Chapter Outline - WOU Homepage · 1 Chapter 11: Properties of Solutions - Their Concentrations and...

5192016

1

Chapter 11 Properties of Solutions -

Their Concentrations and Colligative Properties

111 Energy Changes when Substances Dissolve

112 Vapor Pressure

113 Mixtures of Volatile Substances

114 Colligative Properties of Solutions

115 Osmosis and Osmotic Pressure

116 Using Colligative Properties to Determine Molar Mass

Chapter Outline

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Enthalpy of Solution

Dissolution of Ionic Solids

bull Enthalpy of solution (∆Hsoln) depends on

raquo Energies holding solute ions in crystal lattice

raquo Attractive force holding solvent molecules together

raquo Interactions between solute ions and solvent

molecules

bull ∆Hsoln = ∆Hion-ion + ∆Hdipole-dipole + ∆H ion-dipole

bull When solvent is water

raquo ∆Hsoln = ∆Hion-ion + ∆Hhydration

Enthalpy of Solution for NaCl

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Lattice energy (E) increases as Q

increases andor as r decreases

Lattice energy (U) is the energy required to completely separate one mole of

a solid ionic compound into gaseous ions It is always endothermic

eg MgF2(s) Mg2+(g) + 2F-(g)

Calculating Lattice Energies Using the Born-Haber Cycle

Q1 is the charge on the cation

Q2 is the charge on the anion

d is the distance between the ions

Compound Lattice

Energy (U)

MgF2 2957 Q= +2-1

MgO 3938 Q= +2-2

LiF

LiCl

1036

853

radius F lt

radius Cl

119864119890119897 = 231 119909 10minus19119869 ∙ 11989911989811987611199091198762

119889

5192016

4

Hessrsquo Law is used to calculate Ulattice

The calculation is broken down into a series of steps (cycles)

Steps

1 sublimation of 1 mole of the metal = ΔHsub

2 if present breaking bond of a diatomic = frac12 ΔHBE

3 ionization of the metal(g) atom = IE1 + IE2 etc

4 electron affinity of the nonmetal atom = EA1 + EA2 etc

5 formation of 1 mole of the salt from ions(g) = ΔHlast = -Ulattice

Born-Haber Cycles ndash Calculating Ulattice

M(s) + frac12 X2(g)

M(g) +

M(g) + X(g)

frac12 X2(g)

X(g) M+(g) + X(g) M+(g) +

X-(g) M+(g) +

MX(s)

ΔHsub

frac12 ΔHBE

IE1 + IE2 etc EA1 + EA2 etc

ΔHlast = -Ulattice

ΔHf

In General for M(s) + frac12 X2(g) MX(s)

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ΔHlast = ΔHf minus ΔHsub minus frac12 ΔHBE minus IE minus EA

Rearranging and solving for ΔHlast -

Ulattice = -ΔHlast

ΔHfdeg = ΔHsub + frac12 ΔHBE + IE + EA + ΔHlast

Calculating Ulattice

Na(s) + frac12 Cl2(g)

Na(g) +

Na(g) + Cl(g)

frac12 Cl2(g)

Cl(g) Na+(g) + Cl(g) Na+(g) +

Cl-(g) Na+(g) +

NaCl(s)

ΔHsub

frac12 ΔHBE

IE1 EA1

ΔHlast = -Ulattice

ΔHf

Calculating Ulattice for NaCl(s)

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7

ΔHlast = ΔHf minus ΔHsub minus frac12 ΔHBE minus IE1 minus EA1

Sample Exercise 111

Calculating Lattice Energy

Calcium fluoride occurs in nature as the mineral fluorite which is the

principle source of the worldrsquos supply of fluorine Use the following data to

calculate the lattice energy of CaF2

ΔHsubCa = 168 kJmol

BEF2 = 155 kJmol

EAF = -328 kJmol

IE1Ca = 590 kJmol

IE2Ca = 1145 kJmol

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Lattice Energies and Solubility

119864119890119897 = 231 119909 10minus19119869 ∙ 11989911989811987611199091198762

119889




5192016

9


112 Vapor Pressure





Chapter Outline

Vapor Pressure Pressure exerted by a gas in equilibrium with its liquid

Where are we going with this

Colligative Properties ndash boiling

point elevation freezing point

depression osmosis

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Vapor Pressure of Solutions

bull Vapor Pressure ndash Pressure exerted by a gas

in equilibrium with liquid

ndash Rates of

evaporationcondensation

are equal

The tendency for a liquid to evaporate

increases as -

1 the temperature rises

2 the surface area

increases

3 the intermolecular

forces decrease

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11

Boiling Point

ldquoIf you have a beaker of water open to the

atmosphere the mass of the atmosphere is

pressing down on the surface As heat is

added more and more water evaporates

pushing the molecules of the atmosphere

aside (wsystem lt 0) If enough heat is added

a temperature is eventually reached at which

the vapor pressure of the liquid equals the

atmospheric pressure and the liquid boilsrdquo

the temperature at which the vapor pressure of a liquid is equal to the

external atmospheric pressure of 1 atm

Increasing the external atmospheric pressure increases the boiling point

Decreasing the external atmospheric pressure decreases the boiling point

Normal boiling point

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Location Elevation (ft) bp H2O (oC)

