chapter outline - wou.edu

342019

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

342019

2

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

CompoundLattice

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 ∙ 11989911989811987611199091198762119889

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 necessary breaking bond of a diatomic = f x ΔHBE

3 ionization of the gas-phase metal = IE1 + IE2 +

4 electron affinity of the nonmetal = EA1 + EA2 +

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

Born-Haber Cycles ndash Calculating Ulattice

342019

3

M(s) + n X2(g)

M(g) +

M(g) + X(g)

f X2(g)

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

X-(g)M+(g) +

MX(s)

ΔHsub

f ΔHBE

IE1 + IE2 etcEA1 + EA2 etc

ΔHlast = -Ulattice

ΔHf

ΔHf = ΔHsub + f x ΔHBE + IE + EA + ΔHlast

Calculating Ulattice for NaCl(s)

342019

4

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

342019

5

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


112 Vapor Pressure





Chapter Outline

342019

6

Vapor PressurePressure 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

Vapor Pressure of Solutions

bull Vapor Pressurendash Pressure exerted by a gas

in equilibrium with liquid

ndash Rates of

evaporationcondensation

are equal

342019

7

The tendency for a liquid to evaporate

increases as -

1 the temperature rises

2 the surface area

increases

3 the intermolecular

forces decrease

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

342019

8

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

Diethyl ether - dipole-dipole

interactions

Water - hydrogen

bonding (stronger)

Intermolecular Forces and Vapor Pressure

342019

9

Clausius-Clapeyron EquationVapor 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


CTR

HP

1ln

vap

vap

342019

10

Clausius-Clapeyron Equation

CTR

HP

1ln

vap

vap

How to use when given (P1 T1) and (P2T2)

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

342019

11


112 Vapor Pressure





Chapter Outline

Mixtures of Volatile Substances

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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12

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


112 Vapor Pressure




116 Using Colligative Properties to Determine Molar

Mass

Chapter Outline

342019

13

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 for Solutions

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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14

Homework Problem 1138

Calculating the Vapor Pressure of a Solution of One

or More Nonvolatile Substances

A solution contains 45 moles of H2O 030 moles of sucrose (C12H22O11) and

020 moles of glucose Sucrose and glucose are nonvolatile What is the mole

fraction of water in the solution What is the vapor pressure of the solution at

35 oC given that the vapor pressure of pure water at 35 oC is 422 Torr

Psolution = solventP solvent

χ H2O =mol H2O

total moles

Psolution = H2OP H2O

Psolution = (090)(422 Torr) = 38 Torr

χ H2O =45 moles

45 + 030 + 020 moles= 090



Consequences

in bp and fp of

solution relative to

pure solvent

Vapor pressure

lowering

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15

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

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16


Calculating the Molality of a Solution

What is the molality of a solution prepared by dissolving 782 g of ethylene glycol

HOCH2CH2OH (MW = 6207) in 150 L of water Assume that the density of the water

is 100 gmL

120002 =mol solute

kg solvent

mol solute = 782 g x1 mol

6207 g= 12599 mol

kg solvent = 150 L x1000 mL

Lx10 g

mLx

10 kg

1000 g= 150 kg

120002 =12599 mol

150 kg= 0840120002


112 Vapor Pressure





Mass

Chapter Outline

342019

17

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

342019

18

Sample Exercise 117 Calculating the

Freezing Point Depression 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 = 186 oCm

Tf = Kf m Assume 100 L of each = 1000 mL

kg H2O (solvent) = 1000 mL x100 g

mLx

1 kg1000 g

=100 kg H2O

mol EG (solute) = 1000 mL x1114 g x

ml1 mol6207 g

=1795 mol EG

Tf = Kf m = (186 oCm)(1795 m) = 334 oC

m = 1795 mol100 kg

= 1795 m

New fp = 00 oC ndash 334 oC = -334 oC

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

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19


Values of vanrsquot Hoff Factors

Sample Exercise 119

Using the vanrsquot Hoff FactorSome 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 for

about 6 hours in a brine (salt solution) prepared by dissolving 910 g of salt (NaCl

