chapter 13 lecture notes 1

Chemistry 1312Chemistry 1312

Bruce GnadeMP 3 210MP 3.210

[email protected] Hours MW 9:00 11:00Office Hours – MW 9:00-11:00

Logisticsg

E S h d lExam Schedule• Wed Sept 21 Exam 1 7 to 8:30pm HH 2.402• Wed Oct 12 Exam 2 7 to 8:30pm HH 2.402• Wed Nov 9 Exam 3 7 to 8:30pm HH 2 402• Wed Nov 9 Exam 3 7 to 8:30pm HH 2.402• Wed Nov 30 Exam 4 7 to 8:30pm HH 2.402• Wed Dec 14 Final Exam 7 to 9:45pm (NOTE TIME CHANGE)

Course Evaluation: (i) Quizzes 15%(ii) Midterm Exams (4 x 15%) 60%(iii) Final Exam 25%

LogisticsgHomework assignments (end of chapter problems): • assigned for each chapter from end-of-chapter exercises in Burdge• large number of problems selected to cover majority of important

concepts• these will not be collected or graded• all homework assignments for the next section will be posted the day• all homework assignments for the next section will be posted the day

after the previous exam

2Quizzes (in class): they will be at the beginning of class about 10 15 minutes2Quizzes (in class): they will be at the beginning of class, about 10-15 minutes• one per chapter – date of quiz will be announced one day before• I will drop your 2 lowest quiz scores; the others will be averaged together to

give your quiz average• there will be no makeup quizzes given (you will receive a “zero” for any quiz

you miss)

Peer Instructional Support

It is critical to attend every session—skipping a PLTL session limits the utility of that session for everybody else. We want people who sign up for the program to be fully committed to attending. You are allowed only 2 absences during the whole semester; students in the PLTL

th t i th 2 PLTL i ill t b ll dprogram that miss more than 2 PLTL sessions will not be allowed to drop their 2 lowest quiz grades. Bottom line: only sign up for PLTL if you are committed to attending every session.

To participate in a PLTL group, you will need to apply online. More details of this program will be announced in class. You can learn more about PLTL and the GEMS Center at the following link:

http://www utdallas edu/GEMS/http://www.utdallas.edu/GEMS/

CHEM 1312 Fall 2011 Homework AssignmentsSection 1 [Chemistry (2nd ed): Burdge]Section 1 [Chemistry (2nd ed): Burdge]

•• Chapter 13: •• 1, 3, 4, 7, 10, 12, 16, 18, 22, 24, 26, 32, 36, 37, 40, 43, 46, 50, 52, 56,

58, 60(a,b), 62, 65, 66, 68, 74 – 76, 82, 84, 88, 94, 99, 102, 133• ______________________________________________________

Ch t 14• Chapter 14: •• 1, 2, 5, 8–10, 12, 13, 15, 16, 18, 20, 23, 25, 27, 28, 30, 32, 35, 36, 42,

44, 47, 51, 52, 54, 56, 58, 59, 61, 66, 72, 84, 85, 88, 115, 122, 12344, 47, 51, 52, 54, 56, 58, 59, 61, 66, 72, 84, 85, 88, 115, 122, 123

Physical Properties of Physical Properties of Solutions Solutions –– Next 4 classesNext 4 classesChapter 13Chapter 13

13Chapter 13Chapter 13

13.1 Types of Solutions13.2 A Molecular View of the Solution Process

The Importance of Intermolecular ForcesThe Importance of Intermolecular ForcesEnergy and Entropy in Solution Formation

13.3 Concentration UnitsMolalityPercent by MassComparison of Concentration Units

13.4 Factors that Affect SolubilityTemperatureTemperaturePressure

13.5 Colligative PropertiesVapor-Pressure Lowering

ili i l iBoiling-Point ElevationFreezing-Point DepressionOsmotic PressureElectrolyte Solutionsect o yte So ut o s

13.6 Calculations Using Colligative Properties13.7 Colloids

Types of SolutionsTypes of Solutions13.1

A solution is a homogeneous mixture of two or more substances.

A solution consists of a solvent and one or more solutesA solution consists of a solvent and one or more solutes.

Types of SolutionsTypes of Solutions

Solutions can be classified by the amount of solute dissolved.

An unsaturated solution is one that contains less solute than the solvent has the capacity to dissolve at a specific temperature.


Solutions can be classified by the amount of solute dissolved.

