of the solution is more solution is more negative than that of negative than that of the solvent and solute, the solvent and solute, the enthalpy of the enthalpy of solution is negative. solution is negative.

The solution process is The solution process is exothermicexothermic!!

13

1414

Solids Dissolving in LiquidsSolids Dissolving in LiquidsThe same rules apply. The same rules apply.

• Compare the intermolecular forces.Compare the intermolecular forces.

• II22 is quite soluble in CCl is quite soluble in CCl44, but not very soluble in , but not very soluble in

water. Explain why?water. Explain why?

SupersaturatedSupersaturatedSodium AcetateSodium Acetate

• One application of a One application of a supersaturated supersaturated solution is the sodium solution is the sodium acetate “heat pack.”acetate “heat pack.”

• Sodium acetate has an Sodium acetate has an ENDOthermic heat of ENDOthermic heat of solution.solution.

15

1616

Ionic SolutionsIonic Solutions

• The heat of solution for ionic compounds The heat of solution for ionic compounds is the sum of the lattice energy (+), bonds is the sum of the lattice energy (+), bonds breaking, and the hydration energy (-), breaking, and the hydration energy (-), bonds forming. bonds forming.

• It may be positive (endo) or negative (exo) It may be positive (endo) or negative (exo) depending on the relative magnitudes of depending on the relative magnitudes of these energies. these energies.

1717

Hsol’n can be calc’d using Hess' Law.

821 kJ/mol – 819 kJ/mol = +2 kJ/mol (endo)

1818

Hsol’n can be calc’d using Hess' Law.

1919

Ion Size also Ion Size also determines determines SolubilitySolubility

Remember Coulombs Law

+-

2

(charge n)(charge n)Force of Attraction =

dk

2020

Ionic SolutionsIonic Solutions

• Temperature has a significant effect on Temperature has a significant effect on solubility for salts and is consistent with solubility for salts and is consistent with Le Chatelier's principle. Le Chatelier's principle.

2121

The heat of solution for many salts is positive, endothermic, as seen by the positive slope of the graph.

SupersaturatedSupersaturatedSodium AcetateSodium Acetate

Sodium acetate has an Sodium acetate has an ENDOthermicENDOthermic heat of heat of solution. solution.

NaCHNaCH33COCO22 (s) + (s) + heatheat ----> ----> NaNa++(aq) + CH(aq) + CH33COCO22

--(aq)(aq)

Therefore, formation of solid sodium acetate Therefore, formation of solid sodium acetate from its ions is from its ions is EXOTHERMICEXOTHERMIC..

NaNa++(aq) + CH(aq) + CH33COCO22--(aq) ---> (aq) ---> NaCHNaCH33COCO22 (s) + (s) + heatheat

22

2323

Dissolving GasesDissolving Gases

• Gas solubility decreases with increasing Gas solubility decreases with increasing temperature which means temperature which means ΔΔHHsolutionsolution < 0 , < 0 , exothermic.exothermic.

• Gas solubilities increase with increasing Gas solubilities increase with increasing pressure. pressure.

• Write the general equation for the solubility Write the general equation for the solubility of a gas showing that the process is of a gas showing that the process is exothermic and show how increasing the exothermic and show how increasing the temperature decreases the solubility.temperature decreases the solubility.

Gas + Solvent ⇄ Solution + Heat

2424

Colligative PropertiesColligative Properties

On adding a solute to a solvent, the properties On adding a solute to a solvent, the properties of the solvent are modified.of the solvent are modified.

• Vapor pressure Vapor pressure decreasesdecreases

• Melting point Melting point decreasesdecreases

• Boiling point Boiling point increasesincreases

• Osmosis is possible (osmotic pressure)Osmosis is possible (osmotic pressure)

25

Colligative PropertiesColligative Properties

These changes are called These changes are called COLLIGATIVE PROPERTIES. COLLIGATIVE PROPERTIES.

They depend only on the They depend only on the NUMBER of solute particles NUMBER of solute particles relative to solvent particles, not relative to solvent particles, not on the KIND of solute particles.on the KIND of solute particles.

26

An An IDEAL SOLUTIONIDEAL SOLUTION is one is one where the properties depend only where the properties depend only on the concentration of solute.on the concentration of solute.

Need concentration units to tell us Need concentration units to tell us the number of solute particles per the number of solute particles per solvent particle.solvent particle.

The unit “molarity” does not do The unit “molarity” does not do this!this!