San Francisco sea level 1000

Salt Lake City 4400 956

Denver 5280 950

Mt Everest 29028 765

Diethyl ether - dipole-dipole

interactions

Water - hydrogen

bonding (stronger)

Intermolecular Forces and Vapor Pressure

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Clausius-Clapeyron Equation Vapor Pressure vs Temperature

CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas

Plot of ln(P) vs 1T yields

straight line

bull Slope = minusΔHvapR

bull Intercept = constant


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14

Sample Exercise 112

Calculating Vapor Pressure Using Hvap

The compound with the common name isooctane has the structure

shown below Its normal boiling point is 99 oC and Hvap = 352 kJmol

What is the vapor pressure at 25 oC

Practice Exercise p 473

Pentane (C5H12) gas is used to blow the bubbles in molten polystyrene that turn it into

Styrofoam which is used in coffee cups and other products that have good thermal

insulation properties The normal boiling point of pentane is 36 oC its vapor pressure

at 25 oC is 505 torr What is the enthalpy of vaporization of pentane

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112 Vapor Pressure





Chapter Outline

Fractional Distillation

A method of separating a

mixture of compounds

based on their different

boiling points

Vapor phase enriched in

the more volatile

component

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Solutions of Volatile Compounds

Raoultrsquos Law

bull Total vapor pressure of an ideal solution

depends on how much the partial pressure of

each volatile component contributes to total

vapor pressure of solution

bull Ptotal = 1P1 + 2P2 + 3P3 + hellip

where i = mole fraction of component i and

Pi = equilibrium vapor pressure of pure

volatile component at a given temperature

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17

Sample Exercise 113 (simplified)

Calculating the Vapor Pressure of a Solution

What is the vapor pressure of a solution prepared by dissolving 13 g

of heptane (C7H16) in 87 g of octane (C8H18) at 25 oC

Real vs Ideal Solutions

Deviations from Raoultrsquos Law

Due to differences in solutendashsolvent and

solventndashsolvent interactions

(dashed lines = ideal behavior)

Negative deviations Positive deviations

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18

Concept Test p 478

Which of the following solutions is least likely to follow

Raoultrsquos Law (a) acetoneethanol (b) pentanehexane (c)

pentanolwater


112 Vapor Pressure




116 Using Colligative Properties to Determine Molar

Mass

Chapter Outline

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19

Colligative Properties of Solutions

bull Colligative Properties

ndash Set of properties of a solution relative to the pure

solvent

ndash Due to solutendashsolvent interactions

ndash Depend on concentration of solute particles not

the identity of particles

ndash Include lowering of vapor pressure boiling point

elevation freezing point depression osmosis

and osmotic pressure


bull Raoultrsquos Law

bull Vapor pressure of solution is equal to the vapor

pressure of the pure solvent multiplied by the

mole fraction of solvent in the solution

bull Psolution = solventP solvent

bull Vapor pressure lowering

bull A colligative property of solutions (Section 115)

bull Ideal Solution

bull One that obeys Raoultrsquos law

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20

Sample Exercise 114 Calculating the Vapor

Pressure of a Solution of One or More Nonvolatile

Substances

Most of the worldrsquos supply of maple syrup is produced in quebec Ontario

where the sap from maple trees is evaporated until the concentration of sugar

(mostly sucrose) in the sap reaches at least 66 by weight What is the vapor

pressure in atm of a 66 aqueous solution of sucrose (MW=342) at 100 oC

Assume the solution obeys Raoultsrsquos Law



Consequences

in bp and fp of

solution relative to

pure solvent

Vapor pressure

lowering

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21

Solute Concentration Molality

bull Changes in boiling pointfreezing point

of solutions depends on molality

ndash Preferred concentration unit for properties

involving temperature changes because it

is independent of temperature

ndash For typical solutions molality gt molarity

solute

kg of solvent

nm

Calculating Molality

Starting with

a) Mass of solute

b) Volume of solvent or

c) Volume of solution

5192016

22

Sample Exercise 115 Calculating the

Molality of a Solution

A popular recipe for preparing corned beef calls for preparing a seasoned brine

(salt solution) that contains 050 lbs of kosher salt (NaCl) dissolved in 30 qt of

water Assuming the density of water = 100 gmL what is the molality of NaCl in

this solution for ldquocorningrdquo beef


112 Vapor Pressure





Mass

Chapter Outline

5192016

23

copy 2012 by W W Norton amp Company


Colligative Properties

bull Solution properties that depend on concentration

of solute particles not the identity of particles

Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution


Boiling-Point Elevation

and Freezing-Point Depression

Boiling Point Elevation (ΔTb)

bull ΔTb = Kb∙m

bull Kb = boiling point elevation

constant of solvent m = molality

Freezing Point Depression (ΔTf)