A saturated solution is one that contains the maximum amount of solute that will dissolve in a solvent at a specific temperature.


Supersaturated solutions are generally unstable.

A Molecular View of the Solution ProcessA Molecular View of the Solution Process13.2

Solvation occurs when solute molecules are separated from one another and surrounded by solvent molecules.y

Solvation depends on three types of interactions:

1) Solute-solute interactions

2) S l l i i2) Solvent-solvent interactions

3) Solute-solvent interactions)

The Importance of Intermolecular ForcesThe Importance of Intermolecular Forces

“Like dissolves like”

Two substances with similar type and magnitude of intermolecularTwo substances with similar type and magnitude of intermolecular forces are likely to be soluble in each other.

Toluene, C7H8 Octane, C8H187 8 8 18

Both non-polar liquids,solution forms when mixed

Two liquids are said to be miscible if they are completely soluble in each other in all proportions.




Water, H2O Octane, C8H18

Polar and non-polar liquids,solution does not form when mixed

2 8 18

solution does not form when mixed




Water, H2O

Ethanol, C2H6O

2

Both polar liquids, solution forms when mixed

Worked Example 13.1

Determine for each solute whether the solubility will be greater in water, which is polar, or in benzene (C6H6), which is nonpolar: (a) Br2, (b) sodium iodide (NaI), ( ) b t t hl id d (d) f ld h d (CH O)

Strategy Consider the structure of each solute to determine whether or not it is l F l l l t t t ith L i t t d l th VSEPR

(c) carbon tetrachloride, and (d) formaldehyde (CH2O).

polar. For molecular solutes, start with a Lewis structure and apply the VSEPR theory. We expect polar solutes, including ionic compounds, to be more soluble in water. Nonpolar solutes will be more soluble in benzene.

Solution (a) Bromine is a homonuclear diatomic molecule and is nonpolar. Bromine is more soluble in benzene.

(b) Sodium iodide is ionic and more soluble in water.

Worked Example 13.1 (cont.)

Solution (c) Carbon tetrachloride has the following Lewis structure:

With four electron domains around the central atom, we expect a tetrahedral arrangement. A symmetrical arrangement of identical bonds results in a nonpolar molecule. Carbon tetrachloride is more soluble in benzene.

Think About It Remember that molecular formula alone is not sufficient to determine the shape or polarity of a polyatomic molecule It must be determined by starting with a correct Lewis

(d) Formaldehyde has the following Lewis structure:

molecule. It must be determined by starting with a correct Lewis structure and applying VSEPR theory.

Crossed arrows represent individual bond dipoles This molecule is polar andCrossed arrows represent individual bond dipoles. This molecule is polar and can form hydrogen bonds in water. Formaldehyde is more soluble in water.

Concentration UnitsConcentration Units13.3

The amount of solute relative to the volume of a solution or to the amount of solvent in a solution is called concentration.

Molarity:

moles of solutemolarity = = liters of solution

M

Mole fraction:f

Amoles of Amole fraction of component A = =

sum of moles of all components

Concentration UnitsConcentration Units

Molality (m) is the number of moles of solute dissolved in 1 kg (1000 g) solvent:( g)

moles of solutemolality = =

mass of solvent in kgm

mass of solvent in kg

Percent by Mass:

f l tmass of solutepercent by mass = 100 %mass of solute + mass of solvent

Worked Example 13.2

A solution is made by dissolving 170.1 g of glucose (C6H12O6) in enough water to make a liter of solution. The density of the solution is 1.062 g/mL. Express the

t ti i ( ) l lit (b) t b d ( ) t illi

Strategy Use the molar mass of glucose to determine the number of moles of glucose in a liter of solution. Use the density (in g/L) to calculate the mass of a

concentration in (a) molality, (b) percent by mass, and (c) parts per million.

glucose in a liter of solution. Use the density (in g/L) to calculate the mass of a liter of solution. Subtract the mass of glucose from the mass of solution to determine the mass of water. The molar mass of glucose is 180.2 g/mol.

