solutions chapter 14 why does a raw egg swell or shrink when placed in different solutions?...

SolutionsSolutionsChapter 14Chapter 14

SolutionsSolutionsChapter 14Chapter 14

Why does a raw egg swell or shrink when Why does a raw egg swell or shrink when placed in different solutions?placed in different solutions?

Why does a raw egg swell or shrink when Why does a raw egg swell or shrink when placed in different solutions?placed in different solutions?

Copyright © 1999 by Harcourt Brace & CompanyAll rights reserved.Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida

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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved

Some DefinitionsSome DefinitionsSome DefinitionsSome DefinitionsA solution is a A solution is a

HOMOGENEOUSHOMOGENEOUS mixture of 2 or more mixture of 2 or more substances in a substances in a single phase. single phase.

One constituent is One constituent is usually regarded as usually regarded as the the SOLVENTSOLVENT and and the others as the others as SOLUTESSOLUTES..

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Solutions can be Solutions can be classified as classified as

unsaturatedunsaturated or or

saturatedsaturated..

DefinitionsDefinitionsDefinitionsDefinitions

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Solutions can be Solutions can be classified as classified as unsaturatedunsaturated or or saturatedsaturated..

A saturated solution A saturated solution contains the maximum contains the maximum quantity of solute that quantity of solute that dissolves at that dissolves at that temperature.temperature.


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Solutions can be Solutions can be classified as classified as unsaturatedunsaturated or or saturatedsaturated..

A saturated solution A saturated solution contains the maximum contains the maximum quantity of solute that quantity of solute that dissovles at that dissovles at that temperature.temperature.

SUPERSATURATED SUPERSATURATED SOLUTIONSSOLUTIONS contain contain more than is possible more than is possible and are unstable.and are unstable.


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Energetics of the Solution Energetics of the Solution ProcessProcess

Energetics of the Solution Energetics of the Solution ProcessProcess

If the enthalpy of If the enthalpy of formation of the formation of the solution is more solution is more negative that that of negative that that of the solvent and the solvent and solute, the enthalpy solute, the enthalpy of solution is of solution is negative. negative.

The solution process The solution process is is exothermicexothermic!!

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SupersaturatedSupersaturatedSodium AcetateSodium Acetate

SupersaturatedSupersaturatedSodium AcetateSodium Acetate

• One application of a One application of a supersaturated supersaturated solution is the solution is the sodium acetate sodium acetate “heat pack.”“heat pack.”

• Sodium acetate has Sodium acetate has an an ENDOthermicENDOthermic heat of solution. heat of solution.

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SupersaturatedSupersaturatedSodium AcetateSodium AcetateSupersaturatedSupersaturatedSodium AcetateSodium Acetate

Sodium acetate has an Sodium acetate has an ENDOthermicENDOthermic heat of heat of solution. solution.

NaCHNaCH33COCO22 (s) + (s) + heatheat ----> ----> NaNa++(aq) + CH(aq) + CH33COCO22

--(aq)(aq)

Therefore, formation of solid sodium acetate Therefore, formation of solid sodium acetate from its ions is from its ions is EXOTHERMICEXOTHERMIC..

NaNa++(aq) + CH(aq) + CH33COCO22--(aq) ---> (aq) ---> NaCHNaCH33COCO22 (s) + (s) + heatheat

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Colligative PropertiesColligative PropertiesColligative PropertiesColligative PropertiesOn adding a solute to a solvent, the props. of On adding a solute to a solvent, the props. of

the solvent are modified.the solvent are modified.

• Vapor pressure Vapor pressure decreasesdecreases

• Melting point Melting point decreasesdecreases

• Boiling point Boiling point increasesincreases

• Osmosis is possible (osmotic pressure)Osmosis is possible (osmotic pressure)

These changes are called These changes are called COLLIGATIVE COLLIGATIVE PROPERTIESPROPERTIES. .

They depend only on the They depend only on the NUMBERNUMBER of solute of solute particles relative to solvent particles, not on particles relative to solvent particles, not on the the KINDKIND of solute particles. of solute particles.

On adding a solute to a solvent, the props. of On adding a solute to a solvent, the props. of the solvent are modified.the solvent are modified.