bull ΔTf = Kf∙m

bull Kf = freezing-point depression

constant m = molality

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24

Sample Exercise 116 Calculating the Boiling

Point Elevation of an Aqueous Solution

What is the boiling point of seawater Assume that the

average concentration of sea salt is equivalent to 0560 m

(molal) NaCl and that the average density of seawater is

1025 gmL


Freezing Point Elevation of a Solution

What is the freezing point of an automobile radiator fluid

prepared by mixing equal volumes of ethylene glycol

(MW=6207) and water at a temperature where the density of

ethylene glycol is 1114 gmL and the density of water is 1000

gmL The freezing point depression constant of water kf is

186 oCm

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25


The vanrsquot Hoff Factor

Solutions of Electrolytes

bull Need to correct for number of particles formed when ionic substance dissolves

vanrsquot Hoff Factor (i)

bull number of ions in formula unit

bull eg NaCl i = 2

ΔTb = i∙Kb∙m amp ΔTf = i∙Kf∙m

Deviations from theoretical value due to ion pair formation


Values of vanrsquot Hoff Factors

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26

Sample Exercise 119

Using the vanrsquot Hoff Factor

Some Thanksgiving dinner chefs ldquobrinerdquo their turkeys prior to

roasting them to help the birds retain moisture and to season

the meat Brining involves completely immersing a turkey fro

about 6 hours in a brine (salt solution) prepared by dissolving

20 pounds of salt (NaCl MW=5844) in 20 gallons of water

Suppose a turkey soaking in such a solution is left for 6 hours

on an unheated porch on a day when the air temperature is

18 oF (-8 oC) Are the brine (and turkey) in danger of

freezing The kf of water is 186 oCm assume that i=185 for

this NaCl solution


112 Vapor Pressure




116 Using Colligative Properties to Determine

Molar Mass

52

Chapter Outline

5192016

27

Molar Mass from Colligative Properties

ΔTf = Kfm and ΔTb = Kbm

ΔTf or ΔTb m = ΔTfKf

kg of solvent

moles

MW = g samplemoles

Step 1

solve for molality

Step 2

solve for moles molality

Step 3

calc MW

Sample Exercise 1112 Using Freezing

Point Depression to Determine Molar Mass

Eicosene is a hydrocarbon used as an additive in the high-

pressure fracturing of rock layers to enhance recovery of

petroleum and natural gas a process sometimes called

ldquofrackingrdquo The freezing point of a solution prepared by

dissolving 100 mg of eicosene in 100 g of benzene is 175 oC

lower that the freezing point of pure benzene What is the molar

mass of eicosene (kf for benzene is 490 oCm)

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2

Enthalpy of Solution

Dissolution of Ionic Solids

bull Enthalpy of solution (∆Hsoln) depends on

raquo Energies holding solute ions in crystal lattice

raquo Attractive force holding solvent molecules together

raquo Interactions between solute ions and solvent

molecules

bull ∆Hsoln = ∆Hion-ion + ∆Hdipole-dipole + ∆H ion-dipole

bull When solvent is water

raquo ∆Hsoln = ∆Hion-ion + ∆Hhydration

Enthalpy of Solution for NaCl

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3










Compound Lattice

Energy (U)

MgF2 2957 Q= +2-1

MgO 3938 Q= +2-2

LiF

LiCl

1036

853

radius F lt

radius Cl

119864119890119897 = 231 119909 10minus19119869 ∙ 11989911989811987611199091198762

119889

5192016

4



Steps







M(s) + frac12 X2(g)

M(g) +

M(g) + X(g)

frac12 X2(g)

X(g) M+(g) + X(g) M+(g) +

X-(g) M+(g) +

MX(s)

ΔHsub

frac12 ΔHBE


ΔHlast = -Ulattice

ΔHf


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5



Ulattice = -ΔHlast




Na(g) +

Na(g) + Cl(g)

frac12 Cl2(g)


Cl-(g) Na+(g) +

NaCl(s)

ΔHsub

frac12 ΔHBE

IE1 EA1

ΔHlast = -Ulattice

ΔHf


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6


5192016

7


Sample Exercise 111






BEF2 = 155 kJmol

EAF = -328 kJmol

IE1Ca = 590 kJmol

IE2Ca = 1145 kJmol

5192016

8


119864119890119897 = 231 119909 10minus19119869 ∙ 11989911989811987611199091198762

119889




5192016

9


112 Vapor Pressure





Chapter Outline





depression osmosis

5192016

10




ndash Rates of


are equal


increases as -


2 the surface area

increases


forces decrease

5192016

11

Boiling Point















5192016

12




Denver 5280 950



interactions

Water - hydrogen

bonding (stronger)