170 1 gSolution (a)

1 liter of solution ×

170.1 g180.2 g/mol = 0.9440 mol glucose per liter of solution

1062 gL = 1062 g

1062 g – 170.1 g = 892 g water = 0.892 kg water

L

0 9440 mol glucose0.9440 mol glucose0.892 kg water = 1.06 m


Solution(b) 170.1 g

1062 g solution × 100% = 16.02% glucose by mass

(c)

1062 g solution

170.1 g× 1 000 000 1 602×105 ppm gl cose( )

1062 g solution × 1,000,000 = 1.602×105 ppm glucose

Think About It Pay careful attention to units in problems such as this. Most require conversions between grams and kilograms and/or liters and milliliters.

Worked Example 13.3

“Rubbing alcohol” is a mixture of isopropyl alcohol (C3H7OH) and water that is 70 percent isopropyl alcohol by mass (density = 0.79 g/mL at 20°C). Express th t ti f bbi l h l i ( ) l it d (b) l lit

Strategy (a) Use density to determine the total mass of a liter of solution, and use percent by mass to determine the mass of isopropyl alcohol in a liter of

the concentration of rubbing alcohol in (a) molarity and (b) molality.

use percent by mass to determine the mass of isopropyl alcohol in a liter of solution. Convert the mass of isopropyl alcohol to moles, and divide moles by liters of solution to get molarity.

(b) Subtract the mass of C3H7OH from the mass of solution to get the mass of water. Divide moles of C3H7OH by the mass of water (in kg) to get molality.

The mass of a liter of rubbing alcohol is 790 g, and the molar mass of isopropyl alcohol is 60.09 g/mol.


Solution(a) 790 g solution

L solution ×70 g C3H7OH100 g solution = 553 g C3H7OH

L solutionL solution 100 g solution L solution

553 g C3H7OHL solution ×

1 mol60.09 g C3H7OH = 9.20 mol C3H7OH

L solution = 9.2 M

(b) 790 g solution – 553 g C3H7OH = 237 g water = 0.237 kg water

9 20 mol C H OH

R bbi l h l i 9 2 M d 39 i i l l h l

9.20 mol C3H7OH0.237 kg water = 39 m

Rubbing alcohol is 9.2 M and 39 m in isopropyl alcohol.

Think About It Note the large difference between molarity and molality in this case Molarity and molality are the same (or similar) only for very dilutecase. Molarity and molality are the same (or similar) only for very dilute aqueous solutions.

Factors That Affect SolubilityFactors That Affect Solubility13.4

Temperature affects the solubility of most substances.

Factors That Affect SolubilityFactors That Affect Solubility

Pressure greatly influences the solubility of a gas.

Henry’s law states that the solubility of a gas in a liquid is proportional to the pressure of the gas over the solution.

c = kP

c molar concentration (mol/L)( )

P pressure (atm)

k proportionalityk proportionality constant called Henry’s law constant

Worked Example 13.4

Calculate the concentration of carbon dioxide in a soft drink that was bottled under a partial pressure of 5.0 atm CO2 at 25°C (a) before the bottle is opened and (b) after the soda has gone “flat” at 25°C . The Henry’s law constant for CO2

in water at this temperature is 3.1×10-2 mol/L·atm. Assume that the partial pressure of CO2 in air is 0.0003 atm and that Henry’s law constant for the soft

Strategy Use c = kP and the given Henry’s law constant to solve for the molar concentration (mol/L) of CO2 at 25°C and the two pressures given

drink is the same as that for water.

concentration (mol/L) of CO2 at 25 C and the two pressures given.

Solution (a) c = (3.1×10-2 mol/L·atm)(5.0 atm) = 1.6×10-1 mol/L

(b) c = (3.1×10-2 mol/L·atm)(0.0003 atm) = 9×10-6 mol/L

Think About It With a pressure approximately 15,000 smaller in part (b) than in part (a) we expect the concentration of CO to be approximately 15 000in part (a), we expect the concentration of CO2 to be approximately 15,000 times smaller–and it is.

Colligative PropertiesColligative Properties13.5

Colligative properties are properties that depend on the number of solute particles in solution.p

Colligative properties do not depend on the nature of the solute particlesparticles.

The colligative properties are: l i vapor-pressure lowering

boiling-point elevationg p

freezing-point depression

osmotic pressure

Colligative PropertiesColligative Properties

Raoult’s law states that the partial pressure of a solvent over a solution is given by the vapor pressure of the pure solvent times the g y p p pmole fraction of the solvent in the solution.

