chapter outline - wou homepage · 1 chapter 11: properties of ... calculating lattice energies...

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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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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

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

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4

Hessrsquo Law is used to calculate Ulattice

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

Steps

1 Sublimate the metal = ΔHsub

2 if necessary breaking bond of a diatomic = ΔHBE

3 ionization of the metal = IE1 + IE2 etc

4 electron affinity of the nonmetal = EA1 + EA2 etc

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

Born-Haber Cycles ndash Calculating Ulattice

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5

M(s) + n X2(g)

M(g) +

M(g) + X(g)

n X2(g)

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

X-(g) M+(g) +

MX(s)

ΔHsub

n ΔHBE

IE1 + IE2 etc EA1 + EA2 etc

ΔHlast = -Ulattice

ΔHf

In General for M(s) + n X2(g) MX2n(s)

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

Rearranging and solving for ΔHlast -

Ulattice = -ΔHlast

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

Calculating Ulattice

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6

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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8


112 Vapor Pressure





Chapter Outline

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9

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

Vapor Pressure of Solutions

bull Vapor Pressure ndash Pressure exerted by a gas

in equilibrium with liquid

ndash Rates of

evaporationcondensation

are equal

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10

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

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11

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

Location Elevation (ft) bp H2O (oC)

San Francisco sea level 1000

Salt Lake City 4400 956

Denver 5280 950

Mt Everest 29028 765

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12

Diethyl ether - dipole-dipole

interactions

Water - hydrogen

bonding (stronger)

Intermolecular Forces and Vapor Pressure

Clausius-Clapeyron Equation Vapor Pressure vs Temperature

CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas

5312018

13

Plot of ln(P) vs 1T yields

straight line

bull Slope = minusΔHvapR

bull Intercept = constant


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

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14

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


112 Vapor Pressure





Chapter Outline

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15

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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16

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

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

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17

Real vs Ideal Solutions

Deviations from Raoultrsquos Law

Due to differences in solutendashsolvent and

solventndashsolvent interactions

(dashed lines = ideal behavior)

Negative deviations Positive deviations

Concept Test p 478

Which of the following solutions is least likely to follow

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

pentanolwater

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18


112 Vapor Pressure




116 Using Colligative Properties to Determine Molar

Mass

Chapter Outline

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

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19


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

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

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20



Consequences

in bp and fp of

solution relative to

pure solvent

Vapor pressure

lowering

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

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21

Calculating Molality

Starting with

a) Mass of solute

b) Volume of solvent or

c) Volume of solution

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

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22


112 Vapor Pressure





Mass

Chapter Outline

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

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23


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


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 x 100 g

mL x

1 kg1000 g

=100 kg H2O

mol EG (solute) = 1000 mL x 1114 g x

ml1 mol

6207 g=1795 mol EG

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

m = 1795 mol

100 kg = 1795 m

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

5312018

24

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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25

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 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 103 mL x 100 gmL

119909 1 kg

1000 g= 757 kg H2O

mol NaCl (solute) = 910 g x 1 mol

5844 g= 1552 mol NaCl

m = 1552 mol

757 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


112 Vapor Pressure




116 Using Colligative Properties to Determine

Molar Mass

Chapter Outline

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26

Molar Mass from Colligative Properties

ΔTf = Kfm and ΔTb = Kbm

ΔTf or ΔTb

moles

MW = g samplemoles

Step 1

solve for molality m = ΔTfKf

Step 2

solve for moles molality

kg of solvent

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)

Step 1

solve for molality =

175 oC

490 oCm = 0357 m

Step 2

solve for moles

moles solute = 0357 mol

kg benzene X 000100 kg benzene

= 357 x 10-4 mol eicosene

Step 3

calc MW MW =

100 x 10minus3

g eicosene

357 x 10minus4

mol = 280 gmol

m = ∆T

Kf

5312018

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

5312018

4



Steps







5312018

5

M(s) + n X2(g)

M(g) +

M(g) + X(g)

n X2(g)

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

X-(g) M+(g) +

MX(s)

ΔHsub

n ΔHBE


ΔHlast = -Ulattice

ΔHf




Ulattice = -ΔHlast



5312018

6


Na(g) +

Na(g) + Cl(g)

frac12 Cl2(g)


Cl-(g) Na+(g) +

NaCl(s)

ΔHsub

frac12 ΔHBE

IE1 EA1

ΔHlast = -Ulattice

ΔHf


5312018

7


Sample Exercise 111






BEF2 = 155 kJmol

EAF = -328 kJmol

IE1Ca = 590 kJmol

IE2Ca = 1145 kJmol

5312018

8


112 Vapor Pressure





Chapter Outline

5312018

9





depression osmosis




ndash Rates of


are equal

5312018

10


increases as -


2 the surface area

increases


forces decrease

Boiling Point










5312018

11









Denver 5280 950


5312018

12


interactions

Water - hydrogen

bonding (stronger)



CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas

5312018

13


straight line




Sample Exercise 112





5312018

14







112 Vapor Pressure





Chapter Outline

5312018

15





boiling points


the more volatile

component

5312018

16


Raoultrsquos Law













5312018

17







Concept Test p 478



pentanolwater

5312018

18


112 Vapor Pressure





Mass

Chapter Outline




solvent







5312018

19









bull Ideal Solution




Substances






5312018

20



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering








solute

kg of solvent

nm

5312018

21


Starting with

a) Mass of solute









5312018

22


112 Vapor Pressure





Mass

Chapter Outline






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution

5312018

23





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m











mL x

1 kg1000 g

=100 kg H2O


ml1 mol

6207 g=1795 mol EG

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

m = 1795 mol

100 kg = 1795 m


5312018

24






bull eg NaCl i = 2





5312018

25

Sample Exercise 119







this NaCl solution

Tf = i Kf m


119909 1 kg

1000 g= 757 kg H2O


5844 g= 1552 mol NaCl

m = 1552 mol

757 kg = 205 m

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



112 Vapor Pressure





Molar Mass

Chapter Outline

5312018

26



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW








Step 1


175 oC

490 oCm = 0357 m

Step 2

solve for moles




Step 3

calc MW MW =

100 x 10minus3

g eicosene

357 x 10minus4

mol = 280 gmol

m = ∆T

Kf

5312018

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

5312018

4



Steps







5312018

5

M(s) + n X2(g)

M(g) +

M(g) + X(g)

n X2(g)

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

X-(g) M+(g) +

MX(s)

ΔHsub

n ΔHBE


ΔHlast = -Ulattice

ΔHf




Ulattice = -ΔHlast



5312018

6


Na(g) +

Na(g) + Cl(g)

frac12 Cl2(g)


Cl-(g) Na+(g) +

NaCl(s)

ΔHsub

frac12 ΔHBE

IE1 EA1

ΔHlast = -Ulattice

ΔHf


5312018

7


Sample Exercise 111






BEF2 = 155 kJmol

EAF = -328 kJmol

IE1Ca = 590 kJmol

IE2Ca = 1145 kJmol

5312018

8


112 Vapor Pressure





Chapter Outline

5312018

9





depression osmosis




ndash Rates of


are equal

5312018

10


increases as -


2 the surface area

increases


forces decrease

Boiling Point










5312018

11









Denver 5280 950


5312018

12


interactions

Water - hydrogen

bonding (stronger)



CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas

5312018

13


straight line




Sample Exercise 112





5312018

14







112 Vapor Pressure





Chapter Outline

5312018

15





boiling points


the more volatile

component

5312018

16


Raoultrsquos Law













5312018

17







Concept Test p 478



pentanolwater

5312018

18


112 Vapor Pressure





Mass

Chapter Outline




solvent







5312018

19









bull Ideal Solution




Substances






5312018

20



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering








solute

kg of solvent

nm

5312018

21


Starting with

a) Mass of solute









5312018

22


112 Vapor Pressure





Mass

Chapter Outline






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution

5312018

23





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m











mL x

1 kg1000 g

=100 kg H2O


ml1 mol

6207 g=1795 mol EG

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

m = 1795 mol

100 kg = 1795 m


5312018

24






bull eg NaCl i = 2





5312018

25

Sample Exercise 119







this NaCl solution

Tf = i Kf m


119909 1 kg

1000 g= 757 kg H2O


5844 g= 1552 mol NaCl

m = 1552 mol

757 kg = 205 m

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



112 Vapor Pressure





Molar Mass

Chapter Outline

5312018

26



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW








Step 1


175 oC

490 oCm = 0357 m

Step 2

solve for moles




Step 3

calc MW MW =

100 x 10minus3

g eicosene

357 x 10minus4

mol = 280 gmol

m = ∆T

Kf

5312018

4



Steps







5312018

5

M(s) + n X2(g)

M(g) +

M(g) + X(g)

n X2(g)

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

X-(g) M+(g) +

MX(s)

ΔHsub

n ΔHBE


ΔHlast = -Ulattice

ΔHf




Ulattice = -ΔHlast



5312018

6


Na(g) +

Na(g) + Cl(g)

frac12 Cl2(g)


Cl-(g) Na+(g) +

NaCl(s)

ΔHsub

frac12 ΔHBE

IE1 EA1

ΔHlast = -Ulattice

ΔHf


5312018

7


Sample Exercise 111






BEF2 = 155 kJmol

EAF = -328 kJmol

IE1Ca = 590 kJmol

IE2Ca = 1145 kJmol

5312018

8


112 Vapor Pressure





Chapter Outline

5312018

9





depression osmosis




ndash Rates of


are equal

5312018

10


increases as -


2 the surface area

increases


forces decrease

Boiling Point










5312018

11









Denver 5280 950


5312018

12


interactions

Water - hydrogen

bonding (stronger)



CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas

5312018

13


straight line




Sample Exercise 112





5312018

14







112 Vapor Pressure





Chapter Outline

5312018

15





boiling points


the more volatile

component

5312018

16


Raoultrsquos Law













5312018

17







Concept Test p 478



pentanolwater

5312018

18


112 Vapor Pressure





Mass

Chapter Outline




solvent







5312018

19









bull Ideal Solution




Substances






5312018

20



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering








solute

kg of solvent

nm

5312018

21


Starting with

a) Mass of solute









5312018

22


112 Vapor Pressure





Mass

Chapter Outline






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution

5312018

23





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m











mL x

1 kg1000 g

=100 kg H2O


ml1 mol

6207 g=1795 mol EG

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

m = 1795 mol

100 kg = 1795 m


5312018

24






bull eg NaCl i = 2





5312018

25

Sample Exercise 119







this NaCl solution

Tf = i Kf m


119909 1 kg

1000 g= 757 kg H2O


5844 g= 1552 mol NaCl

m = 1552 mol

757 kg = 205 m

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



112 Vapor Pressure





Molar Mass

Chapter Outline

5312018

26



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW








Step 1


175 oC

490 oCm = 0357 m

Step 2

solve for moles




Step 3

calc MW MW =

100 x 10minus3

g eicosene

357 x 10minus4

mol = 280 gmol

m = ∆T

Kf

5312018

5

M(s) + n X2(g)

M(g) +

M(g) + X(g)

n X2(g)

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

X-(g) M+(g) +

MX(s)

ΔHsub

n ΔHBE


ΔHlast = -Ulattice

ΔHf




Ulattice = -ΔHlast



5312018

6


Na(g) +

Na(g) + Cl(g)

frac12 Cl2(g)


Cl-(g) Na+(g) +

NaCl(s)

ΔHsub

frac12 ΔHBE

IE1 EA1

ΔHlast = -Ulattice

ΔHf


5312018

7


Sample Exercise 111






BEF2 = 155 kJmol

EAF = -328 kJmol

IE1Ca = 590 kJmol

IE2Ca = 1145 kJmol

5312018

8


112 Vapor Pressure





Chapter Outline

5312018

9





depression osmosis




ndash Rates of


are equal

5312018

10


increases as -


2 the surface area

increases


forces decrease

Boiling Point










5312018

11









Denver 5280 950


5312018

12


interactions

Water - hydrogen

bonding (stronger)



CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas

5312018

13


straight line




Sample Exercise 112





5312018

14







112 Vapor Pressure





Chapter Outline

5312018

15





boiling points


the more volatile

component

5312018

16


Raoultrsquos Law













5312018

17







Concept Test p 478



pentanolwater

5312018

18


112 Vapor Pressure





Mass

Chapter Outline




solvent







5312018

19









bull Ideal Solution




Substances






5312018

20



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering








solute

kg of solvent

nm

5312018

21


Starting with

a) Mass of solute









5312018

22


112 Vapor Pressure





Mass

Chapter Outline






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution

5312018

23





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m











mL x

1 kg1000 g

=100 kg H2O


ml1 mol

6207 g=1795 mol EG

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

m = 1795 mol

100 kg = 1795 m


5312018

24






bull eg NaCl i = 2





5312018

25

Sample Exercise 119







this NaCl solution

Tf = i Kf m


119909 1 kg

1000 g= 757 kg H2O


5844 g= 1552 mol NaCl

m = 1552 mol

757 kg = 205 m

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



112 Vapor Pressure





Molar Mass

Chapter Outline

5312018

26



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW








Step 1


175 oC

490 oCm = 0357 m

Step 2

solve for moles




Step 3

calc MW MW =

100 x 10minus3

g eicosene

357 x 10minus4

mol = 280 gmol

m = ∆T

Kf

5312018

6


Na(g) +

Na(g) + Cl(g)

frac12 Cl2(g)


Cl-(g) Na+(g) +

NaCl(s)

ΔHsub

frac12 ΔHBE

IE1 EA1

ΔHlast = -Ulattice

ΔHf


5312018

7


Sample Exercise 111






BEF2 = 155 kJmol

EAF = -328 kJmol

IE1Ca = 590 kJmol

IE2Ca = 1145 kJmol

5312018

8


112 Vapor Pressure





Chapter Outline

5312018

9





depression osmosis




ndash Rates of


are equal

5312018

10


increases as -


2 the surface area

increases


forces decrease

Boiling Point










5312018

11









Denver 5280 950


5312018

12


interactions

Water - hydrogen

bonding (stronger)



CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas

5312018

13


straight line




Sample Exercise 112





5312018

14







112 Vapor Pressure





Chapter Outline

5312018

15





boiling points


the more volatile

component

5312018

16


Raoultrsquos Law













5312018

17







Concept Test p 478



pentanolwater

5312018

18


112 Vapor Pressure





Mass

Chapter Outline




solvent







5312018

19









bull Ideal Solution




Substances






5312018

20



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering








solute

kg of solvent

nm

5312018

21


Starting with

a) Mass of solute









5312018

22


112 Vapor Pressure





Mass

Chapter Outline






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution

5312018

23





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m











mL x

1 kg1000 g

=100 kg H2O


ml1 mol

6207 g=1795 mol EG

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

m = 1795 mol

100 kg = 1795 m


5312018

24






bull eg NaCl i = 2





5312018

25

Sample Exercise 119







this NaCl solution

Tf = i Kf m


119909 1 kg

1000 g= 757 kg H2O


5844 g= 1552 mol NaCl

m = 1552 mol

757 kg = 205 m

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



112 Vapor Pressure





Molar Mass

Chapter Outline

5312018

26



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW








Step 1


175 oC

490 oCm = 0357 m

Step 2

solve for moles




Step 3

calc MW MW =

100 x 10minus3

g eicosene

357 x 10minus4

mol = 280 gmol

m = ∆T

Kf

5312018

7


Sample Exercise 111






BEF2 = 155 kJmol

EAF = -328 kJmol

IE1Ca = 590 kJmol

IE2Ca = 1145 kJmol

5312018

8


112 Vapor Pressure





Chapter Outline

5312018

9





depression osmosis




ndash Rates of


are equal

5312018

10


increases as -


2 the surface area

increases


forces decrease

Boiling Point










5312018

11









Denver 5280 950


5312018

12


interactions

Water - hydrogen

bonding (stronger)



CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas

5312018

13


straight line




Sample Exercise 112





5312018

14







112 Vapor Pressure





Chapter Outline

5312018

15





boiling points


the more volatile

component

5312018

16


Raoultrsquos Law













5312018

17







Concept Test p 478



pentanolwater

5312018

18


112 Vapor Pressure





Mass

Chapter Outline




solvent







5312018

19









bull Ideal Solution




Substances






5312018

20



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering








solute

kg of solvent

nm

5312018

21


Starting with

a) Mass of solute









5312018

22


112 Vapor Pressure





Mass

Chapter Outline






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution

5312018

23





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m











mL x

1 kg1000 g

=100 kg H2O


ml1 mol

6207 g=1795 mol EG

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

m = 1795 mol

100 kg = 1795 m


5312018

24






bull eg NaCl i = 2





5312018

25

Sample Exercise 119







this NaCl solution

Tf = i Kf m


119909 1 kg

1000 g= 757 kg H2O


5844 g= 1552 mol NaCl

m = 1552 mol

757 kg = 205 m

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



112 Vapor Pressure





Molar Mass

Chapter Outline

5312018

26



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW








Step 1


175 oC

490 oCm = 0357 m

Step 2

solve for moles




Step 3

calc MW MW =

100 x 10minus3

g eicosene

357 x 10minus4

mol = 280 gmol

m = ∆T

Kf

5312018

8


112 Vapor Pressure





Chapter Outline

5312018

9





depression osmosis




ndash Rates of


are equal

5312018

10


increases as -


2 the surface area

increases


forces decrease

Boiling Point










5312018

11









Denver 5280 950


5312018

12


interactions

Water - hydrogen

bonding (stronger)



CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas

5312018

13


straight line




Sample Exercise 112





5312018

14







112 Vapor Pressure





Chapter Outline

5312018

15





boiling points


the more volatile

component

5312018

16


Raoultrsquos Law













5312018

17







Concept Test p 478



pentanolwater

5312018

18


112 Vapor Pressure





Mass

Chapter Outline




solvent







5312018

19









bull Ideal Solution




Substances






5312018

20



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering








solute

kg of solvent

nm

5312018

21


Starting with

a) Mass of solute









5312018

22


112 Vapor Pressure





Mass

Chapter Outline






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution

5312018

23





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m











mL x

1 kg1000 g

=100 kg H2O


ml1 mol

6207 g=1795 mol EG

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

m = 1795 mol

100 kg = 1795 m


5312018

24






bull eg NaCl i = 2





5312018

25

Sample Exercise 119







this NaCl solution

Tf = i Kf m


119909 1 kg

1000 g= 757 kg H2O


5844 g= 1552 mol NaCl

m = 1552 mol

757 kg = 205 m

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



112 Vapor Pressure





Molar Mass

Chapter Outline

5312018

26



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW








Step 1


175 oC

490 oCm = 0357 m

Step 2

solve for moles




Step 3

calc MW MW =

100 x 10minus3

g eicosene

357 x 10minus4

mol = 280 gmol

m = ∆T

Kf

5312018

9





depression osmosis




ndash Rates of


are equal

5312018

10


increases as -


2 the surface area

increases


forces decrease

Boiling Point










5312018

11









Denver 5280 950


5312018

12


interactions

Water - hydrogen

bonding (stronger)



CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas

5312018

13


straight line




Sample Exercise 112





5312018

14







112 Vapor Pressure





Chapter Outline

5312018

15





boiling points


the more volatile

component

5312018

16


Raoultrsquos Law













5312018

17







Concept Test p 478



pentanolwater

5312018

18


112 Vapor Pressure





Mass

Chapter Outline




solvent







5312018

19









bull Ideal Solution




Substances






5312018

20



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering








solute

kg of solvent

nm

5312018

21


Starting with

a) Mass of solute









5312018

22


112 Vapor Pressure





Mass

Chapter Outline






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution

5312018

23





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m











mL x

1 kg1000 g

=100 kg H2O


ml1 mol

6207 g=1795 mol EG

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

m = 1795 mol

100 kg = 1795 m


5312018

24






bull eg NaCl i = 2





5312018

25

Sample Exercise 119







this NaCl solution

Tf = i Kf m


119909 1 kg

1000 g= 757 kg H2O


5844 g= 1552 mol NaCl

m = 1552 mol

757 kg = 205 m

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



112 Vapor Pressure





Molar Mass

Chapter Outline

5312018

26



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW








Step 1


175 oC

490 oCm = 0357 m

Step 2

solve for moles




Step 3

calc MW MW =

100 x 10minus3

g eicosene

357 x 10minus4

mol = 280 gmol

m = ∆T

Kf

5312018

10


increases as -


2 the surface area

increases


forces decrease

Boiling Point










5312018

11









Denver 5280 950


5312018

12


interactions

Water - hydrogen

bonding (stronger)



CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas

5312018

13


straight line




Sample Exercise 112





5312018

14







112 Vapor Pressure





Chapter Outline

5312018

15





boiling points


the more volatile

component

5312018

16


Raoultrsquos Law













5312018

17







Concept Test p 478



pentanolwater

5312018

18


112 Vapor Pressure





Mass

Chapter Outline




solvent







5312018

19









bull Ideal Solution




Substances






5312018

20



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering








solute

kg of solvent

nm

5312018

21


Starting with

a) Mass of solute









5312018

22


112 Vapor Pressure





Mass

Chapter Outline






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution

5312018

23





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m











mL x

1 kg1000 g

=100 kg H2O


ml1 mol

6207 g=1795 mol EG

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

m = 1795 mol

100 kg = 1795 m


5312018

24






bull eg NaCl i = 2





5312018

25

Sample Exercise 119







this NaCl solution

Tf = i Kf m


119909 1 kg

1000 g= 757 kg H2O


5844 g= 1552 mol NaCl

m = 1552 mol

757 kg = 205 m

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



112 Vapor Pressure





Molar Mass

Chapter Outline

5312018

26



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW








Step 1


175 oC

490 oCm = 0357 m

Step 2

solve for moles




Step 3

calc MW MW =

100 x 10minus3

g eicosene

357 x 10minus4

mol = 280 gmol

m = ∆T

Kf

5312018

11









Denver 5280 950


5312018

12


interactions

Water - hydrogen

bonding (stronger)



CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas

5312018

13


straight line




Sample Exercise 112





5312018

14







112 Vapor Pressure





Chapter Outline

5312018

15





boiling points


the more volatile

component

5312018

16


Raoultrsquos Law













5312018

17







Concept Test p 478



pentanolwater

5312018

18


112 Vapor Pressure





Mass

Chapter Outline




solvent







5312018

19









bull Ideal Solution




Substances






5312018

20



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering








solute

kg of solvent

nm

5312018

21


Starting with

a) Mass of solute









5312018

22


112 Vapor Pressure





Mass

Chapter Outline






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution

5312018

23





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m











mL x

1 kg1000 g

=100 kg H2O


ml1 mol

6207 g=1795 mol EG

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

m = 1795 mol

100 kg = 1795 m


5312018

24






bull eg NaCl i = 2





5312018

25

Sample Exercise 119







this NaCl solution

Tf = i Kf m


119909 1 kg

1000 g= 757 kg H2O


5844 g= 1552 mol NaCl

m = 1552 mol

757 kg = 205 m

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



112 Vapor Pressure





Molar Mass

Chapter Outline

5312018

26



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW








Step 1


175 oC

490 oCm = 0357 m

Step 2

solve for moles




Step 3

calc MW MW =

100 x 10minus3

g eicosene

357 x 10minus4

mol = 280 gmol

m = ∆T

Kf

5312018

12


interactions

Water - hydrogen

bonding (stronger)