5192016

13


CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas


straight line




5192016

14

Sample Exercise 112










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15


112 Vapor Pressure





Chapter Outline





boiling points


the more volatile

component

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16


Raoultrsquos Law









5192016

17











5192016

18

Concept Test p 478



pentanolwater


112 Vapor Pressure





Mass

Chapter Outline

5192016

19




solvent















bull Ideal Solution


5192016

20



Substances








Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

5192016

21








solute

kg of solvent

nm


Starting with

a) Mass of solute



5192016

22








112 Vapor Pressure





Mass

Chapter Outline

5192016

23






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



5192016

24






1025 gmL








186 oCm

5192016

25







bull eg NaCl i = 2





5192016

26

Sample Exercise 119











this NaCl solution


112 Vapor Pressure





Molar Mass

52

Chapter Outline

5192016

27




kg of solvent

moles

MW = g samplemoles

Step 1

solve for molality

Step 2


Step 3

calc MW










5192016

3










Compound Lattice

Energy (U)

MgF2 2957 Q= +2-1

MgO 3938 Q= +2-2

LiF

LiCl

1036

853

radius F lt

radius Cl

119864119890119897 = 231 119909 10minus19119869 ∙ 11989911989811987611199091198762

119889

5192016

4



Steps







M(s) + frac12 X2(g)

M(g) +

M(g) + X(g)

frac12 X2(g)

X(g) M+(g) + X(g) M+(g) +

X-(g) M+(g) +

MX(s)

ΔHsub

frac12 ΔHBE


ΔHlast = -Ulattice

ΔHf


5192016

5



Ulattice = -ΔHlast




Na(g) +

Na(g) + Cl(g)

frac12 Cl2(g)


Cl-(g) Na+(g) +

NaCl(s)

ΔHsub

frac12 ΔHBE

IE1 EA1

ΔHlast = -Ulattice

ΔHf


5192016

6


5192016

7


Sample Exercise 111






BEF2 = 155 kJmol

EAF = -328 kJmol

IE1Ca = 590 kJmol

IE2Ca = 1145 kJmol

5192016

8


119864119890119897 = 231 119909 10minus19119869 ∙ 11989911989811987611199091198762

119889




5192016

9


112 Vapor Pressure





Chapter Outline





depression osmosis

5192016

10




ndash Rates of


are equal


increases as -


2 the surface area

increases


forces decrease

5192016

11

Boiling Point















5192016

12




Denver 5280 950



interactions

Water - hydrogen

bonding (stronger)


5192016

13


CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas


straight line




5192016

14

Sample Exercise 112










5192016

15


112 Vapor Pressure





Chapter Outline





boiling points


the more volatile

component

5192016

16


Raoultrsquos Law









5192016

17











5192016

18

Concept Test p 478



pentanolwater


112 Vapor Pressure





Mass

Chapter Outline

5192016

19




solvent















bull Ideal Solution


5192016

20



Substances








Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

5192016

21








solute

kg of solvent

nm


Starting with

a) Mass of solute



5192016

22








112 Vapor Pressure





Mass

Chapter Outline

5192016

23






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



5192016

24






1025 gmL








186 oCm

5192016

25







bull eg NaCl i = 2





5192016

26

Sample Exercise 119











this NaCl solution


112 Vapor Pressure





Molar Mass

52

Chapter Outline

5192016

27




kg of solvent

moles

MW = g samplemoles

Step 1

solve for molality

Step 2


Step 3

calc MW










5192016

4



Steps







M(s) + frac12 X2(g)

M(g) +

M(g) + X(g)

frac12 X2(g)

X(g) M+(g) + X(g) M+(g) +

X-(g) M+(g) +

MX(s)

ΔHsub

frac12 ΔHBE


ΔHlast = -Ulattice

ΔHf


5192016

5



Ulattice = -ΔHlast




Na(g) +

Na(g) + Cl(g)

frac12 Cl2(g)


Cl-(g) Na+(g) +

NaCl(s)

ΔHsub

frac12 ΔHBE

IE1 EA1

ΔHlast = -Ulattice

ΔHf


5192016

6


5192016

7


Sample Exercise 111






BEF2 = 155 kJmol

EAF = -328 kJmol

IE1Ca = 590 kJmol

IE2Ca = 1145 kJmol

5192016

8


119864119890119897 = 231 119909 10minus19119869 ∙ 11989911989811987611199091198762

119889




5192016

9


112 Vapor Pressure





Chapter Outline





depression osmosis

5192016

10




ndash Rates of


are equal


increases as -


2 the surface area

increases


forces decrease

5192016

11

Boiling Point















5192016

12




Denver 5280 950



interactions

Water - hydrogen

bonding (stronger)


5192016

13


CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas


straight line




5192016

14

Sample Exercise 112










5192016

15


112 Vapor Pressure





Chapter Outline





boiling points


the more volatile

component

5192016

16


Raoultrsquos Law









5192016

17











5192016

18

Concept Test p 478



pentanolwater


112 Vapor Pressure





Mass

Chapter Outline

5192016

19




solvent















bull Ideal Solution


5192016

20



Substances








Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

5192016

21








solute

kg of solvent

nm


Starting with

a) Mass of solute



5192016

22








112 Vapor Pressure





Mass

Chapter Outline

5192016

23






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



5192016

24






1025 gmL








186 oCm

5192016

25







bull eg NaCl i = 2





5192016

26

Sample Exercise 119











this NaCl solution


112 Vapor Pressure





Molar Mass

52

Chapter Outline

5192016

27




kg of solvent

moles

MW = g samplemoles

Step 1

solve for molality

Step 2


Step 3

calc MW










5192016

5



Ulattice = -ΔHlast




Na(g) +

Na(g) + Cl(g)

frac12 Cl2(g)