P P P P

P1 partial pressure ofl l i

P P1 1 1 P P 2 1

solvent over solution

P° vapor pressure of pure solventpure solvent

χ1 mole fraction of solvent

ΔP l iΔP vapor pressure lowering

χ2 mole fraction of solute

Worked Example 13.5

Calculate the vapor pressure of water over a solution made by dissolving 225 g of glucose in 575 g of water at 35°C. (At 35°C, P° = 42.2 mmHg.)H2O

Strategy Convert the masses of glucose and water to moles, determine the mole fraction of water, and use P1 = χ1P°1 to find the vapor pressure over the solution. The molar masses of glucose and water are 180.2 and 18.02 g/mol, respectively.The molar masses of glucose and water are 180.2 and 18.02 g/mol, respectively.

Solution

225 g glucose 575 g waterThink About It This problem can also be solved using Equation 13 5 to calculate the vapor pressure lowering ΔP225 g glucose

180.2 g/mol = 1.25 mol glucose575 g water18.02 g/mol = 31.9 mol waterand

31.9 mol water 0 962χ =

13.5 to calculate the vapor-pressure lowering, ΔP.

P = χwaterP° = 0.962 × 42.4 mmHg = 40.6 mmHg

1.25 mol glucose + 31.9 mol water = 0.962

H2OH2O

χwater =

The vapor pressure of water over the solution is 40.6 mmHg.


If both components of a solution are volatile, the vapor pressure of the solution is the sum of the individual partial pressures.

oP P oP PoA A AP P o

B B BP P

o oT A A B BP P P

Benzene Toluene


Benzene Toluene

o oT A A B BP P P

An ideal solution obeys Raoult’sAn ideal solution obeys Raoult s law.


Solutions boil at a higher temperature than the pure p psolvent.

ob b bT T T b b b

T K m

ΔTb boiling point elevation

b bT K m

b g p

Kb boiling point elevation constant (°C/m)

m molality


Solutions freeze at a lower temperature than the pure p psolvent.

of f fT T T f f f

f fT Km

ΔTf freezing point depressionp

Kf freezing point depression constant (°C/m)

m molality

Worked Example 13.6

Ethylene glycol [CH2(OH)CH2(OH)] is a common automobile antifreeze. It is water soluble and fairly nonvolatile (b.p. 197°C). Calculate (a) the freezing

i t d (b) th b ili i t f l ti t i i 685 f th l l l i

Strategy Convert grams of ethylene glycol to moles, and divide by the mass of

point and (b) the boiling point of a solution containing 685 g of ethylene glycol in 2075 g of water.

gy g y g y , ywater in kilograms to get molal concentration. Use molal concentrations and ΔTb= Kbm and ΔTf = Kfm, respectively. The molar mass of ethylene glycol (C2H6O2) is 62.07 g/mol. Kf and Kb for water are 1.86°C/m and 0.52°C/m, respectively.

Think About It Because it both lowers the freezing point and raises the boiling point, antifreeze is useful at both temperature

Solution685 g C2H6O262.07 g/mol = 11.04 mol C2H6O2

11.04 mol C2H6O22.075 kg water = 5.32 m C2H6O2and

extremes.

(a) ΔTf = Kfm = (1.86°C/m)(5.32 m) = 9.89°CThe freezing point of the solution is (0 – 9.89)°C = – 9.89°C

g g

(b) ΔTb = Kbm = (0.52°C/m)(5.32 m) = 2.8°CThe boiling point of the solution is (100.0 + 2.8)°C = 102.8°C


Osmosis is the selective passage of solvent molecules through a porous membrane from amolecules through a porous membrane from a more dilute solution to a more concentrated one.


Osmotic pressure () of a solution is the pressure required to stop osmosis.

MRT

Osmotic pressure (atm)

M molarity (moles/L)

R gas constant (0.08206 L·atm/mol·K)

T absolute temperature (kelvins)


Electrolytes undergo dissociation when dissolved in water.

The van’t Hoff factor (i) accounts for this effect.

actual number of particles in solution after dissociationi

pnumber of formulas units initially dissolved in solution

i

f fT iKm

b bT iK m

iMRT


The van’t Hoff factor (i) is 1 for all nonelectrolytes:

1 particle dissolved i = 1

C12H22O11(s) C12H22O11(aq)H2O

For strong electrolytes i should be equal to the number of ions:

1 particle dissolved, i 1

NaCl(s) Na+(aq) + Cl–(aq)H2O

Na2SO4(s) 2Na+(aq) + SO42–(aq)H2O

2 particles dissolved, i = 2

Na2SO4(s) 2Na (aq) + SO4 (aq)

3 particles dissolved, i = 3

ColligativeColligative PropertiesProperties

The van’t Hoff factor (i) is usually smaller than predicted due to the formation of ion pairs.p

An ion pair is made up of one or more cations and one or more anions held together by electrostatic forces.

ion pairion pair


The van’t Hoff factor (i) is usually smaller than predicted due to the formation of ion pairs.p

An ion pair is made up of one or more cations and one or more anions held together by electrostatic forces.