• Vapor pressure Vapor pressure decreasesdecreases

• Melting point Melting point decreasesdecreases

• Boiling point Boiling point increasesincreases

• Osmosis is possible (osmotic pressure)Osmosis is possible (osmotic pressure)

These changes are called These changes are called COLLIGATIVE COLLIGATIVE PROPERTIESPROPERTIES. .

They depend only on the They depend only on the NUMBERNUMBER of solute of solute particles relative to solvent particles, not on particles relative to solvent particles, not on the the KINDKIND of solute particles. of solute particles.

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An An IDEAL SOLUTIONIDEAL SOLUTION is is one where the properties one where the properties depend only on the depend only on the concentration of solute.concentration of solute.

Need conc. units to tell us the Need conc. units to tell us the number of solute particles number of solute particles per solvent particle.per solvent particle.

The unit “molarity” does not The unit “molarity” does not do this!do this!

An An IDEAL SOLUTIONIDEAL SOLUTION is is one where the properties one where the properties depend only on the depend only on the concentration of solute.concentration of solute.

Need conc. units to tell us the Need conc. units to tell us the number of solute particles number of solute particles per solvent particle.per solvent particle.

The unit “molarity” does not The unit “molarity” does not do this!do this!

Concentration UnitsConcentration UnitsConcentration UnitsConcentration Units

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MOLE FRACTION, XMOLE FRACTION, X

For a mixture of A, B, and CFor a mixture of A, B, and C



XA mol fraction A = mol A

mol A + mol B + mol CXA mol fraction A =

mol Amol A + mol B + mol C

Concentration UnitsConcentration UnitsConcentration UnitsConcentration Units

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Concentration UnitsConcentration UnitsConcentration UnitsConcentration UnitsMOLE FRACTION, XMOLE FRACTION, X


MOLALITY, mMOLALITY, m







m of solute = mol solute

kilograms solventm of solute =

mol solutekilograms solvent

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Concentration UnitsConcentration UnitsConcentration UnitsConcentration UnitsMOLE FRACTION, XMOLE FRACTION, X



WEIGHT %WEIGHT % = grams solute per 100 g solution = grams solute per 100 g solution




WEIGHT %WEIGHT % = grams solute per 100 g solution = grams solute per 100 g solution




m of solute = mol solute

kilograms solventm of solute =

mol solutekilograms solvent

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Calculating ConcentrationsCalculating ConcentrationsCalculating ConcentrationsCalculating Concentrations

Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of Hof H22O. Calculate mol fraction, molality, and weight O. Calculate mol fraction, molality, and weight

% of glycol.% of glycol.

Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of Hof H22O. Calculate mol fraction, molality, and weight O. Calculate mol fraction, molality, and weight

% of glycol.% of glycol.

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250. g H250. g H22O = 13.9 molO = 13.9 mol

250. g H250. g H22O = 13.9 molO = 13.9 mol

Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of Hof H22O. Calculate X, m, and % of glycol.O. Calculate X, m, and % of glycol.


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250. g H250. g H22O = 13.9 molO = 13.9 mol

X X glycolglycol = 0.0672 = 0.0672

250. g H250. g H22O = 13.9 molO = 13.9 mol

X X glycolglycol = 0.0672 = 0.0672



Xglycol = 1.00 mol glycol

1.00 mol glycol + 13.9 mol H2OXglycol =

1.00 mol glycol1.00 mol glycol + 13.9 mol H2O

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Calculate molalityCalculate molality




conc (molality) = 1.00 mol glycol0.250 kg H2O

4.00 molalconc (molality) = 1.00 mol glycol0.250 kg H2O

4.00 molal

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Calculate weight %Calculate weight %


Calculate weight %Calculate weight %



conc (molality) = 1.00 mol glycol0.250 kg H2O

4.00 molalconc (molality) = 1.00 mol glycol0.250 kg H2O

4.00 molal

%glycol = 62.1 g

62.1 g + 250. g x 100% = 19.9%%glycol =

62.1 g62.1 g + 250. g

x 100% = 19.9%

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Dissolving Gases & Dissolving Gases & Henry’s LawHenry’s Law

Dissolving Gases & Dissolving Gases & Henry’s LawHenry’s Law

Gas solubility (M) = kGas solubility (M) = kHH • P • Pgasgas

kkHH for O for O22 = 1.66 x 10 = 1.66 x 10-6-6 M/mmHg M/mmHg

When When PPgasgas drops, solubility drops. drops, solubility drops.