MW=5844) in 76 L of water (dH2O = 100 gmL) 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 -8 oC

Are the brine (and turkey) in danger of freezing Kf = 186 oCm assume that i = 185 for

this NaCl solution

Tf = i Kf m

kg H2O (solvent) = 76 x 103mL x100 gmL

1199091 kg

1000 g= 76 kg H2O

mol NaCl (solute) = 910 g x1 mol5844 g

= 156 mol NaCl

m = 156 mol76 kg

= 205 m

Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC

New fp = 00 oC ndash 70 oC = -70 oC BUT the air T is -8 oC so the brine will freeze

342019

20


112 Vapor Pressure




116 Using Colligative Properties to Determine

Molar Mass

Chapter Outline

Molar Mass from Colligative Properties

ΔTf = Kfm and ΔTb = Kbm

ΔTf or ΔTb

moles

MW = g samplemoles

Step 1

solve for molalitym = ΔTfKf

Step 2

solve for molesmolality

kg of solvent

Step 3

calc MW

342019

21

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)

Step 1

solve for molality=

175 oC

490 oCm= 0357 m

Step 2

solve for moles

moles solute = 0357 mol

kg benzeneX 000100 kg benzene

= 357 x 10-4 mol eicosene

Step 3

calc MWMW =

100 x 10minus3g eicosene

357 x 10minus4 mol= 280 gmol

m =∆T

Kf

342019

2

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

CompoundLattice

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 ∙ 11989911989811987611199091198762119889

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 necessary breaking bond of a diatomic = f x ΔHBE

3 ionization of the gas-phase metal = IE1 + IE2 +

4 electron affinity of the nonmetal = EA1 + EA2 +

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

Born-Haber Cycles ndash Calculating Ulattice

342019

3

M(s) + n X2(g)

M(g) +

M(g) + X(g)

f X2(g)

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

X-(g)M+(g) +

MX(s)

ΔHsub

f ΔHBE


ΔHlast = -Ulattice

ΔHf



342019

4


342019

5

Sample Exercise 111






BEF2 = 155 kJmol

EAF = -328 kJmol

IE1Ca = 590 kJmol

IE2Ca = 1145 kJmol


112 Vapor Pressure





Chapter Outline

342019

6





depression osmosis




ndash Rates of


are equal

342019

7


increases as -


2 the surface area

increases


forces decrease

Boiling Point










342019

8







interactions

Water - hydrogen

bonding (stronger)


342019

9


CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas


straight line




CTR

HP

1ln

vap

vap

342019

10


CTR

HP

1ln

vap

vap







342019

11


112 Vapor Pressure





Chapter Outline


Raoultrsquos Law









342019

12






112 Vapor Pressure





Mass

Chapter Outline

342019

13




solvent















bull Ideal Solution


342019

14









χ H2O =mol H2O

total moles



χ H2O =45 moles

45 + 030 + 020 moles= 090



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

342019

15








solute

kg of solvent

nm


Starting with

a) Mass of solute



342019

16





is 100 gmL

120002 =mol solute

kg solvent


6207 g= 12599 mol


Lx10 g

mLx

10 kg

1000 g= 150 kg

120002 =12599 mol

150 kg= 0840120002


112 Vapor Pressure





Mass

Chapter Outline

342019

17






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



342019

18









mLx

1 kg1000 g

=100 kg H2O


ml1 mol6207 g

=1795 mol EG

Tf = Kf m = (186 oCm)(1795 m) = 334 oC

m = 1795 mol100 kg

= 1795 m







bull eg NaCl i = 2



342019

19



Sample Exercise 119







this NaCl solution

Tf = i Kf m


1199091 kg

1000 g= 76 kg H2O


= 156 mol NaCl

m = 156 mol76 kg

= 205 m

Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC


342019

20


112 Vapor Pressure





Molar Mass

Chapter Outline



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW

342019

21








Step 1

solve for molality=

175 oC

490 oCm= 0357 m

Step 2

solve for moles




Step 3

calc MWMW =



m =∆T

Kf

342019

3

M(s) + n X2(g)

M(g) +

M(g) + X(g)

f X2(g)