Concentration UnitsConcentration Units 27

Concentration UnitsConcentration Units

WEIGHT %WEIGHT % = grams solute per 100 g solution= grams solute per 100 g solution

XA mol fraction A = mol A

mol A + mol B + mol C

m of solute = mol solute

kilograms solvent

MOLALITY, mMOLALITY, m

MOLE FRACTION, XMOLE FRACTION, XFor a mixture of A, B, and CFor a mixture of A, B, and C

28

Calculating ConcentrationsCalculating Concentrations

Dissolve 62.1 g (1.00 mol) of ethylene glycol Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of Hin 250. g of H22O. O.

Calculate mole fraction, molality, and Calculate mole fraction, molality, and weight % of glycol.weight % of glycol.


250. g H250. g H22O = 13.9 moleO = 13.9 mole

Dissolve 62.1 g (1.00 mole) of ethylene glycol in 250. g of H2O. Calculate X, m, and % of glycol.

Xglycol = 1.00 mol glycol

1.00 mol glycol + 13.9 mol H2O

X X glycolglycol = 0.0672 = 0.0672

30


Calculate molalityCalculate molality

Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate X, m, and % of glycol.

conc (molality) = 1.00 mol glycol0.250 kg H2 O

4.00 molal

Calculate weight %Calculate weight %

% glycol = 62.1 g

62.1 g + 250. g x 100% = 19.9%

31

Dissolving Gases & Dissolving Gases & Henry’s LawHenry’s Law

Gas solubility (M) = kGas solubility (M) = kHH • P • Pgasgas

kkHH for O for O22 = 1.66 x 10 = 1.66 x 10-6-6 M/mmHg M/mmHg

When When PPgasgas drops, solubility drops. drops, solubility drops.

32

Lake Nyos, CameroonLake Nyos, Cameroon33

3434

COLLIGATIVE PROPERTIESCOLLIGATIVE PROPERTIES

• Changes in Vapor Pressure: Raoult's LawChanges in Vapor Pressure: Raoult's Law

• The presence of a solute in the solvent The presence of a solute in the solvent lowers the vapor pressure of the solvent.lowers the vapor pressure of the solvent.

PPsolventsolvent = X = Xsolventsolvent P Poosolventsolvent

• If the solute is also volatile, a similar If the solute is also volatile, a similar equation applies to the solute.equation applies to the solute.

PPsolutesolute = X = Xsolutesolute P Poosolute solute

• The total pressure for the solution is given The total pressure for the solution is given by:by: PPtotal total = P = Psolventsolvent + P + Psolutesolute

3535

COLLIGATIVE PROPERTIESCOLLIGATIVE PROPERTIES

• If the solute is nonvolatile, the total If the solute is nonvolatile, the total pressure is just the pressure of the solvent pressure is just the pressure of the solvent and is lower than that of the pure solvent. and is lower than that of the pure solvent.

• Study examples and exercises.Study examples and exercises.

Understanding Understanding Colligative PropertiesColligative Properties


To understand colligative properties, study To understand colligative properties, study the LIQUID-VAPOR EQUILIBRIUM for a the LIQUID-VAPOR EQUILIBRIUM for a solution.solution.

H—O

H

HH—O

H H—O

H—O

H

H—O

Hsurface

36


To understand To understand colligative colligative properties, properties, study the study the LIQUID-VAPOR LIQUID-VAPOR EQUILIBRIUM EQUILIBRIUM for a solution.for a solution.

37

Understanding Understanding Colligative Properties: Colligative Properties:

Raoults’s LawRaoults’s LawVP of HVP of H22O over a solution depends on the O over a solution depends on the

number of Hnumber of H22O molecules per solute molecule.O molecules per solute molecule.

PPsolventsolvent proportional toproportional to X Xsolventsolvent

OROR

PPsolventsolvent = X = Xsolventsolvent • P • Poosolventsolvent

VP of solvent over solution = VP of solvent over solution =

(Mol frac solvent)•(VP pure solvent)(Mol frac solvent)•(VP pure solvent)

RAOULT’S LAW 38

Raoult’s LawRaoult’s LawRaoult’s LawRaoult’s LawAn ideal solution is one that obeys Raoult’s law.An ideal solution is one that obeys Raoult’s law.

PPAA = X = XAA • P • PooAA

Because mole fraction of solvent, XBecause mole fraction of solvent, XAA, is always , is always

less than 1, then Pless than 1, then PAA is always less than P is always less than PooAA..