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Lake Nyos, CameroonLake Nyos, Cameroon

Courtesy of George Kling, page 656-657Courtesy of George Kling, page 656-657

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Understanding Understanding Colligative PropertiesColligative Properties


To understand colligative properties, study To understand colligative properties, study the the LIQUID-VAPOR EQUILIBRIUMLIQUID-VAPOR EQUILIBRIUM for a for a solution.solution.

To understand colligative properties, study To understand colligative properties, study the the LIQUID-VAPOR EQUILIBRIUMLIQUID-VAPOR EQUILIBRIUM for a for a solution.solution.

.

H—O

H

HH—O

H H—O

H—O

H

surface

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To understand To understand colligative colligative properties, properties, study the study the LIQUID-VAPOR LIQUID-VAPOR EQUILIBRIUMEQUILIBRIUM for a solution.for a solution.

To understand To understand colligative colligative properties, properties, study the study the LIQUID-VAPOR LIQUID-VAPOR EQUILIBRIUMEQUILIBRIUM for a solution.for a solution.

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VP of HVP of H22O over a solution depends on the O over a solution depends on the

number of Hnumber of H22O molecules per solute molecule.O molecules per solute molecule.

PPsolventsolvent proportional toproportional to X Xsolventsolvent

OROR

PPsolventsolvent = X = Xsolventsolvent • P • Poosolventsolvent

VP of solvent over solution = VP of solvent over solution = (Mol frac solvent)•(VP pure solvent)(Mol frac solvent)•(VP pure solvent)

RAOULT’S LAWRAOULT’S LAW

VP of HVP of H22O over a solution depends on the O over a solution depends on the

number of Hnumber of H22O molecules per solute molecule.O molecules per solute molecule.

PPsolventsolvent proportional toproportional to X Xsolventsolvent

OROR

PPsolventsolvent = X = Xsolventsolvent • P • Poosolventsolvent

VP of solvent over solution = VP of solvent over solution = (Mol frac solvent)•(VP pure solvent)(Mol frac solvent)•(VP pure solvent)

RAOULT’S LAWRAOULT’S LAW

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Raoult’s LawRaoult’s LawRaoult’s LawRaoult’s LawAn ideal solution is one that obeys Raoult’s An ideal solution is one that obeys Raoult’s

law.law.

PPAA = X = XAA • P • PooAA

Because mole fraction of solvent, XBecause mole fraction of solvent, XAA, is , is

always less than 1, then Palways less than 1, then PAA is always less is always less

than Pthan PooAA..

The vapor pressure of solvent over a solution The vapor pressure of solvent over a solution

is always is always LOWEREDLOWERED!!

An ideal solution is one that obeys Raoult’s An ideal solution is one that obeys Raoult’s law.law.

PPAA = X = XAA • P • PooAA

Because mole fraction of solvent, XBecause mole fraction of solvent, XAA, is , is

always less than 1, then Palways less than 1, then PAA is always less is always less

than Pthan PooAA..

The vapor pressure of solvent over a solution The vapor pressure of solvent over a solution

is always is always LOWEREDLOWERED!!

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Raoult’s LawRaoult’s LawRaoult’s LawRaoult’s LawAssume the solution containing 62.1 g of glycol in Assume the solution containing 62.1 g of glycol in

250. g of water is ideal. What is the vapor pressure 250. g of water is ideal. What is the vapor pressure of water over the solution at 30 of water over the solution at 30 ooC? C? (The VP of pure (The VP of pure HH22O is 31.8 mm Hg; see App. E.)O is 31.8 mm Hg; see App. E.)

SolutionSolution

XXglycolglycol = 0.0672 = 0.0672 and so Xand so Xwaterwater = ? = ?

