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Homework Chem340 9
P8.3) An ideal solution is formed by mixing liquids A and B at 298 K. The vapor
pressure of pure A is 180. Torr and that of pure B is 82.1 Torr. If the mole fraction of A
in the vapor is 0.450, what is the mole fraction of A in the solution?
The mol fraction of A in the ideal solution of liquids A and B is given by:
y A p
0.450 82.1 Torr
B
xA
0.272
p A p B p A y A 180 Torr 82.1 Torr 180 Torr 0.450
P8.4) A and B form an ideal solution. At a total pressure of 0.900 bar, yA = 0.450 and xA
= 0.650. Using this information, calculate the vapor pressure of pure A and of pure B.
Ptotal x A Pa* yB Ptotal
Pa*
Ptotal yB Ptotal 0.900 bar 0.450
0.623 bar
xA
0.650
xA
y A PB*
*
Pa* PB* PA y A
0.650
0.450 PB*
*
*
PA 0.450 PB* PA
PB* 0.650 1 0.450
2.27
*
PA 0.450 1 0.650
PB* 1.414 bar
*
P8.5) A and B form an ideal solution at 298 K, with xA = 0.600, PA 105 Torr, and
*
PB 63.5 Torr.
a. Calculate the partial pressures of A and B in the gas phase.
b. A portion of the gas phase is removed and condensed in a separate container.
Calculate the partial pressures of A and B in equilibrium with this liquid sample at
298 K.
a) Calculate the partial pressures of A and B in the gas phase.
*
PA x A PA 0.600 105 Torr = 63.0 Torr
PB 1 x A PB* 0.400 63.5 Torr = 25.4 Torr
b) A portion of the gas phase is removed and condensed in a separate container.
Calculate the partial pressures of A and B in equilibrium with this liquid sample at 298 K.
The composition of the initial gas is given by
yA
PA
63.0 Torr
0.713;
PA PB 88.4 Torr
yB 0.287
For the portion removed, the new x A and xB values are the previous y A and y B values.
*
PA x A PA 0.713 105 Torr = 74.9 Torr
PB 1 x A PB* 0.287 63.5 Torr = 18.2 Torr
P8.7) Assume that 1-bromobutane and 1-chlorobutane form an ideal solution. At 273 K,
*
*
Pchloro 3790 Pa and Pbromo 1394 Pa. When only a trace of liquid is present at 273 K,
ychloro = 0.75.
a. Calculate the total pressure above the solution.
b. Calculate the mole fraction of 1-chlorobutane in the solution.
c. What value would Zchloro have in order for there to be 4.86 mol of liquid and 3.21
mol of gas at a total pressure equal to that in part (a)? [Note: This composition is
different from that of part (a).]
a) Calculate the total pressure above the solution.
ychloro
*
*
*
Pchloro Ptotal Pchloro Pbromo
*
*
Ptotal Pchloro Pbromo
0.75
Ptotal
3790 Pa Ptotal 3790 Pa 1394 Pa
Ptotal 3790 Pa 1394 Pa
3790 Pa 1394 Pa
2651 Pa
3790 Pa 0.75 3790 Pa 1394 Pa
b) Calculate the mole fraction of 1-chlorobutane in the solution.
*
*
Ptotal xchloro Pchloro 1 xchloro Pbromo
xchloro
*
Ptotal Pbromo
2651 Pa 1394 Pa
0.525
*
*
Pchloro Pbromo 3790 Pa 1394 Pa
c) What value would Zchloro have in order that there are 4.86 moles of liquid and 3.21
moles of gas at a total pressure equal to that in part a)? (This composition is different than
that in part a.)
