Ch11 z5e solutions

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Ch11 Solutions pp

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11.1 Solution Composition11.1 Solution Composition Solutions occur in all phasesSolutions occur in all phases The The solventsolvent does the dissolving. does the dissolving. The The solutesolute is dissolved. is dissolved. There are examples of all types of There are examples of all types of

solvents dissolving all types of solutes.solvents dissolving all types of solutes. We will focus on We will focus on aqueousaqueous solutions. solutions.

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Ways of MeasuringWays of Measuring

Molarity = Molarity = moles of solutemoles of solute LitersLiters of solut of solutionion

% mass = % mass = Mass of solute Mass of solute x 100% x 100% MassMass of solut of solutionion

Mole fraction of component A =Mole fraction of component A =

AA = = n nAA

nnA A + n+ nBB

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Molality = moles of solute Kilograms of solvent

Molality is abbreviated m. Normality - read for general info.

Ways of Measuring

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Molarity Mass % Mole Fraction Molality Answers are . . .

Given 1.00 g CGiven 1.00 g C22HH55OH & 100. g HOH & 100. g H22OO

and final volume of 101 mLand final volume of 101 mL 0.215 M 0.990% 0.00389 0.217 m

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11.2 The Energies of Solution Formation 11.2 The Energies of Solution Formation pp

Heat of solution ( Heat of solution ( HHsoln soln ) is the energy ) is the energy

change for change for makingmaking a solution. a solution. Most easily understood if broken into Most easily understood if broken into 3 3

energy stepsenergy steps (that must be added). (that must be added). 1.Break apart sol1.Break apart solventvent 2.Break apart sol2.Break apart soluteute 3. 3. MixingMixing solvent and solute solvent and solute

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1. Break apart Solvent 1. Break apart Solvent pp

Have to Have to overcomeovercome attractive forces. So, attractive forces. So, HH11 > 0 (must put in energy). > 0 (must put in energy).

2. Break apart Solute2. Break apart Solute Have to overcome attractive forces. So, Have to overcome attractive forces. So,

HH22 > 0 > 0

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3. Mixing solvent and solute 3. Mixing solvent and solute pp

HH33 dependsdepends on what you are mixing. on what you are mixing. IfIf molecules can molecules can attractattract each other then each other then

HH3 3 is is large and negativelarge and negative..

If molecules If molecules can’tcan’t attract then attract then HH3 3 isis small small

and negativeand negative (negligible interactions). (negligible interactions). This explains the rule “This explains the rule “Like dissolves LikeLike dissolves Like””

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Figure 11.1The Steps in the Dissolving

Process pp

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Energy

Solute

Solution

H1

H2

H3

Solvent

Solute and Solvent

Size of H3 determines whether a solution will form pp

H3

No solution

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Types of Solvent and solutes Types of Solvent and solutes pp

If If HHsolnsoln is small and is small and positivepositive, a solution , a solution

will still form will still form because of entropybecause of entropy.. This is because there are many more This is because there are many more

ways for them to become mixed than ways for them to become mixed than there is for them to stay separate.there is for them to stay separate.

This why NaCl is so soluble in water This why NaCl is so soluble in water even though its even though its HHsolnsoln is + 3 kJ/mol is + 3 kJ/mol

Memorize Table 11.3 p. 492Memorize Table 11.3 p. 492

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Table 11.3 p.521 pp

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Energetics of Solutions & Solubility pp

The lattice energy of KCl is -715 kJ/mol, & The lattice energy of KCl is -715 kJ/mol, & the enthalpy of hydration is -684 kJ/mol. the enthalpy of hydration is -684 kJ/mol. Calculate the enthalpy of solution per mole Calculate the enthalpy of solution per mole of solid KCl. Steps . . .of solid KCl. Steps . . .

Lattice energy was defined in Chapter 8 as Lattice energy was defined in Chapter 8 as MM++

(g)(g) + X + X--(g)(g) MX MX(s)(s) (memorize this). (memorize this).