Cl-(g) Na+(g) +

NaCl(s)

ΔHsub

frac12 ΔHBE

IE1 EA1

ΔHlast = -Ulattice

ΔHf


5192016

6


5192016

7


Sample Exercise 111






BEF2 = 155 kJmol

EAF = -328 kJmol

IE1Ca = 590 kJmol

IE2Ca = 1145 kJmol

5192016

8


119864119890119897 = 231 119909 10minus19119869 ∙ 11989911989811987611199091198762

119889




5192016

9


112 Vapor Pressure





Chapter Outline





depression osmosis

5192016

10




ndash Rates of


are equal


increases as -


2 the surface area

increases


forces decrease

5192016

11

Boiling Point















5192016

12




Denver 5280 950



interactions

Water - hydrogen

bonding (stronger)


5192016

13


CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas


straight line




5192016

14

Sample Exercise 112










5192016

15


112 Vapor Pressure





Chapter Outline





boiling points


the more volatile

component

5192016

16


Raoultrsquos Law









5192016

17











5192016

18

Concept Test p 478



pentanolwater


112 Vapor Pressure





Mass

Chapter Outline

5192016

19




solvent















bull Ideal Solution


5192016

20



Substances








Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

5192016

21








solute

kg of solvent

nm


Starting with

a) Mass of solute



5192016

22








112 Vapor Pressure





Mass

Chapter Outline

5192016

23






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



5192016

24






1025 gmL








186 oCm

5192016

25







bull eg NaCl i = 2





5192016

26

Sample Exercise 119











this NaCl solution


112 Vapor Pressure





Molar Mass

52

Chapter Outline

5192016

27




kg of solvent

moles

MW = g samplemoles

Step 1

solve for molality

Step 2


Step 3

calc MW










5192016

6


5192016

7


Sample Exercise 111






BEF2 = 155 kJmol

EAF = -328 kJmol

IE1Ca = 590 kJmol

IE2Ca = 1145 kJmol

5192016

8


119864119890119897 = 231 119909 10minus19119869 ∙ 11989911989811987611199091198762

119889




5192016

9


112 Vapor Pressure





Chapter Outline





depression osmosis

5192016

10




ndash Rates of


are equal


increases as -


2 the surface area

increases


forces decrease

5192016

11

Boiling Point















5192016

12




Denver 5280 950



interactions

Water - hydrogen

bonding (stronger)


5192016

13


CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas


straight line




5192016

14

Sample Exercise 112










5192016

15


112 Vapor Pressure





Chapter Outline





boiling points


the more volatile

component

5192016

16


Raoultrsquos Law









5192016

17











5192016

18

Concept Test p 478



pentanolwater


112 Vapor Pressure





Mass

Chapter Outline

5192016

19




solvent















bull Ideal Solution


5192016

20



Substances








Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

5192016

21








solute

kg of solvent

nm


Starting with

a) Mass of solute



5192016

22








112 Vapor Pressure





Mass

Chapter Outline

5192016

23






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



5192016

24






1025 gmL








186 oCm

5192016

25







bull eg NaCl i = 2





5192016

26

Sample Exercise 119











this NaCl solution


112 Vapor Pressure





Molar Mass

52

Chapter Outline

5192016

27




kg of solvent

moles

MW = g samplemoles

Step 1

solve for molality

Step 2


Step 3

calc MW










5192016

7


Sample Exercise 111






BEF2 = 155 kJmol

EAF = -328 kJmol

IE1Ca = 590 kJmol

IE2Ca = 1145 kJmol

5192016

8


119864119890119897 = 231 119909 10minus19119869 ∙ 11989911989811987611199091198762

119889




5192016

9


112 Vapor Pressure





Chapter Outline





depression osmosis

5192016

10




ndash Rates of


are equal


increases as -


2 the surface area

increases


forces decrease

5192016

11

Boiling Point















5192016

12




Denver 5280 950



interactions

Water - hydrogen

bonding (stronger)


5192016

13


CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas


straight line




5192016

14

Sample Exercise 112










5192016

15


112 Vapor Pressure





Chapter Outline





boiling points


the more volatile

component

5192016

16


Raoultrsquos Law









5192016

17











5192016

18

Concept Test p 478



pentanolwater


112 Vapor Pressure





Mass

Chapter Outline

5192016

19




solvent















bull Ideal Solution


5192016

20



Substances








Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

5192016

21








solute

kg of solvent

nm


Starting with

a) Mass of solute



5192016

22








112 Vapor Pressure





Mass

Chapter Outline

5192016

23






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



5192016

24






1025 gmL








186 oCm

5192016

25







bull eg NaCl i = 2





5192016

26

Sample Exercise 119











this NaCl solution


112 Vapor Pressure





Molar Mass

52

Chapter Outline

5192016

27




kg of solvent

moles

MW = g samplemoles

Step 1

solve for molality

Step 2


Step 3

calc MW










5192016

8


119864119890119897 = 231 119909 10minus19119869 ∙ 11989911989811987611199091198762