CTR

HP

1ln

vap

vap

Pre

ssu

re

Temperature

solid

liquid

gas

5312018

13


straight line




Sample Exercise 112





5312018

14







112 Vapor Pressure





Chapter Outline

5312018

15





boiling points


the more volatile

component

5312018

16


Raoultrsquos Law













5312018

17







Concept Test p 478



pentanolwater

5312018

18


112 Vapor Pressure





Mass

Chapter Outline




solvent







5312018

19









bull Ideal Solution




Substances






5312018

20



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering








solute

kg of solvent

nm

5312018

21


Starting with

a) Mass of solute









5312018

22


112 Vapor Pressure





Mass

Chapter Outline






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution

5312018

23





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m











mL x

1 kg1000 g

=100 kg H2O


ml1 mol

6207 g=1795 mol EG

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

m = 1795 mol

100 kg = 1795 m


5312018

24






bull eg NaCl i = 2





5312018

25

Sample Exercise 119







this NaCl solution

Tf = i Kf m


119909 1 kg

1000 g= 757 kg H2O


5844 g= 1552 mol NaCl

m = 1552 mol

757 kg = 205 m

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



112 Vapor Pressure





Molar Mass

Chapter Outline

5312018

26



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW








Step 1


175 oC

490 oCm = 0357 m

Step 2

solve for moles




Step 3

calc MW MW =

100 x 10minus3

g eicosene

357 x 10minus4

mol = 280 gmol

m = ∆T

Kf

5312018

13


straight line




Sample Exercise 112





5312018

14







112 Vapor Pressure





Chapter Outline

5312018

15





boiling points


the more volatile

component

5312018

16


Raoultrsquos Law













5312018

17







Concept Test p 478



pentanolwater

5312018

18


112 Vapor Pressure





Mass

Chapter Outline




solvent







5312018

19









bull Ideal Solution




Substances






5312018

20



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering








solute

kg of solvent

nm

5312018

21


Starting with

a) Mass of solute









5312018

22


112 Vapor Pressure





Mass

Chapter Outline






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution

5312018

23





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m











mL x

1 kg1000 g

=100 kg H2O


ml1 mol

6207 g=1795 mol EG

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

m = 1795 mol

100 kg = 1795 m


5312018

24






bull eg NaCl i = 2





5312018

25

Sample Exercise 119







this NaCl solution

Tf = i Kf m


119909 1 kg

1000 g= 757 kg H2O


5844 g= 1552 mol NaCl

m = 1552 mol

757 kg = 205 m

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



112 Vapor Pressure





Molar Mass

Chapter Outline

5312018

26



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW








Step 1


175 oC

490 oCm = 0357 m

Step 2

solve for moles




Step 3

calc MW MW =

100 x 10minus3

g eicosene

357 x 10minus4

mol = 280 gmol

m = ∆T

Kf

5312018

14







112 Vapor Pressure





Chapter Outline

5312018

15





boiling points


the more volatile

component

5312018

16


Raoultrsquos Law













5312018

17







Concept Test p 478



pentanolwater

5312018

18


112 Vapor Pressure





Mass

Chapter Outline




solvent







5312018

19









bull Ideal Solution




Substances






5312018

20



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering








solute

kg of solvent

nm

5312018

21


Starting with

a) Mass of solute









5312018

22


112 Vapor Pressure





Mass

Chapter Outline






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution

5312018

23





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m











mL x

1 kg1000 g

=100 kg H2O


ml1 mol

6207 g=1795 mol EG

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

m = 1795 mol

100 kg = 1795 m


5312018

24






bull eg NaCl i = 2





5312018

25

Sample Exercise 119







this NaCl solution

Tf = i Kf m


119909 1 kg

1000 g= 757 kg H2O


5844 g= 1552 mol NaCl

m = 1552 mol

757 kg = 205 m

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



112 Vapor Pressure





Molar Mass

Chapter Outline

5312018

26



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW








Step 1


175 oC

490 oCm = 0357 m

Step 2

solve for moles




Step 3

calc MW MW =

100 x 10minus3

g eicosene

357 x 10minus4

mol = 280 gmol

m = ∆T

Kf

5312018

15





boiling points


the more volatile

component

5312018

16


Raoultrsquos Law













5312018

17







Concept Test p 478



pentanolwater

5312018

18


112 Vapor Pressure





Mass

Chapter Outline




solvent







5312018

19









bull Ideal Solution




Substances






5312018

20



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering








solute

kg of solvent

nm

5312018

21


Starting with

a) Mass of solute









5312018

22


112 Vapor Pressure





Mass

Chapter Outline






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution

5312018

23





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m











mL x

1 kg1000 g

=100 kg H2O


ml1 mol

6207 g=1795 mol EG

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

m = 1795 mol