Use Hess’ law in two steps: 1st, take the Use Hess’ law in two steps: 1st, take the solid and convert it into gaseous ions solid and convert it into gaseous ions (opposite of lattice energy) to get . . . (opposite of lattice energy) to get . . .

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Energetics of Solutions & Solubility Energetics of Solutions & Solubility pp

KClKCl(s)(s) K K1+1+(g)(g) + Cl + Cl1-1-

(g)(g)

whose ∆H is the (-) of ∆Hwhose ∆H is the (-) of ∆Hlattice energylattice energy

= = -(-) 715 kJ/mol-(-) 715 kJ/mol Next, hydrate the ions to get a solution:Next, hydrate the ions to get a solution: KK1+1+

(g)(g) + Cl + Cl1-1-(g)(g) K K1+1+

((aqaq)) + Cl + Cl1-1-((aqaq)), where , where

∆H∆Hhydrationhydration = = -684 kJ/mol-684 kJ/mol

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Energetics of Solutions & Solubility pp

Add the energies to get . . . KCl(s) K1+

(g) + Cl1-(g) = -(-) 715 kJ/mol

K1+(g) + Cl1-

(g) K1+(aq)+ Cl1-

(aq) = -684 kJ/mol

KCl(s) K1+(aq)+ Cl1-

(aq) = 31 kJ/mol

∆Hsolution = 31 kJ/mol (endothermic). You have a HW problem on this (#33). You will have a test question on this!!!

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11.3 Factors Affecting Solubility 11.3 Factors Affecting Solubility StructureStructure and Solubility and Solubility

Water soluble molecules Water soluble molecules mustmust have have dipole moments; dipole moments; i.e.i.e., polar bonds., polar bonds.

To be soluble in To be soluble in nonnonpolar solvents the polar solvents the molecules must be molecules must be nonnonpolar.polar.

Read Vitamins A and C discussion on Read Vitamins A and C discussion on figure 11.4 pg. 493 on your own.figure 11.4 pg. 493 on your own.

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Soap

P O-

CH3

CH2CH2

CH2CH2

CH2

CH2

CH2

O-

O-

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Soap

Hydrophobic nonpolar end

P O-

CH3

CH2CH2

CH2CH2

CH2

CH2

CH2

O-

O-

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Soap

Hydrophilic polar end

P O-

CH3

CH2CH2

CH2CH2

CH2

CH2

CH2

O-

O-

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P O-

CH3

CH2CH2

CH2CH2

CH2

CH2

CH2

O-

O-

_

Skeletal structure of Soap

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A drop of grease in water Grease is nonpolar Water is polar Soap lets you dissolve the nonpolar

in the polar.

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Hydrophobic ends dissolve in

grease

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Hydrophilic ends dissolve in water

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Water molecules can surround and dissolve grease.

Helps get grease out of your way.

Figure 23.29Soap Micelles

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Pressure effectsPressure effects Changing the pressure doesn’t affect Changing the pressure doesn’t affect

much the amount of much the amount of solid or liquidsolid or liquid that that dissolvesdissolves

They are basically incompressible.They are basically incompressible. But, changing pressure But, changing pressure does affect does affect

gasesgases (including those dissolved in (including those dissolved in liquids).liquids).

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Dissolving GasesDissolving Gases

Pressure affects the Pressure affects the amount of gas that can amount of gas that can dissolve in a liquid.dissolve in a liquid.

The The dissolveddissolved gas is at gas is at equilibrium with the gas equilibrium with the gas aboveabove the liquid the liquid..

(Mentos video and (Mentos video and http://eepybird.com/http://eepybird.com/)

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The gas is at equilibrium with the dissolved gas in this solution.

The equilibrium is dynamic.

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If you increase the pressure the gas molecules dissolve faster than they escape.

The equilibrium is disturbed.

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The system reaches a new equilibrium with more gas dissolved.