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

X-(g)M+(g) +

MX(s)

ΔHsub

f ΔHBE


ΔHlast = -Ulattice

ΔHf



342019

4


342019

5

Sample Exercise 111






BEF2 = 155 kJmol

EAF = -328 kJmol

IE1Ca = 590 kJmol

IE2Ca = 1145 kJmol


112 Vapor Pressure





Chapter Outline

342019

6





depression osmosis




ndash Rates of


are equal

342019

7


increases as -


2 the surface area

increases


forces decrease

Boiling Point










342019

8







interactions

Water - hydrogen

bonding (stronger)


342019

9


CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas


straight line




CTR

HP

1ln

vap

vap

342019

10


CTR

HP

1ln

vap

vap







342019

11


112 Vapor Pressure





Chapter Outline


Raoultrsquos Law









342019

12






112 Vapor Pressure





Mass

Chapter Outline

342019

13




solvent















bull Ideal Solution


342019

14









χ H2O =mol H2O

total moles



χ H2O =45 moles

45 + 030 + 020 moles= 090



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

342019

15








solute

kg of solvent

nm


Starting with

a) Mass of solute



342019

16





is 100 gmL

120002 =mol solute

kg solvent


6207 g= 12599 mol


Lx10 g

mLx

10 kg

1000 g= 150 kg

120002 =12599 mol

150 kg= 0840120002


112 Vapor Pressure





Mass

Chapter Outline

342019

17






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



342019

18









mLx

1 kg1000 g

=100 kg H2O


ml1 mol6207 g

=1795 mol EG

Tf = Kf m = (186 oCm)(1795 m) = 334 oC

m = 1795 mol100 kg

= 1795 m







bull eg NaCl i = 2



342019

19



Sample Exercise 119







this NaCl solution

Tf = i Kf m


1199091 kg

1000 g= 76 kg H2O


= 156 mol NaCl

m = 156 mol76 kg

= 205 m

Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC


342019

20


112 Vapor Pressure





Molar Mass

Chapter Outline



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW

342019

21








Step 1

solve for molality=

175 oC

490 oCm= 0357 m

Step 2

solve for moles




Step 3

calc MWMW =



m =∆T

Kf

342019

4


342019

5

Sample Exercise 111






BEF2 = 155 kJmol

EAF = -328 kJmol

IE1Ca = 590 kJmol

IE2Ca = 1145 kJmol


112 Vapor Pressure





Chapter Outline

342019

6





depression osmosis




ndash Rates of


are equal

342019

7


increases as -


2 the surface area

increases


forces decrease

Boiling Point










342019

8







interactions

Water - hydrogen

bonding (stronger)


342019

9


CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas


straight line




CTR

HP

1ln

vap

vap

342019

10


CTR

HP

1ln

vap

vap







342019

11


112 Vapor Pressure





Chapter Outline


Raoultrsquos Law









342019

12






112 Vapor Pressure





Mass

Chapter Outline

342019

13




solvent















bull Ideal Solution


342019

14









χ H2O =mol H2O

total moles



χ H2O =45 moles

45 + 030 + 020 moles= 090



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

342019

15








solute

kg of solvent

nm


Starting with

a) Mass of solute



342019

16





is 100 gmL

120002 =mol solute

kg solvent


6207 g= 12599 mol


Lx10 g

mLx

10 kg

1000 g= 150 kg

120002 =12599 mol

150 kg= 0840120002


112 Vapor Pressure





Mass

Chapter Outline

342019

17






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



342019

18









mLx

1 kg1000 g

=100 kg H2O


ml1 mol6207 g

=1795 mol EG

Tf = Kf m = (186 oCm)(1795 m) = 334 oC

m = 1795 mol100 kg

= 1795 m







bull eg NaCl i = 2



342019

19



Sample Exercise 119







this NaCl solution

Tf = i Kf m


1199091 kg

1000 g= 76 kg H2O


= 156 mol NaCl

m = 156 mol76 kg

= 205 m

Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC


342019

20


112 Vapor Pressure





Molar Mass

Chapter Outline



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW

342019

21








Step 1

solve for molality=

175 oC

490 oCm= 0357 m

Step 2

solve for moles




Step 3

calc MWMW =



m =∆T

Kf

342019

5

Sample Exercise 111






BEF2 = 155 kJmol

EAF = -328 kJmol

IE1Ca = 590 kJmol

IE2Ca = 1145 kJmol


112 Vapor Pressure





Chapter Outline

342019

6





depression osmosis




ndash Rates of


are equal

342019

7


increases as -


2 the surface area

increases


forces decrease

Boiling Point










342019

8







interactions

Water - hydrogen

bonding (stronger)