Concentration has an effect on experimentally measured van’t Hoff factors (i).f ( )

Worked Example 13.7

The osmotic pressure of a 0.0100 M potassium iodide (KI) solution at 25°C is 0.465 atm. Determine the experiment van’t Hoff factor for KI at this

t ti

Strategy Use osmotic pressure to calculate the molar concentration of KI, and divide by the nominal concentration of 0.01000 M. R = 0.08206 L·atm/K·mol,

concentration.

divide by the nominal concentration of 0.01000 M. R 0.08206 L atm/K mol, and T = 298 K.

Solution Solving π = MRT for M,Think About It The calculated van’t Hoff factor for KI is 2. The experimental van’t Hoff factor must be less than or equal to the g

M = =πRT = 0.0190 M0.465 atm

(0.08206 L·atm/K·mol)(298 K)

calculated value.

i =

Th i t l ’t H ff f t f KI t thi t ti i 1 90

0.0190 M0.0100 M = 1.90

The experimental van’t Hoff factor for KI at this concentration is 1.90.

Worked Example 13.8

Quinine was the first drug widely used to treat malaria, and it remains the treatment of choice for severe cases. A solution prepared by dissolving 10.0 g of

i i i 50 0 L f th l h f i i t 1 55°C b l th t fquinine in 50.0 mL of ethanol has a freezing point 1.55 C below that of pure ethanol. Determine the molar mass of quinine. (The density of ethanol is 0.789 g/mL.) Assume that quinine is a nonelectrolyte.

Strategy Use ΔTf = Kfm to determine the molal concentration of the solution. Use the density of ethanol to determine the mass of the solvent. The molal concentration of quinine multiplied by the mass of ethanol (in kg) gives moles of q p y ( g) gquinine. The mass of quinine (in grams) divided by moles of quinine gives the molar mass. Kf for ethanol is 1.99°C/m.

S l ti f h l 50 0 L 0 789 / L 39 5 3 95 10 2 kSolution mass of ethanol = 50.0 mL × 0.789 g/mL = 39.5 g or 3.95×10-2 kg

Solving ΔTf = Kfm for molal concentration

m = =ΔTfKf

= 0.779 m1.55°C1.99°C/

m


Solution The solution is 0.779 m in quinine (i.e., 0.779 mol of quinine/kg ethanol solvent.)

(3.95×10-2 kg ethanol)0.779 mol quininekg ethanol

10 0 i i

= 0.00308 mol quinine

molar mass of quinine = = 325 g/mol10.0 g quinine0.00308 mol quinine

Think About It Check the result using the molecular formula of quinine: bout t C ec t e esu t us g t e o ecu a o u a o qu e:C20H24N2O2 (324.4 g/mol). Multistep problems such as this one require careful tracking of units at each step.

Worked Example 13.9

A solution is prepared by dissolving 50.0 g of hemoglobin (Hb) in enough water to make 1.00 L of solution. The osmotic pressure of the solution is measured and f d t b 14 3 H t 25°C C l l t th l f h l bifound to be 14.3 mmHg at 25 C. Calculate the molar mass of hemoglobin. (Assume that there is no change in volume when the hemoglobin is added to water.)

Strategy Use π = MRT to calculate the molarity of the solution. Because the solution volume is 1 L, the molarity is equal to the number of moles of hemoglobin. Dividing the given mass of hemoglobin by the number of moles g g g g ygives the molar mass. R = 0.08206 L·atm/K·mol, T = 298 K, and π = 14.3 mmHg/(760 mmHg/atm) = 1.88×10-2 atm.


Solution Rearranging π = MRT to solve for molarity we get,

M = =π= 7 69×10-4 M1.88×10-2 atm

( / l)( )M

Thus, the solution contains 7.69×10-4 moles of hemoglobin.