This is Henry’s Law.

P= kC

Pressure = constant x

Concentration of gas

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Calculations using Henry’s Law Calculations using Henry’s Law pp

A bottle of soft drink at 25A bottle of soft drink at 25oo Celsius Celsius contains COcontains CO22 at a pressure of 5.0 atm at a pressure of 5.0 atm

over the liquid. over the liquid. Assuming that the partial pressure of Assuming that the partial pressure of

COCO22 in the in the atmosphereatmosphere (that is, in an (that is, in an

open system) is 4.0 x 10open system) is 4.0 x 10-4-4 atm, atm, Calculate the equilibrium concentrations Calculate the equilibrium concentrations

of COof CO22 both before and afterboth before and after the bottle is the bottle is

opened. opened. Henry’s Henry’s constantconstant herehere is 32 L atm/mol. . . is 32 L atm/mol. . .

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Henry’s Law is P = kCHenry’s Law is P = kC For this problem it is PFor this problem it is PCO2CO2 = k = kCO2CO2CCCO2CO2

In the In the unopenedunopened bottle, solve for C bottle, solve for CCO2CO2 to to

get . . . get . . .

CCCO2CO2 = P = PCO2CO2/k/kCO2 CO2 = 5.0 atm/32 L atm/mol= 5.0 atm/32 L atm/mol

= 0.16 mol/L.0.16 mol/L.

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In the In the openedopened bottle the gas will reach bottle the gas will reach equilibrium with the atmosphere equilibrium with the atmosphere

So PSo PCO2CO2 = the partial pressure = the partial pressure

= 4.0 x 10= 4.0 x 10-4-4 atm. atm. CCCO2CO2 = P = PCO2CO2/k/kCO2 CO2 = = 4.0 x 104.0 x 10-4-4 atm /32 L atm /32 L

atm/mol = atm/mol = 1.2 x 101.2 x 10-5-5 mol/L mol/L.. This large change in concentration (from This large change in concentration (from

0.16 mol/L to 1.2 x 100.16 mol/L to 1.2 x 10-5-5 mol/L) explains mol/L) explains why soda goes flat.why soda goes flat.

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Temperature Effects

Increased temperature increases the Increased temperature increases the raterate at which a solid dissolves. at which a solid dissolves.

We canWe cannotnot predict whether it will predict whether it will increase the increase the amountamount of solid that of solid that dissolves (only the rate).dissolves (only the rate).

We must read it from a We must read it from a graphgraph of of experimentalexperimental data. data.

35 20 40 60 80 10035

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Gases are predictable

As temperature increases, solubility decreases.

So, gas molecules can move fast enough to escape.

Causes thermal pollution.

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Figure 11.7 The Solubilities of Several Gases in Water

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11.4 The Vapor Pressures of Solutions

A nonvolatile solute lowers the vapor pressure of the solvent.

The molecules of the solventmust overcome the force of both the other solvent molecules and the solute molecules.

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Raoult’s Law:Raoult’s Law: PPsolusolutiontion = P = Poo

solsolvent vent x x solsolventvent

Form of y = mx + b, plot of PForm of y = mx + b, plot of Psolnsoln and and

solsolventvent is straight line with slope P is straight line with slope Poo

Vapor pressure of the solution = Vapor pressure of the solution = mole fraction of solvent x vapor mole fraction of solvent x vapor

pressure of the pressure of the purepure solvent solvent Applies Applies onlyonly to an to an idealideal solution where solution where

the the solutesolute itself doesn’t contribute to itself doesn’t contribute to the vapor pressure (the vapor pressure (i.ei.e., the solute is ., the solute is nonnonvolatile.volatile.

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Aqueous Solution

Pure water

Pure water has a higher vapor pressure than a solution

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Aqueous Solution

Pure water

Water evaporates faster from pure water than from a solution

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The water condenses faster in the solution so it should all end up there.