342019

9


CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas


straight line




CTR

HP

1ln

vap

vap

342019

10


CTR

HP

1ln

vap

vap







342019

11


112 Vapor Pressure





Chapter Outline


Raoultrsquos Law









342019

12






112 Vapor Pressure





Mass

Chapter Outline

342019

13




solvent















bull Ideal Solution


342019

14









χ H2O =mol H2O

total moles



χ H2O =45 moles

45 + 030 + 020 moles= 090



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

342019

15








solute

kg of solvent

nm


Starting with

a) Mass of solute



342019

16





is 100 gmL

120002 =mol solute

kg solvent


6207 g= 12599 mol


Lx10 g

mLx

10 kg

1000 g= 150 kg

120002 =12599 mol

150 kg= 0840120002


112 Vapor Pressure





Mass

Chapter Outline

342019

17






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



342019

18









mLx

1 kg1000 g

=100 kg H2O


ml1 mol6207 g

=1795 mol EG

Tf = Kf m = (186 oCm)(1795 m) = 334 oC

m = 1795 mol100 kg

= 1795 m







bull eg NaCl i = 2



342019

19



Sample Exercise 119







this NaCl solution

Tf = i Kf m


1199091 kg

1000 g= 76 kg H2O


= 156 mol NaCl

m = 156 mol76 kg

= 205 m

Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC


342019

20


112 Vapor Pressure





Molar Mass

Chapter Outline



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW

342019

21








Step 1

solve for molality=

175 oC

490 oCm= 0357 m

Step 2

solve for moles




Step 3

calc MWMW =



m =∆T

Kf

342019

6





depression osmosis




ndash Rates of


are equal

342019

7


increases as -


2 the surface area

increases


forces decrease

Boiling Point










342019

8







interactions

Water - hydrogen

bonding (stronger)


342019

9


CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas


straight line




CTR

HP

1ln

vap

vap

342019

10


CTR

HP

1ln

vap

vap







342019

11


112 Vapor Pressure





Chapter Outline


Raoultrsquos Law









342019

12






112 Vapor Pressure





Mass

Chapter Outline

342019

13




solvent















bull Ideal Solution


342019

14









χ H2O =mol H2O

total moles



χ H2O =45 moles

45 + 030 + 020 moles= 090



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

342019

15








solute

kg of solvent

nm


Starting with

a) Mass of solute



342019

16





is 100 gmL

120002 =mol solute

kg solvent


6207 g= 12599 mol


Lx10 g

mLx

10 kg

1000 g= 150 kg

120002 =12599 mol

150 kg= 0840120002


112 Vapor Pressure





Mass

Chapter Outline

342019

17






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



342019

18









mLx

1 kg1000 g

=100 kg H2O


ml1 mol6207 g

=1795 mol EG

Tf = Kf m = (186 oCm)(1795 m) = 334 oC

m = 1795 mol100 kg

= 1795 m







bull eg NaCl i = 2



342019

19



Sample Exercise 119







this NaCl solution

Tf = i Kf m


1199091 kg

1000 g= 76 kg H2O


= 156 mol NaCl

m = 156 mol76 kg

= 205 m

Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC


342019

20


112 Vapor Pressure





Molar Mass

Chapter Outline



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW

342019

21








Step 1

solve for molality=

175 oC

490 oCm= 0357 m

Step 2

solve for moles




Step 3

calc MWMW =



m =∆T

Kf

342019

7


increases as -


2 the surface area

increases


forces decrease

Boiling Point










342019

8







interactions

Water - hydrogen

bonding (stronger)