RT 7.69×10 M(0.08206 L·atm/K·mol)(298 K)

50 0molar mass of hemoglobin = 50.0 g7.69×10-4 mol = 6.50×104 g/mol

Think About It Biological molecules can have very high molar massesThink About It Biological molecules can have very high molar masses.

Calculations Using Colligative PropertiesCalculations Using Colligative Properties13.6

Percent dissociation is the percentage of dissolved molecules (or formula units, in the case of an ionic compound) that separate into , p ) pions in a solution.

Strong electrolytes should have complete or 100% dissociationStrong electrolytes should have complete, or 100%, dissociation, however, experimentally determined van’t Hoff factors indicate that this is not the case.

Percent dissociation of a strong electrolyte is more complete at lower concentration.

Percent ionization of weak electrolytes is also dependent on concentrationconcentration.

Worked Example 13.10

A solution that is 0.100 M in hydrofluoric acid (HF) has an osmotic pressure of 2.64 atm at 25°C. Calculate the percent ionization of HF at this concentration.

Strategy Use the osmotic pressure and π = MRT to determine the molar concentration of the particles in solution. Compare the concentration of particles to the nominal concentration (0.100 M) to determine what percentage of the ( ) p goriginal HF molecules are ionized. R = 0.08206 L·atm/K·mol, and T = 298 K.

Solution Rearranging π = MRT to solve for molarity,

M = =πRT = 0.108 M2.64 atm

(0.08206 L·atm/K·mol)(298 K)


Solution The concentration of dissolved particles is 0.108 M. Consider the ionization of HF:

HF(aq) H+(aq) + F-(aq)HF(aq) H (aq) + F (aq)

According to this equation, if x HF molecules ionize, we get x H+ ions and x F-

ions. Thus, the total concentration of particles in solution will be the original , p gconcentration of HF minus x, which gives the concentration of intact HF molecules, plus 2x, which is the concentration of ions (H+ and F-):

Think About It For weak acids, the lower the concentration, the greater the percent ionization. A 0.010 M solution of HF has an osmotic pressure of 0.30 atm, corresponding to 23 percent

(0.100 – x) + 2x = 0.100 + x

Therefore, 0.108 = 0.100 + x and x = 0.008. Because we earlier defined x as the t f HF i i d th t i i ti i i b

ionization. A 0.0010 M solution of HF has an osmotic pressure of 3.8×10-2 atm, corresponding to 56 percent ionization.

amount of HF ionized, the percent ionization is given by

percent ionization = 0.008 M0.100 M ×100% =8%

At this concentration HF is 8 percent ionized.

ColloidsColloids13.7

A colloid is a dispersion of particles of one substance throughout another substance. Colloid particles are much larger than the normal p gsolute molecules.

Categories of colloids:Categories of colloids: aerosols

foams foams

emulsions

sols

gels

ColloidsColloids

Examples of colloids

ColloidsColloids

Colloids with water as the dispersing medium can be categorized as hydrophilic (water loving) or hydrophobic (water fearing). y p ( g) y p ( f g)

Hydrophilic groups on thesurface of a large moleculesurface of a large moleculestabilize the molecule inwater.

ColloidsColloids

Colloids with water as the dispersing medium can be categorized as hydrophilic (water loving) or hydrophobic (water fearing). y p ( g) y p ( f g)

Negative ions are adsorbed onto the surface of hydrophobic colloids.

The repulsion between like charges prevents aggregation of the articles.

ColloidsColloids

Hydrophobic colloids can be stabilized by the presence of hydrophilic groups on their surface.y p g p

ColloidsColloids

Emulsification is the process of stabilizing a colloid that would otherwise not stay dispersed.y p

Key ConceptsKey Concepts13

Types of SolutionsA Molecular View of the Solution Process

The Importance of Intermolecular ForcesThe Importance of Intermolecular ForcesEnergy and Entropy in Solution Formation

Concentration UnitsMolalityPercent by MassComparison of Concentration Units

Factors that Affect SolubilityTemperatureTemperaturePressure

Colligative PropertiesVapor-Pressure Lowering

ili i l iBoiling-Point ElevationFreezing-Point DepressionOsmotic PressureElectrolyte Solutionsect o yte So ut o s

Calculations Using Colligative PropertiesColloids

chapter 13 lecture notes 1

Documents

pan ion pair

henrys law

high molar

affect solubility

solution processthe

sol vent

ion pairs

tf kfm