100 kg = 1795 m


5312018

24






bull eg NaCl i = 2





5312018

25

Sample Exercise 119







this NaCl solution

Tf = i Kf m


119909 1 kg

1000 g= 757 kg H2O


5844 g= 1552 mol NaCl

m = 1552 mol

757 kg = 205 m

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



112 Vapor Pressure





Molar Mass

Chapter Outline

5312018

26



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW








Step 1


175 oC

490 oCm = 0357 m

Step 2

solve for moles




Step 3

calc MW MW =

100 x 10minus3

g eicosene

357 x 10minus4

mol = 280 gmol

m = ∆T

Kf

5312018

16


Raoultrsquos Law













5312018

17







Concept Test p 478



pentanolwater

5312018

18


112 Vapor Pressure





Mass

Chapter Outline




solvent







5312018

19









bull Ideal Solution




Substances






5312018

20



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering








solute

kg of solvent

nm

5312018

21


Starting with

a) Mass of solute









5312018

22


112 Vapor Pressure





Mass

Chapter Outline






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution

5312018

23





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m











mL x

1 kg1000 g

=100 kg H2O


ml1 mol

6207 g=1795 mol EG

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

m = 1795 mol

100 kg = 1795 m


5312018

24






bull eg NaCl i = 2





5312018

25

Sample Exercise 119







this NaCl solution

Tf = i Kf m


119909 1 kg

1000 g= 757 kg H2O


5844 g= 1552 mol NaCl

m = 1552 mol

757 kg = 205 m

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



112 Vapor Pressure





Molar Mass

Chapter Outline

5312018

26



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW








Step 1


175 oC

490 oCm = 0357 m

Step 2

solve for moles




Step 3

calc MW MW =

100 x 10minus3

g eicosene

357 x 10minus4

mol = 280 gmol

m = ∆T

Kf

5312018

17







Concept Test p 478



pentanolwater

5312018

18


112 Vapor Pressure





Mass

Chapter Outline




solvent







5312018

19









bull Ideal Solution




Substances






5312018

20



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering








solute

kg of solvent

nm

5312018

21


Starting with

a) Mass of solute









5312018

22


112 Vapor Pressure





Mass

Chapter Outline






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution

5312018

23





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m











mL x

1 kg1000 g

=100 kg H2O


ml1 mol

6207 g=1795 mol EG

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

m = 1795 mol

100 kg = 1795 m


5312018

24






bull eg NaCl i = 2





5312018

25

Sample Exercise 119







this NaCl solution

Tf = i Kf m


119909 1 kg

1000 g= 757 kg H2O


5844 g= 1552 mol NaCl

m = 1552 mol

757 kg = 205 m

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



112 Vapor Pressure





Molar Mass

Chapter Outline

5312018

26



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW








Step 1


175 oC

490 oCm = 0357 m

Step 2

solve for moles




Step 3

calc MW MW =

100 x 10minus3

g eicosene

357 x 10minus4

mol = 280 gmol

m = ∆T

Kf

5312018

18


112 Vapor Pressure





Mass

Chapter Outline




solvent







5312018

19









bull Ideal Solution




Substances






5312018

20



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering








solute

kg of solvent

nm

5312018

21


Starting with

a) Mass of solute









5312018

22


112 Vapor Pressure





Mass

Chapter Outline






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution

5312018

23





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m











mL x

1 kg1000 g

=100 kg H2O


ml1 mol

6207 g=1795 mol EG

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

m = 1795 mol

100 kg = 1795 m


5312018

24






bull eg NaCl i = 2





5312018

25

Sample Exercise 119







this NaCl solution

Tf = i Kf m


119909 1 kg

1000 g= 757 kg H2O


5844 g= 1552 mol NaCl

m = 1552 mol

757 kg = 205 m

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



112 Vapor Pressure





Molar Mass

Chapter Outline

5312018

26



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW








Step 1


175 oC

490 oCm = 0357 m

Step 2

solve for moles




Step 3

calc MW MW =

100 x 10minus3

g eicosene

357 x 10minus4

mol = 280 gmol

m = ∆T

Kf

5312018

19









bull Ideal Solution




Substances






5312018

20



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering








solute

kg of solvent

nm

5312018

21


Starting with

a) Mass of solute









5312018

22


112 Vapor Pressure





Mass

Chapter Outline






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution

5312018

23





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m











mL x

1 kg1000 g

=100 kg H2O


ml1 mol

6207 g=1795 mol EG

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

m = 1795 mol

100 kg = 1795 m


5312018

24






bull eg NaCl i = 2





5312018

25

Sample Exercise 119







this NaCl solution

Tf = i Kf m


119909 1 kg

1000 g= 757 kg H2O


5844 g= 1552 mol NaCl

m = 1552 mol

757 kg = 205 m

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



112 Vapor Pressure





Molar Mass

Chapter Outline

5312018

26



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW








Step 1


175 oC

490 oCm = 0357 m

Step 2

solve for moles




Step 3

calc MW MW =

100 x 10minus3

g eicosene

357 x 10minus4

mol = 280 gmol

m = ∆T

Kf

5312018

20



Consequences

in bp and fp of


pure solvent

Vapor pressure

lowering








solute

kg of solvent

nm

5312018

21


Starting with

a) Mass of solute









5312018

22


112 Vapor Pressure





Mass

Chapter Outline






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution

5312018

23





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m











mL x

1 kg1000 g

=100 kg H2O


ml1 mol

6207 g=1795 mol EG

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

m = 1795 mol

100 kg = 1795 m


5312018

24






bull eg NaCl i = 2





5312018

25

Sample Exercise 119







this NaCl solution

Tf = i Kf m


119909 1 kg

1000 g= 757 kg H2O


5844 g= 1552 mol NaCl

m = 1552 mol

757 kg = 205 m

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



112 Vapor Pressure





Molar Mass

Chapter Outline

5312018

26



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW








Step 1


175 oC

490 oCm = 0357 m

Step 2

solve for moles




Step 3

calc MW MW =

100 x 10minus3

g eicosene

357 x 10minus4

mol = 280 gmol

m = ∆T

Kf

5312018

21


Starting with

a) Mass of solute









5312018

22


112 Vapor Pressure





Mass

Chapter Outline






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution

5312018

23





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m











mL x

1 kg1000 g

=100 kg H2O


ml1 mol

6207 g=1795 mol EG

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

m = 1795 mol

100 kg = 1795 m


5312018

24






bull eg NaCl i = 2





5312018

25

Sample Exercise 119







this NaCl solution

Tf = i Kf m


119909 1 kg

1000 g= 757 kg H2O


5844 g= 1552 mol NaCl

m = 1552 mol

757 kg = 205 m

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



112 Vapor Pressure





Molar Mass

Chapter Outline

5312018

26



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW








Step 1


175 oC

490 oCm = 0357 m

Step 2

solve for moles




Step 3

calc MW MW =

100 x 10minus3

g eicosene

357 x 10minus4

mol = 280 gmol

m = ∆T

Kf

5312018

22


112 Vapor Pressure





Mass

Chapter Outline






Previous example

vapor pressure

lowering

Consequences

change in bp

and fp of solution

5312018

23





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m











mL x

1 kg1000 g

=100 kg H2O


ml1 mol

6207 g=1795 mol EG

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

m = 1795 mol

100 kg = 1795 m


5312018

24






bull eg NaCl i = 2





5312018

25

Sample Exercise 119







this NaCl solution

Tf = i Kf m


119909 1 kg

1000 g= 757 kg H2O


5844 g= 1552 mol NaCl

m = 1552 mol

757 kg = 205 m

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



112 Vapor Pressure





Molar Mass

Chapter Outline

5312018

26



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW








Step 1


175 oC

490 oCm = 0357 m

Step 2

solve for moles




Step 3

calc MW MW =

100 x 10minus3

g eicosene

357 x 10minus4

mol = 280 gmol

m = ∆T

Kf

5312018

23





bull ΔTb = Kb∙m




bull ΔTf = Kf∙m











mL x

1 kg1000 g

=100 kg H2O


ml1 mol

6207 g=1795 mol EG

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

m = 1795 mol

100 kg = 1795 m


5312018

24






bull eg NaCl i = 2





5312018

25

Sample Exercise 119







this NaCl solution

Tf = i Kf m


119909 1 kg

1000 g= 757 kg H2O


5844 g= 1552 mol NaCl

m = 1552 mol

757 kg = 205 m

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



112 Vapor Pressure





Molar Mass

Chapter Outline

5312018

26



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW








Step 1


175 oC

490 oCm = 0357 m

Step 2

solve for moles




Step 3

calc MW MW =

100 x 10minus3

g eicosene

357 x 10minus4

mol = 280 gmol

m = ∆T

Kf

5312018

24






bull eg NaCl i = 2





5312018

25

Sample Exercise 119







this NaCl solution

Tf = i Kf m


119909 1 kg

1000 g= 757 kg H2O


5844 g= 1552 mol NaCl

m = 1552 mol

757 kg = 205 m

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



112 Vapor Pressure





Molar Mass

Chapter Outline

5312018

26



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW








Step 1


175 oC

490 oCm = 0357 m

Step 2

solve for moles




Step 3

calc MW MW =

100 x 10minus3

g eicosene

357 x 10minus4

mol = 280 gmol

m = ∆T

Kf

5312018

25

Sample Exercise 119







this NaCl solution

Tf = i Kf m


119909 1 kg

1000 g= 757 kg H2O


5844 g= 1552 mol NaCl

m = 1552 mol

757 kg = 205 m

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



112 Vapor Pressure





Molar Mass

Chapter Outline

5312018

26



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW








Step 1


175 oC

490 oCm = 0357 m

Step 2

solve for moles




Step 3

calc MW MW =

100 x 10minus3

g eicosene

357 x 10minus4

mol = 280 gmol

m = ∆T

Kf

5312018

26



ΔTf or ΔTb

moles

MW = g samplemoles

Step 1


Step 2


kg of solvent

Step 3

calc MW








Step 1


175 oC

490 oCm = 0357 m

Step 2

solve for moles




Step 3

calc MW MW =

100 x 10minus3

g eicosene

357 x 10minus4

mol = 280 gmol

m = ∆T

Kf

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