342019

9


CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas


straight line




CTR

HP

1ln

vap

vap

342019

10


CTR

HP

1ln

vap

vap







342019

11


112 Vapor Pressure





Chapter Outline


Raoultrsquos Law









342019

12






112 Vapor Pressure





Mass

Chapter Outline

342019

13




solvent















bull Ideal Solution


342019

14









χ H2O =mol H2O

total moles



χ H2O =45 moles

45 + 030 + 020 moles= 090



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

342019

15








solute

kg of solvent

nm


Starting with

a) Mass of solute



342019

16





is 100 gmL

120002 =mol solute

kg solvent


6207 g= 12599 mol


Lx10 g

mLx

10 kg

1000 g= 150 kg

120002 =12599 mol

150 kg= 0840120002


112 Vapor Pressure





Mass

Chapter Outline

342019

17






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



342019

18









mLx

1 kg1000 g

=100 kg H2O


ml1 mol6207 g

=1795 mol EG

Tf = Kf m = (186 oCm)(1795 m) = 334 oC

m = 1795 mol100 kg

= 1795 m







bull eg NaCl i = 2



342019

19



Sample Exercise 119







this NaCl solution

Tf = i Kf m


1199091 kg

1000 g= 76 kg H2O


= 156 mol NaCl

m = 156 mol76 kg

= 205 m

Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC


342019

20


112 Vapor Pressure





Molar Mass

Chapter Outline



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW

342019

21








Step 1

solve for molality=

175 oC

490 oCm= 0357 m

Step 2

solve for moles




Step 3

calc MWMW =



m =∆T

Kf

342019

8







interactions

Water - hydrogen

bonding (stronger)