Because XBecause Xglycolglycol + X + Xwaterwater = 1 = 1

XXwaterwater = 1.000 - 0.0672 = 0.9328 = 1.000 - 0.0672 = 0.9328

PPwaterwater = X = Xwaterwater • P • Poowaterwater = (0.9382)(31.8 mm Hg) = (0.9382)(31.8 mm Hg)

PPwaterwater = 29.7 mm Hg = 29.7 mm Hg

Assume the solution containing 62.1 g of glycol in Assume the solution containing 62.1 g of glycol in 250. g of water is ideal. What is the vapor pressure 250. g of water is ideal. What is the vapor pressure of water over the solution at 30 of water over the solution at 30 ooC? C? (The VP of pure (The VP of pure HH22O is 31.8 mm Hg; see App. E.)O is 31.8 mm Hg; see App. E.)

SolutionSolution

XXglycolglycol = 0.0672 = 0.0672 and so Xand so Xwaterwater = ? = ?

Because XBecause Xglycolglycol + X + Xwaterwater = 1 = 1

XXwaterwater = 1.000 - 0.0672 = 0.9328 = 1.000 - 0.0672 = 0.9328

PPwaterwater = X = Xwaterwater • P • Poowaterwater = (0.9382)(31.8 mm Hg) = (0.9382)(31.8 mm Hg)

PPwaterwater = 29.7 mm Hg = 29.7 mm Hg

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Raoult’s LawRaoult’s LawRaoult’s LawRaoult’s LawFor a 2-component system where A is the For a 2-component system where A is the

solvent and B is the solutesolvent and B is the solute

PPAA = VP lowering = X = VP lowering = XBBPPooAA

VP lowering is proportional to mol frac VP lowering is proportional to mol frac solutesolute!!

For very dilute solutions, For very dilute solutions, PPAA = K•molality = K•molalityBB where K is a proportionality constant.where K is a proportionality constant.

This helps explain changes in melting and This helps explain changes in melting and boiling points.boiling points.

For a 2-component system where A is the For a 2-component system where A is the solvent and B is the solutesolvent and B is the solute

PPAA = VP lowering = X = VP lowering = XBBPPooAA

VP lowering is proportional to mol frac VP lowering is proportional to mol frac solutesolute!!

For very dilute solutions, For very dilute solutions, PPAA = K•molality = K•molalityBB where K is a proportionality constant.where K is a proportionality constant.

This helps explain changes in melting and This helps explain changes in melting and boiling points.boiling points.

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Changes in Freezing and Changes in Freezing and Boiling Points of SolventBoiling Points of SolventChanges in Freezing and Changes in Freezing and Boiling Points of SolventBoiling Points of Solvent

See Figure 14.13See Figure 14.13

VP solventafter addingsolute

VP Pure solvent

BP puresolvent

BP solution

1 atm

P

T


VP Pure solvent

BP puresolvent

BP solution

1 atm

P

T

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The boiling point of a The boiling point of a solution is higher than that solution is higher than that

of the pure solvent.of the pure solvent.

The boiling point of a The boiling point of a solution is higher than that solution is higher than that

of the pure solvent.of the pure solvent.

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Elevation of Boiling Point Elevation of Boiling Point Elevation of Boiling Point Elevation of Boiling Point Elevation in BP = Elevation in BP = ttBPBP = K = KBP BP • m• m(where K(where KBPBP is characteristic of solvent) is characteristic of solvent)


VP Pure solvent

BP puresolvent

BP solution

1 atm

P

T


VP Pure solvent

BP puresolvent

BP solution

1 atm

P

T

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Change in Boiling Point Change in Boiling Point Change in Boiling Point Change in Boiling Point

Dissolve 62.1 g of glycol (1.00 mol) in 250. g Dissolve 62.1 g of glycol (1.00 mol) in 250. g of water. What is the BP of the solution?of water. What is the BP of the solution?

KKBPBP = +0.512 = +0.512 ooC/molal for water (see Table C/molal for water (see Table 14.3).14.3).

