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5 Simple mixtures

Answers to discussion questions

D5.1 At equilibrium, the chemical potentials of any component in both the liquid and vapor phases must be

equal. This is justified by the requirement that, for systems at equilibrium under constant temperature and

pressure conditions, with no additional work, G= 0 [see Section 3.5(e) and the answer to Discussion

question 3.3]. HereG = i(v) i(1), for all components, i , of the solution; hence their chemical

potentials must be equal in the liquid and vapor phases.

D5.3 All of the colligative properties are a function of the concentration of the solute, which implies that the

concentration can be determined by a measurement of these properties. See eqns 5.33, 5.34, 5.36, 5.37,

and 5.40. Knowing the mass of the solute in solution then allows for a calculation of its molar mass. For

example, the mole fraction of the solute is related to its mass as follows:

xB = mB/MBmB/MB+ mA/MA

.

The only unknown in this expression is MB which is easily solved for. See Example 5.4 for the details

of how molar mass is determined from osmotic pressure.

D5.5 A regular solution has excess entropy of zero, but an excess enthalpy that is non-zero and dependent on

composition, perhaps in the manner of eqn 5.30. We can think of a regular solution as one in which the

different molecules of the solution are distributed randomly, as in an ideal solution, but have different

energies of interaction with each other.

Solutions to exercises

E5.1(a) Let A denote acetone and C chloroform. The total volume of the solution is

V =nAVA + nCVC.

VA and Vcare given; hence we need to determinenA and nC in 1.000 kg of the solution with the stated

mole fraction. The total mass of the sample is m = nAMA + nCMC (a). We also know that

xA =nA

nA+ nCimplies that (xA 1)nA+ xAnC =0

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SIMPLE MIXTURES 95

and hence that

xCnA+ xAnC =0. (b)

On solving (a) and (b), we find

nA =

xA

xC

nC, nC =

mxC

xAMA+ xCMC.

SincexC =0.4693,xA =1 xC =0.5307,

nC =(0.4693)(1000 g)

[(0.5307)(58.08)+(0.4693)(119.37)] gmol1 =5.404 mol,

nA =

0.5307

0.4693

(5.404)mol = 6.111 mol.

The total volume,V = nAVA+ nBVB, is therefore

V =(6.111 mol)(74.166cm3 mol1)+(5.404 mol)(80.235cm3 mol1)

= 886.8 cm3 .

E5.2(a) Let A denote water and B ethanol. The total volume of the solution isV = nAVA+ nBVB

We are given VA, we need to determine nA and nB in order to solve forVB.

Assume we have 100 cm3 of solution; then the mass of solution is

m = d V = (0.914 g cm3)(100 cm3)= 91.4 g

of which 45.7 g is water and 45.7 g ethanol.

100 cm3 =

45.7 g

18.02 g mol1

(17.4 cm3 mol1)+

45.7g

46.07 g mol1

VB

=44.13 cm3 +0.9920 molVB,

VB =55.87 cm3

0.9920 mol= 56.3 cm3 mol1 .

E5.3(a) Check whetherpB

xBis equal to a constant (KB)

x 0.005 0.012 0.019

p/x 6.4103 6.4103 6.4103 kPa

Hence,KB 6.4103 kPa .

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96 SOLUTIONS MANUAL

E5.4(a) In Exercise 5.3(a), the Henrys law constant was determined for concentrations expressed in mole

fractions. Thus the concentration in molality must be converted to mole fraction.

m(GeCl4) = 1000 g, corresponding to

n(GeCl4)=1000 g

214.39 g mol1 = 4.664 mol, n(HCl) = 0.10 mol.

Therefore,x=0.10 mol

(0.10 mol)+ (4.664 mol)=0.0210.

FromKB =6.4 103 kPa (Exercise 5.3(a)), p = (0.02106.4103 kPa) = 1.3102 kPa .

E5.5(a) We assume that the solvent, benzene, is ideal and obeys Raoults law.

Let B denote benzene and A the solute; then

pB = xBpB and xB =

nB

nA+ nB.

HencepB =nBp

B

nA + nB, which solves to

nA =nB(p

B pB)

pB.

Then, sincenA = mAMA

, wheremA is the mass of A present,

MA =mApB

nB(pB pB)

=mAMBpB

mB(pB pB)

.