342019

9


CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas


straight line




CTR

HP

1ln

vap

vap

342019

10


CTR

HP

1ln

vap

vap







342019

11


112 Vapor Pressure





Chapter Outline


Raoultrsquos Law









342019

12






112 Vapor Pressure





Mass

Chapter Outline

342019

13




solvent















bull Ideal Solution


342019

14









χ H2O =mol H2O

total moles



χ H2O =45 moles

45 + 030 + 020 moles= 090



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

342019

15








solute

kg of solvent

nm


Starting with

a) Mass of solute



342019

16





is 100 gmL

120002 =mol solute

kg solvent


6207 g= 12599 mol


Lx10 g

mLx

10 kg

1000 g= 150 kg

120002 =12599 mol

150 kg= 0840120002


112 Vapor Pressure





Mass

Chapter Outline

342019

17






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



342019

18









mLx

1 kg1000 g

=100 kg H2O


ml1 mol6207 g

=1795 mol EG

Tf = Kf m = (186 oCm)(1795 m) = 334 oC

m = 1795 mol100 kg

= 1795 m







bull eg NaCl i = 2



342019

19



Sample Exercise 119







this NaCl solution

Tf = i Kf m


1199091 kg

1000 g= 76 kg H2O


= 156 mol NaCl

m = 156 mol76 kg

= 205 m

Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC


342019

20


112 Vapor Pressure





Molar Mass

Chapter Outline



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW

342019

21








Step 1

solve for molality=

175 oC

490 oCm= 0357 m

Step 2

solve for moles




Step 3

calc MWMW =



m =∆T

Kf

342019

9


CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas


straight line




CTR

HP

1ln

vap

vap

342019

10


CTR

HP

1ln

vap

vap







342019

11


112 Vapor Pressure





Chapter Outline


Raoultrsquos Law









342019

12






112 Vapor Pressure





Mass

Chapter Outline

342019

13




solvent















bull Ideal Solution


342019

14









χ H2O =mol H2O

total moles



χ H2O =45 moles

45 + 030 + 020 moles= 090



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

342019

15








solute

kg of solvent

nm


Starting with

a) Mass of solute



342019

16





is 100 gmL

120002 =mol solute

kg solvent


6207 g= 12599 mol


Lx10 g

mLx

10 kg

1000 g= 150 kg

120002 =12599 mol

150 kg= 0840120002


112 Vapor Pressure





Mass

Chapter Outline

342019

17






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



342019

18









mLx

1 kg1000 g

=100 kg H2O


ml1 mol6207 g

=1795 mol EG

Tf = Kf m = (186 oCm)(1795 m) = 334 oC

m = 1795 mol100 kg

= 1795 m







bull eg NaCl i = 2



342019

19



Sample Exercise 119







this NaCl solution

Tf = i Kf m


1199091 kg

1000 g= 76 kg H2O


= 156 mol NaCl

m = 156 mol76 kg

= 205 m

Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC


342019

20


112 Vapor Pressure





Molar Mass

Chapter Outline



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW

342019

21








Step 1

solve for molality=

175 oC

490 oCm= 0357 m

Step 2

solve for moles




Step 3

calc MWMW =



m =∆T

Kf

342019

10


CTR

HP

1ln

vap

vap







342019

11


112 Vapor Pressure





Chapter Outline


Raoultrsquos Law









342019

12






112 Vapor Pressure





Mass

Chapter Outline

342019

13




solvent















bull Ideal Solution


342019

14









χ H2O =mol H2O

total moles



χ H2O =45 moles

45 + 030 + 020 moles= 090



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

342019

15








solute

kg of solvent

nm


Starting with

a) Mass of solute



342019

16





is 100 gmL

120002 =mol solute

kg solvent


6207 g= 12599 mol


Lx10 g

mLx

10 kg

1000 g= 150 kg

120002 =12599 mol

150 kg= 0840120002


112 Vapor Pressure





Mass

Chapter Outline

342019

17






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



342019

18









mLx

1 kg1000 g

=100 kg H2O


ml1 mol6207 g

=1795 mol EG

Tf = Kf m = (186 oCm)(1795 m) = 334 oC

m = 1795 mol100 kg

= 1795 m







bull eg NaCl i = 2



342019

19



Sample Exercise 119







this NaCl solution

Tf = i Kf m


1199091 kg

1000 g= 76 kg H2O


= 156 mol NaCl

m = 156 mol76 kg

= 205 m

Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC


342019

20


112 Vapor Pressure





Molar Mass

Chapter Outline



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW

342019

21








Step 1

solve for molality=

175 oC

490 oCm= 0357 m

Step 2

solve for moles




Step 3

calc MWMW =



m =∆T

Kf

342019

11


112 Vapor Pressure





Chapter Outline


Raoultrsquos Law









342019

12






112 Vapor Pressure