SolutionSolution

1.1. Calculate solution molality = 4.00 mCalculate solution molality = 4.00 m

2.2. ttBPBP = K = KBPBP • m • m

ttBPBP = +0.512 = +0.512 ooC/molal (4.00 molal)C/molal (4.00 molal)

ttBPBP = +2.05 = +2.05 ooCC

BP = 102.05 BP = 102.05 ooCC

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Change in Freezing Point Change in Freezing Point Change in Freezing Point Change in Freezing Point

The freezing point of a solution is The freezing point of a solution is LOWERLOWER than that of the pure solvent.than that of the pure solvent.

FP depression = FP depression = ttFPFP = K = KFPFP•m•m

Pure waterPure water Ethylene glycol/water Ethylene glycol/water solutionsolution

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Freezing Point DepressionFreezing Point DepressionFreezing Point DepressionFreezing Point DepressionConsider equilibrium at melting pointConsider equilibrium at melting point

Liquid solvent <------> Solid solventLiquid solvent <------> Solid solvent

•• Rate at which molecules go from S to L depends Rate at which molecules go from S to L depends only on the nature of the solid.only on the nature of the solid.

•• BUT — rate for L ---> S depends on how much is BUT — rate for L ---> S depends on how much is dissolved. This rate is SLOWED for the same dissolved. This rate is SLOWED for the same reason VP is lowered.reason VP is lowered.

•• Therefore, to bring S ---> L and L ---> S rates into Therefore, to bring S ---> L and L ---> S rates into equilibrium for a solution, T must be lowered.equilibrium for a solution, T must be lowered.

Thus, FP for solution < FP for solventThus, FP for solution < FP for solvent

FP depression = FP depression = ttFPFP = K = KFPFP•m•m

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Calculate the FP of a 4.00 molal glycol/water Calculate the FP of a 4.00 molal glycol/water solution.solution.

KKFPFP = -1.86 = -1.86 ooC/molal (Table 14.4)C/molal (Table 14.4)

SolutionSolution

ttFPFP = K = KFPFP • m • m

= (-1.86 = (-1.86 ooC/molal)(4.00 m)C/molal)(4.00 m)

ttFP FP = -7.44 = -7.44 ooCC

Freezing Point DepressionFreezing Point DepressionFreezing Point DepressionFreezing Point Depression

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How much NaCl must be dissolved in 4.00 How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 kg of water to lower FP to -10.00 ooC?.C?.

SolutionSolution

Calc. required molalityCalc. required molality

ttFPFP = K = KFPFP • m • m

-10.00 -10.00 ooC = (-1.86 C = (-1.86 ooC/molal) • ConcC/molal) • Conc

Conc = 5.38 molal Conc = 5.38 molal


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How much NaCl must be dissolved in 4.00 kg of water to lower How much NaCl must be dissolved in 4.00 kg of water to lower

FP to -10.00 FP to -10.00 ooC?C?..

SolutionSolution

Conc req’d = 5.38 molalConc req’d = 5.38 molal

This means we need 5.38 mol of dissolved This means we need 5.38 mol of dissolved particles per kg of solvent. particles per kg of solvent.

Recognize that m represents the total conc. Recognize that m represents the total conc. of all dissolved particles.of all dissolved particles.

Recall that 1 mol NaCl(aq) Recall that 1 mol NaCl(aq) --> 1 mol Na--> 1 mol Na++(aq) + 1 mol (aq) + 1 mol

ClCl--(aq)(aq)


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How much NaCl must be dissolved in 4.00 kg of water to lower How much NaCl must be dissolved in 4.00 kg of water to lower

FP to -10.00 FP to -10.00 ooC?C?..

SolutionSolution

Conc req’d = 5.38 molalConc req’d = 5.38 molal

We need 5.38 mol of dissolved particles per We need 5.38 mol of dissolved particles per kg of solvent. kg of solvent.

NaCl(aq) --> NaNaCl(aq) --> Na++(aq) + Cl(aq) + Cl--(aq)(aq)

To get 5.38 mol/kg of particles we needTo get 5.38 mol/kg of particles we need

5.38 mol / 2 = 2.69 mol NaCl / kg5.38 mol / 2 = 2.69 mol NaCl / kg2.69 mol NaCl / kg ---> 157 g NaCl / kg2.69 mol NaCl / kg ---> 157 g NaCl / kg

(157 g NaCl / kg)•(4.00 kg) = (157 g NaCl / kg)•(4.00 kg) = 629 g NaCl629 g NaCl


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Boiling Point Elevation and Boiling Point Elevation and Freezing Point DepressionFreezing Point Depression

Boiling Point Elevation and Boiling Point Elevation and Freezing Point DepressionFreezing Point Depression

t = K • m • it = K • m • iA generally useful equation A generally useful equation

i = van’t Hoff factor = number of particles i = van’t Hoff factor = number of particles produced per formula unit.produced per formula unit.