*
*
Pc* Ptotal Pchloro Pbromo
hloro
ychloro
*
*
Ptotal Pchloro Pbromo
xchloro
3790 Pa 2651 Pa 3790 Pa 1394 Pa
0.750
2651 Pa 3790 Pa 1394 Pa
*
Pchloro
*
ychloro Pbromo
*
*
Pbromo Pchloro ychloro
0.750 1394 Pa
0.525
3790 Pa + 1394 Pa 3790 Pa 0.750
tot
tot
nliq Z chloro xchloro nvapor ychloro Z chloro
Z chloro
tot
tot
nvapor ychloro nliq xchloro
n
tot
vapor
n
tot
liq
3.21 mol 0.750 + 4.86 mol 0.525
0.614
3.21 mol + 4.86 mol
P8.10) At 31.2C, pure propane and n-butane have vapor pressures of 1200 and 200
Torr, respectively.
a. Calculate the mole fraction of propane in the liquid mixture that boils at 31.2C at
a pressure of 760 Torr.
b. Calculate the mole fraction of propane in the vapor that is in equilibrium with the
liquid of part (a).
a) At the boiling point the vapor pressure of the mixture is equal to the external pressure:
ptot = pext = 760 Torr
Again, solving p tot x prop p 1 x prop p for xA yields for the mol fraction of
prop
but
propane in the solution:
x prop
p
p
tot
prop
p
760 Torr 200 Torr 0.560
but
1200 Torr 200 Torr
p but
b) The mol fraction of propane in the gas phase is:
ybut
p p tot p p
but
but prop
p tot p p
but
prop
200 Torr 760 Torr 200 Torr 1200 Torr 0.116
760 Torr 200 Torr 1200 Torr
y prop 1 ybut 0.884
P8.18) The partial molar volumes of water and ethanol in a solution with x H O 0.60 at
2
25C are 17.0 and 57.0 cm3 mol1, respectively. Calculate the volume change on mixing
sufficient ethanol with 2 mol of water to give this concentration. The densities of water
and ethanol are 0.997 and 0.7893 g cm3, respectively, at this temperature.
V nH 2O V H 2O nEt V Et
V H 2O 17.0 cm3 mol-1 and V Et 57.0 cm3 mol-1
nH 2O 2.00 and xH 2O
nH 2 O
nH 2O nEt
0.600
2 mol
0.600; nEt 1.333
2 mol nEt
The total mixed volume is given by
Vmixed nH 2O V H 2O nEt V Et
2.00 mol 17.0 cm3 mol-1 1.333 mol 57.0 cm3 mol-1
109.98 cm3
Vunmixed nH 2O
M H 2O
H O
nEt
2
M Et
Et
2.00 mol 18.02 g mol
1cm -3
0.997 g
+1.333 mol 46.07 g mol 1
1cm -3
0.7873 g
36.15 cm3 + 78.00 cm3
114.15 cm3
V Vmixed Vunmixed 109.98 cm3 114.15 cm3 4.2 cm3
P8.20) The heat of fusion of water is 6.008 103 J mol1 at its normal melting point of
273.15 K. Calculate the freezing point depression constant Kf.
Kf
2
RM solventT fusion
H fusion
8.314 J mol-1K -1 18.02 10-3 kg mol-1 273.15 K
2
6.008 103 J mol-1
K f 1.86 K kg mol-1
P8.22) A sample of glucose (C6H12O6) of mass 1.25 g is placed in a test tube of radius
1.00 cm. The bottom of the test tube is a membrane that is semipermeable to water. The
tube is partially immersed in a beaker of water at 298 K so that the bottom of the test tube
is only slightly below the level of the water in the beaker. The density of water at this
temperature is 997 kg m3. After equilibrium is reached, how high is the water level of
the water in the tube above that in the beaker? What is the value of the osmotic pressure?
You may find the approximation ln
1
x useful.