Mass

Chapter Outline

342019

13




solvent















bull Ideal Solution


342019

14









χ H2O =mol H2O

total moles



χ H2O =45 moles

45 + 030 + 020 moles= 090



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

342019

15








solute

kg of solvent

nm


Starting with

a) Mass of solute



342019

16





is 100 gmL

120002 =mol solute

kg solvent


6207 g= 12599 mol


Lx10 g

mLx

10 kg

1000 g= 150 kg

120002 =12599 mol

150 kg= 0840120002


112 Vapor Pressure





Mass

Chapter Outline

342019

17






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



342019

18









mLx

1 kg1000 g

=100 kg H2O


ml1 mol6207 g

=1795 mol EG

Tf = Kf m = (186 oCm)(1795 m) = 334 oC

m = 1795 mol100 kg

= 1795 m







bull eg NaCl i = 2



342019

19



Sample Exercise 119







this NaCl solution

Tf = i Kf m


1199091 kg

1000 g= 76 kg H2O


= 156 mol NaCl

m = 156 mol76 kg

= 205 m

Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC


342019

20


112 Vapor Pressure





Molar Mass

Chapter Outline



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW

342019

21








Step 1

solve for molality=

175 oC

490 oCm= 0357 m

Step 2

solve for moles




Step 3

calc MWMW =



m =∆T

Kf

342019

12






112 Vapor Pressure





Mass

Chapter Outline

342019

13




solvent















bull Ideal Solution


342019

14









χ H2O =mol H2O

total moles



χ H2O =45 moles

45 + 030 + 020 moles= 090



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

342019

15








solute

kg of solvent

nm


Starting with

a) Mass of solute



342019

16





is 100 gmL

120002 =mol solute

kg solvent


6207 g= 12599 mol


Lx10 g

mLx

10 kg

1000 g= 150 kg

120002 =12599 mol

150 kg= 0840120002


112 Vapor Pressure





Mass

Chapter Outline

342019

17






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



342019

18









mLx

1 kg1000 g

=100 kg H2O


ml1 mol6207 g

=1795 mol EG

Tf = Kf m = (186 oCm)(1795 m) = 334 oC

m = 1795 mol100 kg

= 1795 m







bull eg NaCl i = 2



342019

19



Sample Exercise 119







this NaCl solution

Tf = i Kf m


1199091 kg

1000 g= 76 kg H2O


= 156 mol NaCl

m = 156 mol76 kg

= 205 m

Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC


342019

20


112 Vapor Pressure





Molar Mass

Chapter Outline



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW

342019

21








Step 1

solve for molality=

175 oC

490 oCm= 0357 m

Step 2

solve for moles




Step 3

calc MWMW =



m =∆T

Kf

342019

13




solvent















bull Ideal Solution


342019

14









χ H2O =mol H2O

total moles



χ H2O =45 moles

45 + 030 + 020 moles= 090



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

342019

15








solute

kg of solvent

nm


Starting with

a) Mass of solute



342019

16





is 100 gmL

120002 =mol solute

kg solvent


6207 g= 12599 mol


Lx10 g

mLx

10 kg

1000 g= 150 kg

120002 =12599 mol

150 kg= 0840120002


112 Vapor Pressure





Mass

Chapter Outline

342019

17






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



342019

18









mLx

1 kg1000 g

=100 kg H2O


ml1 mol6207 g

=1795 mol EG

Tf = Kf m = (186 oCm)(1795 m) = 334 oC

m = 1795 mol100 kg

= 1795 m







bull eg NaCl i = 2



342019

19



Sample Exercise 119







this NaCl solution

Tf = i Kf m


1199091 kg

1000 g= 76 kg H2O


= 156 mol NaCl

m = 156 mol76 kg

= 205 m

Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC


342019

20


112 Vapor Pressure





Molar Mass

Chapter Outline



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW

342019

21








Step 1

solve for molality=

175 oC

490 oCm= 0357 m

Step 2

solve for moles




Step 3

calc MWMW =



m =∆T

Kf

342019

14









χ H2O =mol H2O

total moles



χ H2O =45 moles

45 + 030 + 020 moles= 090



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering

342019

15








solute

kg of solvent

nm


Starting with

a) Mass of solute



342019

16





is 100 gmL

120002 =mol solute

kg solvent


6207 g= 12599 mol


Lx10 g

mLx

10 kg

1000 g= 150 kg

120002 =12599 mol

150 kg= 0840120002


112 Vapor Pressure





Mass

Chapter Outline

342019

17






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



342019

18









mLx

1 kg1000 g

=100 kg H2O


ml1 mol6207 g

=1795 mol EG

Tf = Kf m = (186 oCm)(1795 m) = 334 oC

m = 1795 mol100 kg

= 1795 m







bull eg NaCl i = 2



342019

19



Sample Exercise 119







this NaCl solution

Tf = i Kf m


1199091 kg

1000 g= 76 kg H2O


= 156 mol NaCl