119889




5192016

9


112 Vapor Pressure





Chapter Outline





depression osmosis

5192016

10




ndash Rates of


are equal


increases as -


2 the surface area

increases


forces decrease

5192016

11

Boiling Point















5192016

12




Denver 5280 950



interactions

Water - hydrogen

bonding (stronger)


5192016

13


CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas


straight line




5192016

14

Sample Exercise 112










5192016

15


112 Vapor Pressure





Chapter Outline





boiling points


the more volatile

component

5192016

16


Raoultrsquos Law









5192016

17











5192016

18

Concept Test p 478



pentanolwater


112 Vapor Pressure





Mass

Chapter Outline

5192016

19




solvent















bull Ideal Solution


5192016

20



Substances








Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

5192016

21








solute

kg of solvent

nm


Starting with

a) Mass of solute



5192016

22








112 Vapor Pressure





Mass

Chapter Outline

5192016

23






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



5192016

24






1025 gmL








186 oCm

5192016

25







bull eg NaCl i = 2





5192016

26

Sample Exercise 119











this NaCl solution


112 Vapor Pressure





Molar Mass

52

Chapter Outline

5192016

27




kg of solvent

moles

MW = g samplemoles

Step 1

solve for molality

Step 2


Step 3

calc MW










5192016

9


112 Vapor Pressure





Chapter Outline





depression osmosis

5192016

10




ndash Rates of


are equal


increases as -


2 the surface area

increases


forces decrease

5192016

11

Boiling Point















5192016

12




Denver 5280 950



interactions

Water - hydrogen

bonding (stronger)


5192016

13


CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas


straight line




5192016

14

Sample Exercise 112










5192016

15


112 Vapor Pressure





Chapter Outline





boiling points


the more volatile

component

5192016

16


Raoultrsquos Law









5192016

17











5192016

18

Concept Test p 478



pentanolwater


112 Vapor Pressure





Mass

Chapter Outline

5192016

19




solvent















bull Ideal Solution


5192016

20



Substances








Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

5192016

21








solute

kg of solvent

nm


Starting with

a) Mass of solute



5192016

22








112 Vapor Pressure





Mass

Chapter Outline

5192016

23






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



5192016

24






1025 gmL








186 oCm

5192016

25







bull eg NaCl i = 2





5192016

26

Sample Exercise 119











this NaCl solution


112 Vapor Pressure





Molar Mass

52

Chapter Outline

5192016

27




kg of solvent

moles

MW = g samplemoles

Step 1

solve for molality

Step 2


Step 3

calc MW










5192016

10




ndash Rates of


are equal


increases as -


2 the surface area

increases


forces decrease

5192016

11

Boiling Point















5192016

12




Denver 5280 950



interactions

Water - hydrogen

bonding (stronger)


5192016

13


CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas


straight line




5192016

14

Sample Exercise 112










5192016

15


112 Vapor Pressure





Chapter Outline





boiling points


the more volatile

component

5192016

16


Raoultrsquos Law









5192016

17











5192016

18

Concept Test p 478



pentanolwater


112 Vapor Pressure





Mass

Chapter Outline

5192016

19




solvent















bull Ideal Solution


5192016

20



Substances








Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

5192016

21








solute

kg of solvent

nm


Starting with

a) Mass of solute



5192016

22








112 Vapor Pressure





Mass

Chapter Outline

5192016

23






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



5192016

24






1025 gmL








186 oCm

5192016

25







bull eg NaCl i = 2





5192016

26

Sample Exercise 119











this NaCl solution


112 Vapor Pressure





Molar Mass

52

Chapter Outline

5192016

27




kg of solvent

moles

MW = g samplemoles

Step 1

solve for molality

Step 2


Step 3

calc MW










5192016

11

Boiling Point















5192016

12




Denver 5280 950



interactions

Water - hydrogen

bonding (stronger)


5192016

13


CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas


straight line




5192016

14

Sample Exercise 112










5192016

15


112 Vapor Pressure





Chapter Outline





boiling points


the more volatile

component

5192016

16


Raoultrsquos Law









5192016

17











5192016

18

Concept Test p 478



pentanolwater


112 Vapor Pressure





Mass

Chapter Outline

5192016

19




solvent















bull Ideal Solution


5192016

20



Substances








Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

5192016

21








solute

kg of solvent

nm


Starting with

a) Mass of solute



5192016

22








112 Vapor Pressure





Mass

Chapter Outline

5192016

23






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



5192016

24






1025 gmL








186 oCm

5192016

25







bull eg NaCl i = 2





5192016

26

Sample Exercise 119











this NaCl solution


112 Vapor Pressure





Molar Mass

52

Chapter Outline

5192016

27




kg of solvent

moles

MW = g samplemoles

Step 1

solve for molality

Step 2


Step 3

calc MW










5192016

12




Denver 5280 950



interactions

Water - hydrogen

bonding (stronger)