Aqueous Solution

Pure water

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What is the percent composition of a What is the percent composition of a pentane-hexane pentane-hexane solutionsolution (not the gas) (not the gas) that has a vapor pressure of 350 torr at that has a vapor pressure of 350 torr at 25ºC ?25ºC ?

The vapor pressures at 25ºC areThe vapor pressures at 25ºC are

• pentane 511 torrpentane 511 torr

• hexane 150 torr.hexane 150 torr. Try first, then see hints next slide:Try first, then see hints next slide:

Review Question Review Question pp

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% Comp. Of pentane-hexane % Comp. Of pentane-hexane solutionsolution with with VVpp

solnsoln = 350 and V = 350 and Vpppenpen = 511, V = 511, Vpp

hexhex 150? 150? All are molecules, so no colligative All are molecules, so no colligative

property problem.property problem. Use Use can do w/ P since proportional to can do w/ P since proportional to

moles)moles) Answers are . . .Answers are . . . Hexane - 22.7%Hexane - 22.7% (150/(511+150) x 100%) (150/(511+150) x 100%) Pentane - 77.3%Pentane - 77.3% (100 - 22.7%) (100 - 22.7%)

Review Question Review Question pp

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What is the composition (hexane & What is the composition (hexane & pentane pressures) in torr of the pentane pressures) in torr of the vaporvapor??

Remember, % Comp. of pentane-Remember, % Comp. of pentane-hexane is hexane 22.7%, pentane hexane is hexane 22.7%, pentane 77.3% and V77.3% and Vpp

solutionsolution is 350 torr. Use mole is 350 torr. Use mole fraction.fraction.

Answers are . . .Answers are . . . Hexane 79 torrHexane 79 torr and and pentane 271 torrpentane 271 torr

Vapor composition hexane = (.227)(350) Vapor composition hexane = (.227)(350) = 79 torr, so that of pentane = 271= 79 torr, so that of pentane = 271

Review Question pp

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Liquid-liquid solutions where Liquid-liquid solutions where bothboth are are volatile.volatile.

Modify Raoult’s Law to . . . Modify Raoult’s Law to . . . PPtotaltotal = P= PAA + P + PBB = = AAPPAA

00 + + BBPPBB00

PPtotaltotal = vapor pressure of mixture= vapor pressure of mixture PPAA

00= vapor pressure of pure A (etc.)= vapor pressure of pure A (etc.) If this equation works then the solution If this equation works then the solution

isis ideal. ideal. Solvent and solute are alike (as are the Solvent and solute are alike (as are the

interactions between all species)..interactions between all species)..

Ideal/Non-ideal solutionsIdeal/Non-ideal solutions

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Test-type Question (will be on our test!) pp

Pentane (C5H12) & Hexane (C6H14) form an ideal solution with Vp

25° of 511 & 150. Torr respectively. A solution is made from 25 ml pentane (d = 0.63 g/mL) & 45 mL hexane (d = 0.66 g/mL)

(a) What is Vp of the resulting solution? (b) What is composition by mole fraction

of pentane in the vapor that is in equilibrium with this solution?

Reference: Z7e p. 521, # 49

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Test-type Question pp


25° of 511 & 150. Torr respectively. A solution is made from 25 ml pen (d = 0.63 g/mL) & 45 mL hex (d = 0.66 g/ML

(a) What is Vp of the resulting solution?

• Use density to get number of moles. • Then find of liquid pentane & hexane. • Then use Raoult’s law to get P in torr of

both partial and total pressure

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Test-type Question rf. Z7e #49 pp

25 ml Pen x 0.63g/ml x 1 mol/72.15 g = 0.22 mol Pen (and 0.34 mol Hex)

penliq = 0.22/0.56 = 0.39 and

hexliq = 1.00 - 0.39 = 0.61

Ppen = (penliq)(Ppen

o) = (0.39)(511 torr) = 2.0 x 102 torr;

ditto for Phex= 92 torr

Ptotal = 292 torr = 290 torr (sig figs)

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(b) What is composition by mole fraction of pentane in the vapor that is in equilibrium with this solution?