From the data

MA =(19.0 g)(78.11 g mol1)(51.5 kPa)

(500g)(53.351.5) kPa= 82gmol1

E5.6(a) MB =mass of B

hB[B = compound].

nB =mass of CCl4 bB [bB =molality of B].

bB =T

Kf[5.37]; thus

MB =mass of BKf

mass of CCl4 TKf =30 K/(mol kg

1)[Table 5.2],

MB =(100g)(30Kkgmol1)

(0.750 kg)(10.5K)= 381g mol1 .

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SIMPLE MIXTURES 97

E5.7(a) T =KfbB[5.37], bB =nB

mass of water

nB

V

[dilute solution].

103 kg m3[density of solution density of water].

nB V

RT[5.40], T Kf

RT

withKf =1.86 K/(mol kg1)[Table 5.2].

T (1.86K kgmol1)(120103 Pa)

(8.314JK1 mol1)(300K)(103 kg m3)=0.089 K.

Therefore, the solution will freeze at about 0.09C .

COMMENT. Osmotic pressures are inherently large. Even dilute solutions with small freezing point

depressions have large osmotic pressures.

E5.8(a) mixG = nRT{xAlnxA+ xBlnxB}[5.18], xA =xB =0.5, n=pV

RT.

Therefore,

mixG = (pV)

1

2ln

1

2+

1

2ln

1

2

= pVln 2

= (1.0)(1.013105 Pa)(5.0103 m3)(ln 2)

= 3.510

2

J = 0.35 kJ .

mixS= nR{xAlnxA + xBlnxB} =mixG

T[5.19] =

0.35 kJ

298K= +1.2 J K1 .

E5.9(a) mixS= nR

J

xJ lnxJ[5.19].

Therefore, for molar amounts,

mixS= R

J

xJlnxJ

= R[(0.782 ln 0.782)+(0.209 ln 0.209)+(0.009 ln 0.009)+(0.0003 ln 0.0003)]

=0.564R = +4.7 J K

1

mol

1

.

E5.10(a) Hexane and heptane form nearly ideal solutions; therefore eqn 5.19 applies.

mixS= nR(xAlnxA + xBlnxB)[5.19].

We need to differentiate eqn 5.19 with respect toxAand look for the value ofxAat which the derivative

is zero. SincexB =1 xA, we need to differentiate

mixS= nR{xAlnxA+ (1 xA) ln(1 xA)}.

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98 SOLUTIONS MANUAL

This gives using d lnxdx

=1

x

dmixS

dxA= nR{lnxA+ 1ln(1xA)1} = nR ln

xA

1 xA

which is zero when xA =1

2. Hence, the maximum entropy of mixing occurs for the preparation of a

mixture that contains equal mole fractions of the two components.

(a)n(Hex)

n(Hep)=1 =

m(Hex)

M(Hex)

m(Hep)

M(Hep)

.

(b)

m(Hex)

m(Hep) =

M(Hex)

M(Hep) =

86.17 g mol1

100.20 g mol1 = 0.8600 .

E5.11(a) With concentrations expressed in molalities, Henrys law [5.26] becomespB = bBK.

Solving forb, the solubility, we have bB =pB

K.

(a) pB =0.10 atm = 10.1 kPa.

b =10.1 kPa

3.01103 kPa kg mol1 = 3.4 mmol kg1 .

(b) pB =1.00 atm = 101.3 kPa.

b =

101.3 kPa

3.01103 kPa kg mol1 = 34 mmol kg1

.

E5.12(a) As in Exercise 5.11(a), we have

bB =pB

K=

5.0101.3 kPa

3.01103 kPa kg mol1 = 0.17 mol kg1 .

Hence, the molality of the solution is about 0.17 mol kg1 and, since molalities and molar con-

centrations for dilute aqueous solutions are approximately equal, the molar concentration is about

0.17 mol dm3.

E5.13(a) The solubility in grams of anthracene per kg of benzene can be obtained from its mole fraction with use

of the equation.

lnxB =fusH

R

1

T

1

T

[5.39; B, the solute, is anthracene]

=

28.8103 Jmol1

8.314 J K1 mol1

1

490.15 K

1

298.15K

= 4.55.

Therefore,xB =e4.55 =0.0106.

SincexB 1, x(anthracene)n(anthracene)

n(benzene).