The vapor pressure of solvent over a solution is The vapor pressure of solvent over a solution is always LOWERED!always LOWERED!

39

Raoult’s LawRaoult’s LawAssume the solution containing 62.1 g of glycol in 250. g Assume the solution containing 62.1 g of glycol in 250. g

of water is ideal. of water is ideal.

What is the vapor pressure of water over the solution at What is the vapor pressure of water over the solution at 30 30 ooC? C?

(The VP of pure H(The VP of pure H22O is 31.8 mm Hg; see App.)O is 31.8 mm Hg; see App.)

SolutionSolutionXglycol = 0.0672 and so Xwater = ?

Because Xglycol + Xwater = 1

Xwater = 1.000 - 0.0672 = 0.9328

Pwater = Xwater • Powater = (0.9382)(31.8 mm Hg)

PPwaterwater = 29.7 mm Hg = 29.7 mm Hg

40

Raoult’s LawRaoult’s LawOr (see next slide):Or (see next slide):

ΔΔPPAA = VP lowering = X = VP lowering = XBBPPooAA

VP lowering is proportional to mole fraction of VP lowering is proportional to mole fraction of the the solutesolute!!

For very dilute solutions, For very dilute solutions,

ΔΔPPAA = K•molality = K•molalityBB

where K is a proportionality constant.where K is a proportionality constant.This helps explain changes in melting and This helps explain changes in melting and

boiling points.boiling points.

4242

( ) (1 )

1

so 1

solv solv solv

solv sol solv solv sol solv

solv solute

solv solute

solv solute solv

P P P

P P P P

P P

See Exercise 14.6, p. 575See Exercise 14.6, p. 575

Changes in Freezing and Changes in Freezing and Boiling Points of SolventBoiling Points of SolventChanges in Freezing and Changes in Freezing and Boiling Points of SolventBoiling Points of Solvent

See Figure 14.13See Figure 14.13

VP solventafter addingsolute

VP Pure solvent

BP puresolvent

BP solution

1 atm

P

T

43

4444

Boiling Point ElevationBoiling Point Elevation

• If a solute is added to the pure If a solute is added to the pure solvent at its normal boiling point, solvent at its normal boiling point, the equilibrium vapor pressure will the equilibrium vapor pressure will decrease and the liquid will no longer decrease and the liquid will no longer boil. boil.

• To reach the new boiling point the To reach the new boiling point the temperature must be increased, thus temperature must be increased, thus boiling point elevation. boiling point elevation.

ΔΔttbpbp = K = Kbpbp m msolutesolute

4545

Figure 14.13

The boiling point of a The boiling point of a solution is higher than that solution is higher than that

of the pure solvent.of the pure solvent.

46

4747

Elevation of Boiling Point Elevation of Boiling Point Elevation of Boiling Point Elevation of Boiling Point

Elevation in BP = Elevation in BP = ttBPBP = K = KBP BP • m• m(where K(where KBPBP is characteristic of solvent) is characteristic of solvent)

VP solventafter addingsolute

VP Pure solvent

BP puresolvent

BP solution

1 atm

P

T

47

Change in Boiling PointChange in Boiling Point

Dissolve 62.1 g of glycol (1.00 mol) in 250. g of Dissolve 62.1 g of glycol (1.00 mol) in 250. g of water. What is the BP of the solution?water. What is the BP of the solution?

KKBPBP = +0.512 = +0.512 ooC/molal for water C/molal for water (Table 14.4).(Table 14.4).

SolutionSolution

1.1. Calculate solution molality = 4.00 mCalculate solution molality = 4.00 m

2.2. DtDtBPBP = K = KBPBP • m • m

DtDtBPBP = +0.512 = +0.512 ooC/molal (4.00 molal)C/molal (4.00 molal)

DtDtBPBP = +2.05 = +2.05 ooCC

BP = 102.05 BP = 102.05 ooCC

48

mol solute 14

kg solv .250molal

4949

Freezing Point DepressionFreezing Point Depression

• The freezing point is lowered by the The freezing point is lowered by the presence of a solute since these presence of a solute since these particles cannot form the solid and particles cannot form the solid and some of them are occupying the low some of them are occupying the low energy slots needed to form the solid energy slots needed to form the solid solvent.solvent.

ΔΔttfpfp = K = Kfpfp m msolutesolute

• Note the K is a negative value.Note the K is a negative value.