CompoundCompound Theoretical Value of iTheoretical Value of i

glycolglycol 11

NaClNaCl 22

CaClCaCl22 33

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OsmosisOsmosisOsmosisOsmosis

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The semipermeable membrane should The semipermeable membrane should allow only the movement of solvent allow only the movement of solvent molecules.molecules.

Therefore, solvent molecules move from Therefore, solvent molecules move from pure solvent to solution.pure solvent to solution.

Solvent Solution

Semipermeable membrane

Solvent Solution

Semipermeable membrane

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The semipermeable membrane should The semipermeable membrane should allow only the movement of solvent allow only the movement of solvent molecules.molecules.

Therefore, solvent molecules move from Therefore, solvent molecules move from pure solvent to solution.pure solvent to solution.

Solvent Solution

OsmoticPressure

Solvent Solution

OsmoticPressure

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Equilibrium is reached when pressure produced by extra Equilibrium is reached when pressure produced by extra solution — solution —

the the OSMOTIC PRESSURE, OSMOTIC PRESSURE,

= cRT = cRT (where c is conc. in mol/L)(where c is conc. in mol/L)counterbalances pressure of solvent molecules moving thru counterbalances pressure of solvent molecules moving thru

the membrane.the membrane.

Solvent Solution

OsmoticPressure

Solvent Solution

OsmoticPressure

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OsmosisOsmosisOsmosisOsmosis• Osmosis of solvent Osmosis of solvent

from one solution to from one solution to another can continue another can continue until the solutions are until the solutions are ISOTONICISOTONIC — they — they have the same have the same concentration.concentration.

• Osmotic pressure in Osmotic pressure in living systems: living systems: FIGURE 14.16FIGURE 14.16

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Osmosis Osmosis Calculating a Molar MassCalculating a Molar Mass


Dissolve 35.0 g of hemoglobin in enough water to Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L of solution. make 1.00 L of solution. measured to be 10.0 measured to be 10.0 mm Hg at 25 mm Hg at 25 C. Calc. molar mass of C. Calc. molar mass of hemoglobin.hemoglobin.

SolutionSolution

(a)(a) Calc. Calc. in atmospheres in atmospheres

= 10.0 mmHg • (1 atm / 760 mmHg)= 10.0 mmHg • (1 atm / 760 mmHg)

= 0.0132 atm= 0.0132 atm

(b)(b) Calc. concentrationCalc. concentration

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Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L of solution. of solution. measured to be 10.0 mm Hg at 25 measured to be 10.0 mm Hg at 25 C. Calc. C. Calc. molar mass of hemoglobin.molar mass of hemoglobin.

SolutionSolution

(b)(b) Calc. concentration from Calc. concentration from = cRT = cRT

Conc = 0.0132 atm

(0.0821 L • atm/K • mol)(298K)Conc =

0.0132 atm(0.0821 L • atm/K • mol)(298K)

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Conc = 5.39 x 10Conc = 5.39 x 10-4-4 mol/L mol/L

(c)(c) Calc. molar massCalc. molar mass

Molar mass = 35.0 g / 5.39 x 10Molar mass = 35.0 g / 5.39 x 10-4-4 mol/L mol/L

Molar mass = 65,100 g/molMolar mass = 65,100 g/mol

Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L

of solution. of solution. measured to be 10.0 mm Hg at 25 measured to be 10.0 mm Hg at 25 C. Calc. C. Calc. molar mass of hemoglobin.molar mass of hemoglobin.

SolutionSolution

(b)(b) Calc. concentration from Calc. concentration from = cRT = cRT

Conc = 0.0132 atm

(0.0821 L • atm/K • mol)(298K)Conc =

0.0132 atm(0.0821 L • atm/K • mol)(298K)

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