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SIMPLE MIXTURES 99

Therefore, in 1 kg of benzene,

n(anthr.) x(anthr.)

1000 g

78.11 g mol1

(0.0106)(12.80 mol)= 0.136 mol.

Themolality of thesolutionis therefore0.136 molkg1. SinceM=178 g mol1, 0.136 molcorresponds

to 24 g anthracene in 1 kg of benzene.

E5.14(a) The best value of the molar mass is obtained from values of the data extrapolated to zero concentration,

since it is under this condition that eqn 5.40 applies.

V =nBRT[5.40], so =mRT

MV=

cRT

M, c =

m

V.

= gh [hydrostatic pressure], so h = RTgM

c.

dm3 Figure 5.1

Hence, plothagainstcand identify the slope asRT

gM. Figure 5.1 shows the plot of the data.

The slope of the line is 0.29 cm/(g dm3), so

RT

gM=

0.29 cm

g dm3 =0.29 cmdm3 g1 =0.29102 m4 kg1.

Therefore,

M =RT

(g)(0.29102 m4 kg1)

=(8.314 J K1 mol1)(298.15K)

(1.004103 kg m3)(9.81m s2)(0.29102 m4 kg1)= 87 kg mol1 .

E5.15(a) ForA (Raoults law basis; concentration in mole fraction)

aA =pA

pA[5.43] =

250 Torr

300 Torr= 0.833 ; A =

aA

xA=

0.833

0.90= 0.93 .

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100 SOLUTIONS MANUAL

ForB (Henrys law basis; concentration in mole fraction)

aB =pB

KB[5.50] =

25 Torr

200 Torr= 0.125 ; B =

aB

xB=

0.125

0.10= 1.25 .

ForB (Henrys law basis; concentration in molality)

An equation analogous to eqn 5.50 is used, aB =pB

KBwith a modified Henrys law constant KBthat

corresponds to the pressure of B in the limit of very low molalities,

pB =bB

b KB,

is analogous to pB = xBKB. Since xBand bBare related as bB =xB

MAxA, KBand KBare related as

KB = xAMAbKB.

We also needMA,

MA =xB

xAbB=

0.10

(0.90)(2.22 mol kg1)=0.050 kg mol1.

Then,KB =(0.90)(0.050 kgmol1)(1molkg1)(200 Torr) =9.0 Torr

andaB =25 Torr

9.0 Torr= 2.8 , B =

aBbBb

= 2.82.22

= 1.25 .

COMMENT. The two methods for the solute B give different values for the activities. This is reasonable

since the chemical potentials in the reference states and are different.

Question. What are the activity and activity coefficient of B in the Raoults law basis?

E5.16(a) In an ideal dilute solution the solvent (CCl4) obeys Raoults law and the solute (Br2) obeys Henrys law;

hence

p(CCl4)= x(CCl4)p(CCl4)[5.24] =(0.950)(33.85Torr)= 32.2 Torr ,

p(Br2)= x(Br2)K(Br2)[5.26] =(0.050)(122.36Torr) = 6.1 Torr ,

p(Total)= (32.2+6.1)Torr = 38.3 Torr .

The composition of the vapor in equilibrium with the liquid is

y(CCl4)=p(CCl4)

p(Total)=

32.2 Torr

38.3 Torr= 0.841 ,

y(Br2)=p(Br2)

p(Total)=

6.1 Torr

38.3 Torr= 0.16 .

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SIMPLE MIXTURES 101

E5.17(a) Let A =acetone and M =methanol.

yA =pA

pA+ pM[Daltons law] =

pA

101.3 kPa=0.516.

pA =52.3 kPa, pM =49.0 kPa.

pA =pA

pA[5.43] =

52.3 kPa

105 kPa= 0.499 , aM =

pM

pM=

49.0 kPa

73.5 kPa= 0.668 .

A =aA

xA=

0.499

0.400= 1.25 , M =

aM

xM=

0.668

0.600= 1.11 .

E5.18(a) I =1

2

(bi/b

)z2i[5.70]

and for an MpXqsalt,(b+/b) =p(b/b), (b/b

) = q(b/b), so

I =1

2(pz2++ qz

2)

b

b

.

I(KCl) =1

2(11+11)

b

b

=

b

b

.

I(CuSO4) =1

2(122 +122)

b

b

= 4

b

b

.