5192016

13


CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas


straight line




5192016

14

Sample Exercise 112










5192016

15


112 Vapor Pressure





Chapter Outline





boiling points


the more volatile

component

5192016

16


Raoultrsquos Law









5192016

17











5192016

18

Concept Test p 478



pentanolwater


112 Vapor Pressure





Mass

Chapter Outline

5192016

19




solvent















bull Ideal Solution


5192016

20



Substances








Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

5192016

21








solute

kg of solvent

nm


Starting with

a) Mass of solute



5192016

22








112 Vapor Pressure





Mass

Chapter Outline

5192016

23






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



5192016

24






1025 gmL








186 oCm

5192016

25







bull eg NaCl i = 2





5192016

26

Sample Exercise 119











this NaCl solution


112 Vapor Pressure





Molar Mass

52

Chapter Outline

5192016

27




kg of solvent

moles

MW = g samplemoles

Step 1

solve for molality

Step 2


Step 3

calc MW










5192016

13


CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas


straight line




5192016

14

Sample Exercise 112










5192016

15


112 Vapor Pressure





Chapter Outline





boiling points


the more volatile

component

5192016

16


Raoultrsquos Law









5192016

17











5192016

18

Concept Test p 478



pentanolwater


112 Vapor Pressure





Mass

Chapter Outline

5192016

19




solvent















bull Ideal Solution


5192016

20



Substances








Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

5192016

21








solute

kg of solvent

nm


Starting with

a) Mass of solute



5192016

22








112 Vapor Pressure





Mass

Chapter Outline

5192016

23






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



5192016

24






1025 gmL








186 oCm

5192016

25







bull eg NaCl i = 2





5192016

26

Sample Exercise 119











this NaCl solution


112 Vapor Pressure





Molar Mass

52

Chapter Outline

5192016

27




kg of solvent

moles

MW = g samplemoles

Step 1

solve for molality

Step 2


Step 3

calc MW










5192016

14

Sample Exercise 112










5192016

15


112 Vapor Pressure





Chapter Outline





boiling points


the more volatile

component

5192016

16


Raoultrsquos Law









5192016

17











5192016

18

Concept Test p 478



pentanolwater


112 Vapor Pressure





Mass

Chapter Outline

5192016

19




solvent















bull Ideal Solution


5192016

20



Substances








Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

5192016

21








solute

kg of solvent

nm


Starting with

a) Mass of solute



5192016

22








112 Vapor Pressure





Mass

Chapter Outline

5192016

23






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



5192016

24






1025 gmL








186 oCm

5192016

25







bull eg NaCl i = 2





5192016

26

Sample Exercise 119











this NaCl solution


112 Vapor Pressure





Molar Mass

52

Chapter Outline

5192016

27




kg of solvent

moles

MW = g samplemoles

Step 1

solve for molality

Step 2


Step 3

calc MW










5192016

15


112 Vapor Pressure





Chapter Outline





boiling points


the more volatile

component

5192016

16


Raoultrsquos Law









5192016

17











5192016

18

Concept Test p 478



pentanolwater


112 Vapor Pressure





Mass

Chapter Outline

5192016

19




solvent















bull Ideal Solution


5192016

20



Substances








Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

5192016

21








solute

kg of solvent

nm


Starting with

a) Mass of solute



5192016

22








112 Vapor Pressure





Mass

Chapter Outline

5192016

23






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



5192016

24






1025 gmL








186 oCm

5192016

25







bull eg NaCl i = 2





5192016

26

Sample Exercise 119











this NaCl solution


112 Vapor Pressure





Molar Mass

52

Chapter Outline

5192016

27




kg of solvent

moles

MW = g samplemoles

Step 1

solve for molality

Step 2


Step 3

calc MW










5192016

16


Raoultrsquos Law









5192016

17











5192016

18

Concept Test p 478



pentanolwater


112 Vapor Pressure





Mass

Chapter Outline

5192016

19




solvent















bull Ideal Solution


5192016

20



Substances








Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

5192016

21








solute

kg of solvent

nm


Starting with

a) Mass of solute



5192016

22








112 Vapor Pressure





Mass

Chapter Outline

5192016

23






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



5192016

24






1025 gmL








186 oCm

5192016

25







bull eg NaCl i = 2





5192016

26

Sample Exercise 119











this NaCl solution


112 Vapor Pressure





Molar Mass

52

Chapter Outline

5192016

27




kg of solvent

moles

MW = g samplemoles

Step 1

solve for molality

Step 2


Step 3

calc MW










5192016

17











5192016

18

Concept Test p 478



pentanolwater


112 Vapor Pressure





Mass

Chapter Outline

5192016

19




solvent















bull Ideal Solution


5192016

20



Substances








Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