• Hint: Since partial pressure of gas is proportional to number of moles of gas present, use mol pentane in vapor ÷ total mol vapor to get

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(b) What is composition by mole fraction of pentane in the vapor that is in equilibrium with this solution?

The answer is . . . pen

v = Ppen/Ptotal = 2.0 x 102 torr/290 torr = 0.69.

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DeviationsDeviations to non-ideality to non-ideality

If Solvent has a strong affinity for solute If Solvent has a strong affinity for solute (H bonding).(H bonding).

LowersLowers solvent’s ability to escape. solvent’s ability to escape. Lower vapor pressure than expected.Lower vapor pressure than expected. If (-) deviation from Raoult’s law then . . .If (-) deviation from Raoult’s law then . . . HHsoln soln is large and (-), so is large and (-), so exothermicexothermic..

An An EndothermicEndothermic HHsolnsoln indicates indicates positivepositive

deviation.deviation.

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Colligative PropertiesColligative Properties Dissolved particlesDissolved particles affect vapor affect vapor

pressure, so they affect phase changes.pressure, so they affect phase changes. Colligative properties depend Colligative properties depend only on only on

the numberthe number - - not the kindnot the kind of solute of solute particles presentparticles present

The The 33 colligative properties are colligative properties are BpBp, , FpFp and and osmotic pressureosmotic pressure

Useful to determine molar mass (Lab 11)Useful to determine molar mass (Lab 11) M = (KM = (Kfpfp x g x gsolutesolute)/(kg)/(kgsolventsolvent x ∆T x ∆Tfpfp))

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Colligative Properties Colligative Properties pp

See SE 11.6 p. 500 (See SE 11.6 p. 500 (this type will be on test)this type will be on test)

Predict the VPredict the Vpp of a solution of 35.0 g of a solution of 35.0 g NaNa22SOSO44 (M = 142 g/mol) & 175 g H (M = 142 g/mol) & 175 g H220, 0, where Vwhere Vpp

HH22OO = 23.76 torr at 25 = 23.76 torr at 25oo C. . . C. . . Strategy: use Raoult’s Law Strategy: use Raoult’s Law

PPsolnsoln = = HH22OOPPooHH22OO

HH22OO = n = nH2OH2O /(n /(nH2OH2O + n + nsolutesolute)) But . . . Since the But . . . Since the solutesolute falls apart into falls apart into

3 particles, the n3 particles, the nsolutesolute has to be the has to be the moles of Namoles of Na22SOSO44 x 3x 3..

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Colligative Properties Colligative Properties pp

See SE 11.6 p. 532 (See SE 11.6 p. 532 (this type will be on test)this type will be on test)

nnH2OH2O = 175 g/18.0 g/mol = 9.72 mol H = 175 g/18.0 g/mol = 9.72 mol H22OO nnNa2SO4Na2SO4 = 35.0 g/142 g/mol = 0.246 = 35.0 g/142 g/mol = 0.246 nnsolutesolute = = 3 x 3 x nnNa2SO4Na2SO4 = 0.738 = 0.738 H2OH2O = 9.72/(0.738 + 9.72) = 0.929 = 9.72/(0.738 + 9.72) = 0.929 PPsolnsoln = = H2OH2OPPoo

H2O H2O

= (0.929)(23.76 torr) = = (0.929)(23.76 torr) = 22.1 22.1 torrtorr

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11.5 Boiling point Elevation & Freezing Point 11.5 Boiling point Elevation & Freezing Point DepressionDepression

Because a nonvolatile solute lowers the vapor pressure it raises the boiling point.

Stays liquid longer because intermolecular forces between solute/solvent help hold the liquid together.

The equation is: T = Kbmsolute (for a molecular solute; otherwise use van’t Hoff factor, i, for electrolyte solutions).