I =I(KCL)+ I(CuSO4) =

b

b

(KCL)+4

b

b

(CuSO4)

=(0.10)+ (4)(0.20) = 0.90 .

COMMENT. Note that the strength of a solution of more than one electrolyte may be calculated by summing

the ionic strengths of each electrolyte considered as a separate solution, as in the solution to this exercise,

by summing the product 1

2

bi

b

z

2i

for each individual ion as in the definition ofI [5.71].

E5.19(a) I =I(KNO3)=

b

b

(KNO3)=0.150.

Therefore, the ionic strengths of the added salts must be 0.100.

(a) I(Ca(NO3)2) =1

2

(22 +2) b

b= 3

b

b .

Therefore, the solution should be made1

3 0.100 mol kg1 =0.0333 mol kg1 in Ca(NO3)2. The

mass that should be added to 500 g of the solution is therefore (0.500 kg) (0.0333 molkg1)

(164 g mol1) = 2.73 g .

(b) I(NaC1)=

b

b

; therefore, withb= 0.100 mol kg1,

(0.500 kg)(0.100 mol kg1)(58.4 g mol1)= 2.92 g .

(We are neglecting the fact that the mass of solution is slightly different from the mass of solvent.)

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E5.20(a) Theseconcentrationsaresufficiently dilutefor theDebyeHckel limiting lawto givea good approximate

value for the mean ionic activity coefficient. Hence

log = |z+z|AI1/2[5.69].

I =1

2

i

z2i

bi

b

[5.71] =

1

2[(40.010)+ (10.020)+(10.030)+ (10.30)

= 0.060 .

log = 210.509(0.060)1/2 = 0.2494; =0.563 = 0.56 .

E5.21(a) log = A|z+z|I

1/2

1+BI1/2 [5.72].

Solving forB,

B =

1

I1/2 +

A|z+z|

log

.

For HBr,I =

b

b

and |z+z+| =1; so

B =

b

(b/b)1/2 +

0.509

log

.

Hence, draw up the following table.

(b/b) 5.0103 10.0103 120.0103

0.930 0.907 0.879

B 2.01 2.01 2.02

The constancy ofBindicates that the mean ionic activity coefficient of HBr obeys the extended Debye

Hckel law very well.

Solutions to problems

Solutions to numerical problems

P5.1

pA = yApandpB =yBp(Daltons law). Hence, draw up the following table.

pA/kPa 0 1.399 3.566 5.044 6.996 7.940 9.211 10.105 11.287 12.295

xA 0 0.0898 0.2476 0.3577 0.5194 0.6036 0.7188 0.8019 0.9105 1

yA 0 0.0410 0.1154 0.1762 0.2772 0.3393 0.4450 0.5435 0.7284 1

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SIMPLE MIXTURES 103

pB/kPa 0 4.209 8.487 11.487 15.462 18.243 23.582 27.334 32.722 36.066

xB 0 0.0895 0.1981 0.2812 0.3964 0.4806 0.6423 0.7524 0.9102 1

yB 0 0.2716 0.4565 0.5550 0.6607 0.7228 0.8238 0.8846 0.9590 1

The data are plotted in Fig. 5.2.

Figure 5.2

We can assume, at the lowest concentrations of both A and B, that Henrys law will hold. The Henrys

law constants are then given by

KA =pA

xA= 15.58 kPa from the point atxA =0.0898.

KB =pB

xB= 47.03 kPa from the point atxB =0.0895.

P5.3 Vsalt =

V

b

H2O

mol1 [Problem 5.2]

=69.38(b0.070)cm3 mol1 withb b/(mol kg1).

Therefore, atb = 0.050 mol kg1, Vsalt = 1.4 cm3 mol1 .

The total volume at this molality is

V =(1001.21)+ (34.69)(0.02)2 cm3 =1001.22 cm3.

Hence, as in Problem 5.2,

V(H2O) =(1001.22 cm3) (0.050 mol)(1.4 cm3mol1)

55.49 mol= 18.04 cm2 mol1 .

Question. What meaning can be ascribed to a negative partial molar volume?

P5.5 Let E denote ethanol and W denote water; then

V =nEVE + nWVW[5.3].

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For a 50 per cent mixture by mass, mE = mW, implying that

nEME =nWMW, ornW =nEME

MW.