m = 156 mol76 kg

= 205 m

Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC


342019

20


112 Vapor Pressure





Molar Mass

Chapter Outline



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW

342019

21








Step 1

solve for molality=

175 oC

490 oCm= 0357 m

Step 2

solve for moles




Step 3

calc MWMW =



m =∆T

Kf

342019

15








solute

kg of solvent

nm


Starting with

a) Mass of solute



342019

16





is 100 gmL

120002 =mol solute

kg solvent


6207 g= 12599 mol


Lx10 g

mLx

10 kg

1000 g= 150 kg

120002 =12599 mol

150 kg= 0840120002


112 Vapor Pressure





Mass

Chapter Outline

342019

17






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



342019

18









mLx

1 kg1000 g

=100 kg H2O


ml1 mol6207 g

=1795 mol EG

Tf = Kf m = (186 oCm)(1795 m) = 334 oC

m = 1795 mol100 kg

= 1795 m







bull eg NaCl i = 2



342019

19



Sample Exercise 119







this NaCl solution

Tf = i Kf m


1199091 kg

1000 g= 76 kg H2O


= 156 mol NaCl

m = 156 mol76 kg

= 205 m

Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC


342019

20


112 Vapor Pressure





Molar Mass

Chapter Outline



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW

342019

21








Step 1

solve for molality=

175 oC

490 oCm= 0357 m

Step 2

solve for moles




Step 3

calc MWMW =



m =∆T

Kf

342019

16





is 100 gmL

120002 =mol solute

kg solvent


6207 g= 12599 mol


Lx10 g

mLx

10 kg

1000 g= 150 kg

120002 =12599 mol

150 kg= 0840120002


112 Vapor Pressure





Mass

Chapter Outline

342019

17






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



342019

18









mLx

1 kg1000 g

=100 kg H2O


ml1 mol6207 g

=1795 mol EG

Tf = Kf m = (186 oCm)(1795 m) = 334 oC

m = 1795 mol100 kg

= 1795 m







bull eg NaCl i = 2



342019

19



Sample Exercise 119







this NaCl solution

Tf = i Kf m


1199091 kg

1000 g= 76 kg H2O


= 156 mol NaCl

m = 156 mol76 kg

= 205 m

Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC


342019

20


112 Vapor Pressure





Molar Mass

Chapter Outline



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW

342019

21








Step 1

solve for molality=

175 oC

490 oCm= 0357 m

Step 2

solve for moles




Step 3

calc MWMW =



m =∆T

Kf

342019

17






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m



342019

18









mLx

1 kg1000 g

=100 kg H2O


ml1 mol6207 g

=1795 mol EG

Tf = Kf m = (186 oCm)(1795 m) = 334 oC

m = 1795 mol100 kg

= 1795 m







bull eg NaCl i = 2



342019

19



Sample Exercise 119







this NaCl solution

Tf = i Kf m


1199091 kg

1000 g= 76 kg H2O


= 156 mol NaCl

m = 156 mol76 kg

= 205 m

Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC


342019

20


112 Vapor Pressure





Molar Mass

Chapter Outline



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW

342019

21








Step 1

solve for molality=

175 oC

490 oCm= 0357 m

Step 2

solve for moles




Step 3

calc MWMW =



m =∆T

Kf

342019

18









mLx

1 kg1000 g

=100 kg H2O


ml1 mol6207 g

=1795 mol EG

Tf = Kf m = (186 oCm)(1795 m) = 334 oC

m = 1795 mol100 kg

= 1795 m







bull eg NaCl i = 2



342019

19



Sample Exercise 119







this NaCl solution

Tf = i Kf m


1199091 kg

1000 g= 76 kg H2O


= 156 mol NaCl

m = 156 mol76 kg

= 205 m

Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC


342019

20


112 Vapor Pressure





Molar Mass

Chapter Outline



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW

342019

21








Step 1

solve for molality=

175 oC

490 oCm= 0357 m

Step 2

solve for moles




Step 3

calc MWMW =



m =∆T

Kf

342019

19



Sample Exercise 119







this NaCl solution

Tf = i Kf m


1199091 kg

1000 g= 76 kg H2O


= 156 mol NaCl

m = 156 mol76 kg

= 205 m

Tf = i Kf m = (185)(186 oCm)(205 m) = 70 oC


342019

20


112 Vapor Pressure





Molar Mass

Chapter Outline



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW

342019

21








Step 1

solve for molality=

175 oC

490 oCm= 0357 m

Step 2

solve for moles




Step 3

calc MWMW =



m =∆T

Kf

342019

20


112 Vapor Pressure





Molar Mass

Chapter Outline



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW

342019

21








Step 1

solve for molality=

175 oC

490 oCm= 0357 m

Step 2

solve for moles




Step 3

calc MWMW =



m =∆T

Kf

342019

21








Step 1

solve for molality=

175 oC

490 oCm= 0357 m

Step 2

solve for moles




Step 3

calc MWMW =



m =∆T

Kf

chapter outline - wou.edu

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