5192016

21








solute

kg of solvent

nm


Starting with

a) Mass of solute



5192016

22








112 Vapor Pressure





Mass

Chapter Outline

5192016

23






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



5192016

24






1025 gmL








186 oCm

5192016

25







bull eg NaCl i = 2





5192016

26

Sample Exercise 119











this NaCl solution


112 Vapor Pressure





Molar Mass

52

Chapter Outline

5192016

27




kg of solvent

moles

MW = g samplemoles

Step 1

solve for molality

Step 2


Step 3

calc MW










5192016

18

Concept Test p 478



pentanolwater


112 Vapor Pressure





Mass

Chapter Outline

5192016

19




solvent















bull Ideal Solution


5192016

20



Substances








Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

5192016

21








solute

kg of solvent

nm


Starting with

a) Mass of solute



5192016

22








112 Vapor Pressure





Mass

Chapter Outline

5192016

23






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



5192016

24






1025 gmL








186 oCm

5192016

25







bull eg NaCl i = 2





5192016

26

Sample Exercise 119











this NaCl solution


112 Vapor Pressure





Molar Mass

52

Chapter Outline

5192016

27




kg of solvent

moles

MW = g samplemoles

Step 1

solve for molality

Step 2


Step 3

calc MW










5192016

19




solvent















bull Ideal Solution


5192016

20



Substances








Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

5192016

21








solute

kg of solvent

nm


Starting with

a) Mass of solute



5192016

22








112 Vapor Pressure





Mass

Chapter Outline

5192016

23






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



5192016

24






1025 gmL








186 oCm

5192016

25







bull eg NaCl i = 2





5192016

26

Sample Exercise 119











this NaCl solution


112 Vapor Pressure





Molar Mass

52

Chapter Outline

5192016

27




kg of solvent

moles

MW = g samplemoles

Step 1

solve for molality

Step 2


Step 3

calc MW










5192016

20



Substances








Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

5192016

21








solute

kg of solvent

nm


Starting with

a) Mass of solute



5192016

22








112 Vapor Pressure





Mass

Chapter Outline

5192016

23






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



5192016

24






1025 gmL








186 oCm

5192016

25







bull eg NaCl i = 2





5192016

26

Sample Exercise 119











this NaCl solution


112 Vapor Pressure





Molar Mass

52

Chapter Outline

5192016

27




kg of solvent

moles

MW = g samplemoles

Step 1

solve for molality

Step 2


Step 3

calc MW










5192016

21








solute

kg of solvent

nm


Starting with

a) Mass of solute



5192016

22








112 Vapor Pressure





Mass

Chapter Outline

5192016

23






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



5192016

24






1025 gmL








186 oCm

5192016

25







bull eg NaCl i = 2





5192016

26

Sample Exercise 119











this NaCl solution


112 Vapor Pressure





Molar Mass

52

Chapter Outline

5192016

27




kg of solvent

moles

MW = g samplemoles

Step 1

solve for molality

Step 2


Step 3

calc MW










5192016

22








112 Vapor Pressure





Mass

Chapter Outline

5192016

23






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



5192016

24






1025 gmL








186 oCm

5192016

25







bull eg NaCl i = 2





5192016

26

Sample Exercise 119











this NaCl solution


112 Vapor Pressure





Molar Mass

52

Chapter Outline

5192016

27




kg of solvent

moles

MW = g samplemoles

Step 1

solve for molality

Step 2


Step 3

calc MW










5192016

23






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



5192016

24






1025 gmL








186 oCm

5192016

25







bull eg NaCl i = 2





5192016

26

Sample Exercise 119











this NaCl solution


112 Vapor Pressure





Molar Mass

52

Chapter Outline

5192016

27




kg of solvent

moles

MW = g samplemoles

Step 1

solve for molality

Step 2


Step 3

calc MW










5192016

24






1025 gmL








186 oCm

5192016

25







bull eg NaCl i = 2





5192016

26

Sample Exercise 119











this NaCl solution


112 Vapor Pressure





Molar Mass

52

Chapter Outline

5192016

27




kg of solvent

moles

MW = g samplemoles

Step 1

solve for molality

Step 2


Step 3

calc MW










5192016

25







bull eg NaCl i = 2





5192016

26

Sample Exercise 119











this NaCl solution


112 Vapor Pressure





Molar Mass

52

Chapter Outline

5192016

27




kg of solvent

moles

MW = g samplemoles

Step 1

solve for molality

Step 2


Step 3

calc MW










5192016

26

Sample Exercise 119











this NaCl solution


112 Vapor Pressure





Molar Mass

52

Chapter Outline

5192016

27




kg of solvent

moles

MW = g samplemoles

Step 1

solve for molality

Step 2


Step 3

calc MW










5192016

27




kg of solvent

moles

MW = g samplemoles

Step 1

solve for molality

Step 2


Step 3

calc MW










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Chapter Outline - WOU Homepage · 1 Chapter 11: Properties of Solutions - Their Concentrations and...

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