Hence,V =nEVE+nEMEVW

MW

which solves tonE =V

VE +MEVW

MW

, nW =MEV

VEMW + MEVW.

Furthermore,xE =nE

nE + nW=

1

1+ME

MW

.

SinceME =46.07 g mol1andMW =18.02 g mol1, MEMW

=2.557. Therefore

xE =0.2811, xW =1xE =0.7189.

At this composition

VE =56.0 cm3mol1, VW =17.5 cm

3 mol1[Fig.5.1 of the text].

Therefore,nE =100 cm3

(56.0 cm3 mol1)+ (2.557)(17.5 cm3 mol1)=0.993 mol,

nW = (2.557)(0.993 mol)=2.54mol.

The fact that these amounts correspond to a mixture containing 50 per cent by mass of both components

is easily checked as follows:

mE = nEME =(0.993 mol)(46.07 g mol1)=45.7g ethanol,

mW = nWMW = (2.54 mol)(18.02 g mol1)=45.7g water.

At 20C the densities of ethanol and water are,

E =0.789 g cm3,W =0.997 g cm

3. Hence,

VE =mE

E=

45.7 g

0.789 g cm3 = 57.9 cm3 of ethanol,

VW = mWW

= 45.7 g0.997 g cm3

= 45.8 cm3 of water.

The change in volume upon adding a small amount of ethanol can be approximated by

V =

dV

VEdnE VEnE

where we have assumed that both VEand VWare constant over this small range ofnE. Hence

V (56.0 cm3mol1)

(1.00 cm3)(0.789 g cm3)

(46.07 g mol1)

= +0.96 cm3 .

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SIMPLE MIXTURES 105

P5.7bB =

T

Kf

=0.0703K

1.86 K/(mol kg1

)

=0.0378 molkg1.

Since the solution molality is nominally 0.0096 mol kg1 in Th(NO3)4, each formula unit supplies0.0378

0.0096 4 ions . (More careful data, as described in the original reference gives 5 to 6.)

P5.9 Thedata areplotted in Figure5.3. Theregions where thevaporpressurecurves show approximate straight

lines are denoted R for Raoult and H for Henry. A and B denote acetic acid and benzene respectively.

H

A

R H

R

B

40.0

26.7

13.3

kPa

00 0.2 0.4 0.6

xA

0.8 1.0

Henry

Henry

Raoult

Raoult

Extrapolate

for KB

Figure 5.3

As in Problem 5.8, we need to formA =pA

xApA

and B =pB

xBpB

for theRaoults lawactivitycoefficients

and B =pB

xBKfor the activity coefficient of benzene on a Henrys law basis, with Kdetermined by

extrapolation. We use pA =7.3 kPa, pB =35.2 kPa, and K

B =80.0 kPa to draw up the following

table.

xA 0 0.2 0.4 0.6 0.8 1.0

pA/kPa 0 2.7 4.0 5.1 6.7 7.3

pB/kPa 35.2 30.4 25.3 20.0 12.4 0

aA(R) 0 0.36 0.55 0.69 0.91 1.00[pA/pA]

aB(R) 1.00 0.86 0.72 0.57 0.35 0[pB/pB]

A(R) 1.82 1.36 1.15 1.14 1.00[pA/xApA]

B(R) 1.00 1.08 1.20 1.42 1.76 [pB/xBpB]

aB(H) 0.44 0.38 0.32 0.25 0.16 0[pB/KB]

B(H) 0.44 0.48 0.53 0.63 0.78 1.00[pB/xBKB]

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GEis defined as [Section 5.4]

GE = mixG(actual) mixG(ideal)= nRT(xAln aA + xBln aB) nRT(xAlnxA+ xBlnxB)

and, witha = x,

GE = nRT(xAln A+ xAln B).

Forn =1, we can draw up the following table from the information above and RT =2.69 kJ mol1.

xA 0 0.2 0.4 0.6 0.8 1.0

xAln A 0 0.12 0.12 0.08 0.10 0

xBlnB 0 0.06 0.11 0.14 0.11 0GE/(kJ mol1) 0 0.48 0.62 0.59 0.56 0

P5.11 (a) The volume of an ideal mixture is

Videal =n1Vm,1 + n2Vm,2

so the volume of a real mixture is

V = Videal+ VE.