T is the change in the boiling point Kb is a constant determined by the solvent (look

up in a table). msolute is the molality of the solute

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Freezing point Depression Because a nonvolatile solute lowers the

vapor pressure of the solution it lowers the freezing point.

Stays liquid longer (before freezing) because IMF of solute/solution prevents forming orderly crystal

The equation is: T = Kfmsolute

T is the change in the freezing point Kf is a constant determined by the solvent msolute is the molality of the solute

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1 atm

Vapor Pressure of solution

Vapor Pressure of pure water

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1 atm

Freezing and boiling points of water

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1 atm

Freezing and boiling points of solution

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1 atm

TfTb

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Figure 11.14 p. 536 The Development of Osmotic Pressure pp

AP requires you to diagram the change when adding solute to a solvent

The effect of adding a solute is to extend the range of The effect of adding a solute is to extend the range of a solvent.a solvent.

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11.6 Osmotic Pressure

Osmosis: The flow of solvent into the solution through a semi permeable membrane.Osmotic Pressure: The excess hydrostatic pressure on the solution compared to the pure solvent.Osmotic Pressure π = MRTWatch units for R!

Figure 11.16Osmotic Pressure.

Net transfer of solvent molecules until

hydrostatic pressure equalizes the solvent

flow in both directions

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Figure 11.18Osmosis at Equilibrium

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If the external pressure is larger than the osmotic pressure, reverse osmosis occurs.

One application is desalination of seawater.

Figure 11.17Osmosis

When pressure exceeds this then reverse osmosis

occurs.

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Review Problems Bp, Fp, πReview Problems Bp, Fp, π

Try SE 11.8 p. 537 Try SE 11.8 p. 537 (& Text # 60)(& Text # 60)



Try SE 11.12 (& Try SE 11.12 (& Text # 70)Text # 70)

#60 = 498 g/mol#60 = 498 g/mol #65 = 456 g/mol#65 = 456 g/mol #68 = 27 000 g/mol#68 = 27 000 g/mol

#70 = Dissolve 18 g #70 = Dissolve 18 g in 1 L of solutionin 1 L of solution

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11.7 Colligative Properties of 11.7 Colligative Properties of Electrolyte SolutionsElectrolyte Solutions

Since colligative properties only depend Since colligative properties only depend on the number of particles:on the number of particles:

Ionic compounds should have a bigger Ionic compounds should have a bigger effect because . . .effect because . . .

When they dissolve they dissociate.When they dissolve they dissociate. Individual Na and Cl ions fall apart.Individual Na and Cl ions fall apart. 1 mole of NaCl makes 2 moles of ions.1 mole of NaCl makes 2 moles of ions. 1 mole Al(NO1 mole Al(NO33))33 makes 4 moles ions. makes 4 moles ions.

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Electrolytes have a bigger impact on on Electrolytes have a bigger impact on on melting and freezing points per mole melting and freezing points per mole because they make more pieces.because they make more pieces.

Relationship is expressed using the van’t Relationship is expressed using the van’t Hoff factorHoff factor i i

i = i = Moles of particles in solutionMoles of particles in solution

Moles of solute dissolvedMoles of solute dissolved The The expectedexpected value can be determined value can be determined

from the above formula.

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The actual value is usually less because: At any given instant some of the ions in

solution will be paired. Ion pairing increases with concentration. i decreases with increased

concentration. We can change our generic formulae to

H = iKm

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Colligative Properties of ElectrolytesColligative Properties of Electrolytes

T = T = iiKKbbmmsolutesolute == Bp elevationBp elevation

T = T = iiKKffmmsolutesolute = Fp depression= Fp depression π = π = iMiMRTRT = Osmotic Pressure = Osmotic Pressure 11.8 - Tyndall effect = the scattering of 11.8 - Tyndall effect = the scattering of

light by particles (read on own).light by particles (read on own). End, Chapter 11.End, Chapter 11.

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Ch11 z5e solutions

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