Change in Freezing Point Change in Freezing Point

The freezing point of a solution is The freezing point of a solution is LOWER LOWER than that of the pure solvent.than that of the pure solvent.

FP depression = FP depression = ΔΔttFPFP = K = KFPFP•m•m

Pure water Ethylene glycol/water solution

50


Consider equilibrium at melting pointConsider equilibrium at melting point

Liquid solvent <------> Solid solventLiquid solvent <------> Solid solvent

•• Rate at which molecules go from S to L Rate at which molecules go from S to L depends only on the nature of the solid.depends only on the nature of the solid.

•• BUT — rate for L ---> S depends on how BUT — rate for L ---> S depends on how much is dissolved. This rate is SLOWED for much is dissolved. This rate is SLOWED for the same reason VP is lowered.the same reason VP is lowered.

•• Therefore, to bring S ---> L and L ---> S rates Therefore, to bring S ---> L and L ---> S rates into equilibrium for a solution, T must be into equilibrium for a solution, T must be lowered.lowered.

Thus, FP for solution < FP for solventThus, FP for solution < FP for solvent

FP depression = FP depression = ΔΔttFPFP = K = KFPFP•m•m

51

Calculate the FP of a 4.00 molal Calculate the FP of a 4.00 molal glycol/water solution.glycol/water solution.

KKFPFP = -1.86 = -1.86 ooC/molal (Table 14.4, p. 577)C/molal (Table 14.4, p. 577)

SolutionSolution

ΔΔttFPFP = K = KFPFP • m • m

= (-1.86 = (-1.86 ooC/molal)(4.00 m)C/molal)(4.00 m)

ΔΔttFP FP = -7.44 = -7.44 ooCC

Freezing Point DepressionFreezing Point Depression 52

5353

Colligative Properties Of Colligative Properties Of Ionic SolutionsIonic Solutions

• Ionic compounds dissociate completely into Ionic compounds dissociate completely into ions in water. ions in water.

• All calculations involving water and an ionic All calculations involving water and an ionic solute must account for the total number of solute must account for the total number of particles present. particles present.

• This factor is called the van't Hoff factor, i. This factor is called the van't Hoff factor, i.

How much NaCl must be dissolved How much NaCl must be dissolved in 4.00 kg of water to lower FP to in 4.00 kg of water to lower FP to -10.00 -10.00 ooC?.C?.

SolutionSolutionCalculate the required molality.Calculate the required molality.

ΔΔttFPFP = K = KFPFP • m • m

-10.00 -10.00 ooC = (-1.86 C = (-1.86 ooC/molal) • MolalityC/molal) • Molality Concentration = 5.38 molal Concentration = 5.38 molal

Freezing Point DepressionFreezing Point Depression 54

How much NaCl must be dissolved in 4.00 kg of water How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 to lower FP to -10.00 ooC?C?..

SolutionSolutionConcentration required = 5.38 molalConcentration required = 5.38 molal


This means we need 5.38 mol of dissolved This means we need 5.38 mol of dissolved particles per kg of solvent. particles per kg of solvent.

Recognize that m represents the total conc. of all Recognize that m represents the total conc. of all dissolved particles.dissolved particles.

Recall that Recall that

1 mol NaCl(aq) 1 mol NaCl(aq) 1 mol Na 1 mol Na++(aq) + 1 mol Cl(aq) + 1 mol Cl--(aq)(aq)

55

How much NaCl must be dissolved in 4.00 kg of How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 water to lower FP to -10.00 ooC?.C?.

SolutionSolution

Concentration required = 5.38 molalConcentration required = 5.38 molal

We need 5.38 mol of dissolved particles per kg We need 5.38 mol of dissolved particles per kg of solvent. of solvent.

NaCl(aq) --> NaNaCl(aq) --> Na++(aq) + Cl(aq) + Cl--(aq)(aq)


To get 5.38 mol/kg of particles we needTo get 5.38 mol/kg of particles we need

5.38 mol / 2 = 2.69 mol NaCl / kg5.38 mol / 2 = 2.69 mol NaCl / kg

2.69 mol NaCl / kg ---> 157 g NaCl / kg2.69 mol NaCl / kg ---> 157 g NaCl / kg

(157 g NaCl / kg)•(4.00 kg) = (157 g NaCl / kg)•(4.00 kg) = 629 g NaCl629 g NaCl

56

Boiling Point Elevation and Boiling Point Elevation and Freezing Point DepressionFreezing Point Depression