We have an expression for excess molar volume in terms of mole fractions. To compute partial molar

volumes, we need an expression for the actual excess volume as a function of moles.

VE = (n1+ n2)VEm =

n1n2

n1+ n2

a0+

a1(n1 n2)

n1+ n2

soV = n1Vm,1+ n2Vm,2 +n1n2

n1+ n2

a0+

a1(n1 n2)

n1+ n2

.

The partial molar volume of propionic acid is

V1 =

V

n1

p,T,n2

= Vm,1+a0n

22

(n1+ n2)2

+a1(3n1 n2)n

22

(n1+ n2)3

,

V1 = Vm,1 + a0x22 + a1(3x1 x2)x

22 .

That of oxane is

V2 =Vm,2 + a0x21 +a1(x13x2)x

21 .

(b) We need the molar volumes of the pure liquids,

Vm,1 =M1

1=

74.08 g mol1

0.97174 g cm3 =76.23 cm3mol1

andVm,2 =86.13 g mol1

0.86398 g cm3 =99.69 cm3mol1.

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SIMPLE MIXTURES 107

In an equimolar mixture, the partial molar volume of propionic acid is

V1 =76.23+(2.4697)(0.500)2 +(0.0608) [3(0.5)0.5] (0.5)2cm3mol1

= 75.63 cm3 mol1

and that of oxane is

V2 =99.69+ (2.4697)(0.500)2 +(0.0608) [0.53(0.5)] (0.5)2 cm3 mol1

= 99.06 cm3 mol1 .

P5.13 Henrys law constant is the slope of a plot ofpBversusxBin the limit of zero xB(Fig. 5.4). The partial

pressures of CO2are almost but not quite equal to the total pressures reported above.

pCO2 =pyCO2 =p(1ycyc).

Linear regression of the low-pressure points gives KH = 371 bar .

0.0 0.1 0.2 0.30

20

40

60

80

Figure 5.4

The activity of a solute is

aB =pB

KH=xBB

so the activity coefficient is

B =pB

xBKH=

yBp

xBKH

where the last equality applies Daltons lawof partial pressures to thevapor phase. A spreadsheet applied

this equation to the above data to yield

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p/bar ycyc xcyc CO2

10.0 0.0267 0.9741 1.01

20.0 0.0149 0.9464 0.99

30.0 0.0112 0.9204 1.00

40.0 0.00947 0.892 0.99

60.0 0.00835 0.836 0.98

80.0 0.00921 0.773 0.94

P5.15 GE = RTx(1 x){0.48570.1077(2x1)+0.0191(2x1)2}

withx= 0.25 gives GE =0.1021RT. Therefore, since

mixG(actual)= mixG(ideal)+ nGE

,

mixG= nRT(xAlnxA+ xBlnxB)+nGE = nRT(0.25 ln 0.25+0.75 ln 0.75)+ nGE

= 0.562nRT+0.1021nRT = 0.460nRT.

Sincen =4 mol andRT = (8.314 JK1 mol1)(303.15 K) =2.52 kJ mol1,

mixG= (0.460)(4mol)(2.52 kJ mol1)= 4.6 kJ .

Solutions to theoretical problems

P5.17 A =GnA

nB

[5.4] =oA+ nA

(nGE)

nB

[oAis ideal value = A+ RTlnxA],

nGE

nA

nB

= GE +n

GE

nA

nB

= GE +n

xA

nA

B

GE

xA

B

= GE +nxB

n

GE

xA

B

[xA/nA =xB/n]

= gRTxA(1xA)+ (1 xA)gRT(12xA)

= gRT(1xA)2 = gRTx2B.

Therefore, A =A+ RTlnxA+ gRTx2B .

P5.19 nAdVA + nBdVB =0 [Example 5.1].

HencenA

nBdVA = dVB.

Therefore, by integration,

VB(xA) VB(0) =

VA(xA)VA(0)

nA

nBdVA =

VA(xA)VA(0)

xAdVA

1 xA[nA =xAn, nB =xBn].

Therefore,VB(xA,xB) = VB(0, 1)

VA(xA)

VA(0)

xAdVA

1 xA.

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SIMPLE MIXTURES 109

We should now plot xA/(1 xA)against VA and estimate the integral. For the present purpose we

integrate up toVA(0.5, 0.5) = 74.06 cm3

mol1

[Fig. 5.5], and use the data to construct the followingtable.

VA(cm3 mol1) 74.11 73.96 73.50 72.74

xA 0.60 0.40 0.20 0

xA/(1 xA) 1.50 0.67 0.25 0

The points are plotted in Fig. 5.5, and the area required is 0.30. Hence,

1.5

1.0

0.5

072 73 74 75

xA

/(1xA

)

VA,m/cm3 mol1

( )

Figure 5.5

V(CHCl3; 0.5, 0.5) =80.66 cm3 mol1 0.30 cm3 mol1

= 80.36 cm3 mol1 .

P5.21 = ln aA

r.

Therefore, d = 1

rd ln aA+

1

r2ln aAdr,

d ln aA =1

rln aAdr rd.

From the GibbsDuhem equation, xAdA = xBdB = 0, which implies that (since = +

RTln a, dA = RTd ln aA, dB =RTd ln aB)

d ln aB = xA

xBd ln aA =

d ln aA

r

= 1

r2ln aAdr=d[from(b)] =

1

rdr=dr=d[from(a)]

= d ln r+d.

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Subtract d ln rfrom both sides, to obtain

d lnaB

r=( 1) d ln r+d =

(1)

rdr+d.

Then, by integration and noting that lnaB

r

r=0

=lnBxB

r

r=0

=ln (B)r=0 =ln 1= 0,

lnaB

r= (0)=

r0

1

r

dr .

P5.23 A(s) A(l).

A(s) = A(l)+RTln aA

andfusG = A(l)

A(s)= RT ln aA.

Hence, ln aA =fusG

RT.

d ln aA

dT=

1

R

d

dT

fusG

T

=

fusH

RT2 [GibbsHelmholtz eqn].

ForT = Tf

T, dT = dTand

d ln aA

dT=

fusH

RT2

fusH

RT2f

.

ButKf =RT2

fMA

fusH.

Therefore,

d ln aA

dT=

MA

Kfand d lnaA =

MAdT

Kf.

According to the GibbsDuhem equation

nAdA+ nBdB =0

which implies that

nAd ln aA + nBd ln aB =0[= +RTln a]

and hence that d ln aA = nB

nAd ln aB.

Hence,d ln aB

dT=

nAMA

nBKf=

1

bBKf[fornAMA =1kg]

We know from the GibbsDuhem equation that

xAd ln aA+ xBd ln aB =0

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SIMPLE MIXTURES 111

and hence that d ln aA = xB

xA

d ln aB.

Therefore ln aA = xB

xAd ln aB

The osmotic coefficient was defined in Problem 5.21 as

= 1

rln aA =

xA

xBln aA.

Therefore,

=xA

xB

xB

xAd ln aB =

1

b

b0

b d ln aB =1

b

b0

b d ln b =1

b

b0

b d ln b+1

b

b0

b d ln

=1+1

b

b

0

b d ln .

From the DebyeHckel limiting law,

ln = Ab1/2 [A =2.303A].

Hence, d ln = 1

2Ab1/2dband so

=1+1

b

1

2A b

0

b1/2 db= 11

2

A

b

2

3b3/2 = 1

1

3A 1/2 .

COMMENT. For the depression of the freezing point in a 1,1-electrolyte

ln aA =fusG

RT+

fusG

RT

and hencer =fusH

R

1

T

1

T

.

Therefore, =fusHxA

RxB

1

T

1

T

=

fusHxA

RxB

T T

TT

fusHxAT

RxBT2

fusHT

RbBT2MA

where =2. Therefore, sinceKf =MRT2

fusH

,

=T

2bBKf.

Solutions to applications

P5.25 In this case it is convenient to rewrite the Henrys law expression as

mass of N2 = pN2 mass of H2OKN2 .

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(1) AtpN2 =0.784.0 atm=3.1 atm,

mass of N2 =3.1atm100 g H2O0.18g N2/(g H2O atm) = 56g N2 .

(2) AtpN2 =0.78 atm, mass of N2 = 14 g N2 .

(3) In fatty tissue the increase in N2concentration from 1 atm to 4 atm is

4(5614)g N2 = 1.7102g N2 .

P5.27 (a) i = 1 only, N1 =4, K1 =1.0107 dm3 mol1,

[A]=

410dm3 mol1

1+10dm3mol1 [A].