ΔΔt = K • m • it = K • m • iA generally useful equation A generally useful equation i = van’t Hoff factor = number of particles i = van’t Hoff factor = number of particles

produced per formula unit.produced per formula unit.CompoundCompound Theoretical Value of iTheoretical Value of i glycolglycol 11 NaClNaCl 22

CaClCaCl22 33

57

OsmosisOsmosis58

Salt water Pure water

5959

• Osmosis occurs when a molecule moves from a Osmosis occurs when a molecule moves from a region of high concentration to lower region of high concentration to lower concentration through a semipermeable concentration through a semipermeable membrane. membrane.

6060

Osmotic pressure is defined by:

= cRT

where c is the molarity of the solute and R is the gas constant.

.08206 atm×LR

mol×K

6161

OsmosisOsmosisOsmosisOsmosis

The semipermeable membrane should The semipermeable membrane should allow only the movement of solvent allow only the movement of solvent molecules.molecules.

Therefore, solvent molecules move Therefore, solvent molecules move from pure solvent to solution.from pure solvent to solution.

62

Solvent Solution

Semipermeable membrane


The semipermeable membrane The semipermeable membrane should allow only the movement of should allow only the movement of solvent molecules.solvent molecules.

Therefore, solvent molecules move Therefore, solvent molecules move from pure solvent to solution.from pure solvent to solution.

63

Solvent Solution

OsmoticPressure

6464


Equilibrium is reached when pressure Equilibrium is reached when pressure produced by extra solution — produced by extra solution —

the the OSMOTIC PRESSURE, pOSMOTIC PRESSURE, p

p = cRTp = cRT (where c is conc. in mol/L)(where c is conc. in mol/L)

counterbalances pressure of solvent counterbalances pressure of solvent molecules moving thru the membrane.molecules moving thru the membrane.

Solvent Solution

OsmoticPressure

OsmosisOsmosis65

OsmosisOsmosis• Osmosis of solvent Osmosis of solvent

from one solution from one solution to another can to another can continue until the continue until the solutions are solutions are ISOTONICISOTONIC — — they have the same they have the same concentration.concentration.

66

Osmosis Osmosis Calculating a Molar MassCalculating a Molar Mass

Dissolve 35.0 g of hemoglobin in enough water Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L of solution. to make 1.00 L of solution. pp measured to be measured to be 10.0 mm Hg at 25 10.0 mm Hg at 25 °°C. Calculate molar mass C. Calculate molar mass of hemoglobin.of hemoglobin.

SolutionSolution

(a) Calculate (a) Calculate pp in atmospheres in atmospheres

pp = 10.0 mmHg • (1 atm / 760 mmHg) = 10.0 mmHg • (1 atm / 760 mmHg)

= 0.0132 atm= 0.0132 atm(b)(b) Calculate the concentrationCalculate the concentration

67

Calculating a Molar MassCalculating a Molar Mass

Concentration = 5.39 x 10Concentration = 5.39 x 10-4-4 mol/L mol/L(c)(c) Calculate the molar massCalculate the molar mass Molar mass = 35.0 g / 5.39 x 10Molar mass = 35.0 g / 5.39 x 10-4-4 mol/L mol/L Molar mass = 65,100 g/molMolar mass = 65,100 g/mol

Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L of solution. measured to be 10.0 mm Hg at 25 C. Calculate molar mass of hemoglobin.

SolutionSolution

(b) Calculate concentration from = cRT

Conc = 0.0132 atm

(0.0821 L • atm/K • mol)(298K)

68

6969

Reverse Osmosis

7070

COLLOIDSCOLLOIDS

• Colloids are a suspension of very small Colloids are a suspension of very small particles that do not settle out. particles that do not settle out. – (milk, jello, ..) (milk, jello, ..)

7171

7272

Hydrophilic and hydrophobic colloids exist and emulsions make use of molecules that contain both.

7373

Soap and SurfactantsSoap and Surfactants

H2O

H2O

H2OH2O

H2O

H2O

H2O

H2O H2O

H2O

Dirt

C

O

O-

H

O

H

Nonpolar tail Polar head

7474

Soap and SurfactantsSoap and Surfactants

7575

DetergentDetergent Fabric SoftenerFabric Softener

Documents

1 SOLUTIONS AND THEIR BEHAVIOR CHAPTER 14 CHAPTER 14