The plot is shown in Fig. 5.6(a).

Figure 5.6(a)

(b) i = 1; N1 =4, N2 =2; K1 =1.0105 dm3 mol1 =0.10 dm3 mol1,

K2 =2.0106 dm3 mol1 =2.0 dm3 mol1.

[A]=

40.10 dm3mol1

1+0.10 dm3 mol1 [A]+

22.0 dm3 mol1

1+2.0 dm3 mol1 [A].

The plot is shown in Fig. 5.6(b).

P5.28 By the vant Hoff equation [5.40],

= [B]RT =cRT

M.

Division by the standard acceleration of free fall,g, gives

8=

c(R/g)T

M.

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SIMPLE MIXTURES 113

Figure 5.6(b)

(a) This expression may be written in the form

=cRT

M,

which has the same form as the vant Hoff equation, but the unit of osmotic pressure () is now

force/area

length/time2 =

(mass length)/(area time2)

length/time2 =

mass

area.

This ratio can be specified in g cm2. Likewise, the constant of proportionality (R) would have the

units ofR/g,

energyK 1 mol1

length/time2 =

(mass length2/time2)K1 mol1

length/time2 =mass lengthK1mol1.

This result may be specified in g cm K1 mol1 .

R =R

g=

8.31447J K1 mol1

9.80665m s2

=0.847840kg m K1 mol1

103

gkg

102

cmm

R =84784.0 g cm K1 mol1 .

In the following we will drop the primes giving

=cRT

M

and use theunits of g cm2 and theRunits g cm K1mol1.

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(b) By extrapolating the low concentration plot of/cversus c (Fig. 5.7 (a)) to c = 0 we find the

intercept 230g cm2/g cm

3. In this limit vant Hoff equation is valid so

RT

M=intercept orM =

RT

intercept,

M =(84784.0gcmK1 mol1) (298.15K)

(230gcm2)/(g cm3),

M=1.1105 g mol1 .

500

450

400

350

300

250

200

0.000 0.010 0.020 0.030 0.040

Figure 5.7(a)

(c) The plot of/cversus c for the full concentration range (Fig. 5.7(b)) is very nonlinear. We may

conclude that the solvent is good. This may be due to the nonpolar nature of both solvent and solute.

(d) /c = (RT/M)(1+Bc+ Cc2).

SinceRT/Mhas been determined in part (b)by extrapolation to c =0, it is best to determine thesecond and third virial coefficients with the linear regressionfit

(/c)/(RT/M)1

c= B +Cc,

R =0.9791.

B =21.4 cm3 g1,

C =211cm6 g2,

standard deviation =2.4 cm3 g1.

standard deviation =15 cm6 g2.

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SIMPLE MIXTURES 115

7000

6000

5000

4000

3000

2000

1000

0

0.3000.2500.2000.1500.1000.0500.00

Figure 5.7(b)

(e) Using 1/4 forgand neglecting terms beyond the second power, we may write

c

1/2=

RT

M

1/2(1+

1

2Bc).

We can solve forB; theng(B)2 =C,

c

1/2RT

M

1/21=

1

2Bc.

RT/Mhas been determined above as 230 g cm 2/g cm3. We may analytically solve for B from

one of the data points, say,/c = 430 g cm2/g cm3 atc =0.033g cm3.

430gcm2/g cm3

230gcm2/g cm3

1/21=

1

2B (0.033 g cm3).

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B =2 (1.3671)

0.033 g cm3

=22.2 cm3 g1.

C = g(B)2 =0.25 (22.2 cm3 g1)2 =123 cm6 g2.

Better values ofB andC can be obtained by plotting

c

1/2/

RT

M

1/2againstc. This plot is

shown in Fig. 5.7(c). The slope is 14.03 cm3 g1. B =2 slope = 28.0 cm3 g1 . C is then

196 cm6 g2 . The intercept of this plot should theoretically be 1.00, but it is in fact 0.916 with a

standard deviation of 0.066. The overall consistency of the values of the parameters confirms thatg

is roughly 1/4 as assumed.

6.0

5.0

4.0

3.0

2.0

1.0

0.0

0.00 0.05 0.10 0.15 0.20 0.25 0.30

n

Figure 5.7(c)

05 Atkins Chap05

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