1 x
*
*
Vm RT ln xsolvent Vm RT ln
nsolvent
0
nsolvent nsucrose
Ah
*
ghVm RT ln
Ah
M
M
nsucrose
*
ghVm RT ln
1
0
nsucrose M
1
Ah
Expanding the argument of the logarithmic term in a Taylor series, ln
*
ghVm RT
h
1
x
1 x
nsucrose M
0
Ah
RTnsucrose M
*
2 AgVm
RTnsucrose
Ag
1.25 10-3 kg
0.18016 kg mol-1
2.37 m
997 kg m-3 3.14 x10-4 m 2 9.81 m s -2
8.314 J mol-1K -1 298 K
gh 997 kg m -3 9.81 m s-2 2.37 m = 2.32 104 Pa
P8.24) To extend the safe diving limit, both oxygen and nitrogen must be reduced in the
breathing mixture. One way to do this is to mix oxygen with helium. Assume mixture
of a 10.% oxygen and 90.0% helium,. Assuming Henrys law behavior, calculate the levels
of oxygen and helium in the blood of a diver at 100 m. Assume T = 298 K.
p gh 101325Pa 1000kg m 3 9.81m s 2 100m 101325Pa 1.081106 Pa
pO2 10% 1.081 106 Pa 1.08 105 Pa
xO2
PO2
O
kH2
1.08 105 Pa
2.18 105
4
5
4.95 10 bar 10 Pa / bar
P8.32) Calculate the activity and activity coefficient for CS2 at xCS 0.7220 using the
2
data in Table 8.3 for both a Raoults law and a Henrys law standard state.
R
aCS2
R
CS2
PCS2
*
CS2
P
R
aCS2
xCS2
446.9 Torr
0.8723
512.3 Torr
0.8723
1.208
0.7220
H
aCS2
H
CS2
PCS2
k H ,CS2
H
aCS2
xCS2
446.9 Torr
0.2223
2010 Torr
0.2223
0.3079
0.7220
P8.36) Assume sucrose and water form an ideal solution. What is the equilibrium vapor
pressure of a solution of 2.0 grams of sucrose (molecular weight 342g mol1) at T = 293
K if the vapor pressure of pure water at 293 K is 17.54 Torr. What is the osmotic pressure
of the sucrose solution versus pure water? Assume 100.0 g of water.
The equilibrium water pressure is given by:
p solvent
n solute
x solvent 1 x solute 1
*
n solute n solvate
p solvent
m sucrose / M sucrose
n solute
p*
p solvent p *
solvent 1
solvent 1
n solute n solvate
m water / M water m sucrose / M sucrose
2.0 g / 342 g mol -1
17.54 Torr 1
100 g / 18.02 g mol -1 2.0 g / 342 g mol -1
17.52 Torr
The osmotic pressure is:
n solute RT 2.0 g / 342 g mol -1 8.314472 J K mol -1 293 K
142464.3 Pa 1068.57 Torr
V
0.0001 m 3
P8.45) The concentrations in moles per kilogram of water for the dominant salts in sea
water are:
Ion
Mol kg1
Cl
Na+
Mg2+
SO2
4
Ca2+
0.546
0.456
0.053
0.028
0.010
Calculate the osmotic pressure exerted by sea water at T = 298 K. Suppose sea water is
separated from pure water by a membrane that is permeable to water but impermeable to
the ions in sea water. Assuming the density of sea water is about equal to pure water at T
= 298 K, calculate the column of sea water that would be supported by osmotic pressure.
For 1 kg of sea water (1 L) the osmotic pressure is:
n solutes RT 0.546 0.456 0.053 0.028 0.010 mol 8.314472 J K mol -1 298 K
V
0.001 m 3
2708139.93 Pa 27.08 bar
The column of sea water that would be supported by this osmotic pressure can be
calculated as follows:
p 1 atm g h
h
1 atm 2708139.93 Pa 101325 Pa 265.73 m 0.266 km
g
1000 kg m 9.81 m s
Q1 (modified from P8.15)
-3
-2
(a) Plot P vs. XA , extrapolate to XA=0, P*B=266.8 Torr and
XA=1, P*A=91.9 Torr.
260
240
P=266.78547-174.83655xXA
220
P/Torr
200
180
160
XA=0, P* =266.8 Torr
B
140
XA=1, P* =91.9 Torr
A
120
100
0.0
0.1
0.2
0.3
0.4
0.5
XA
0.6
0.7
0.8
0.9
1.0
(b)
aA
PA y A Ptotal 0.0175 257.9 Torr
0.0820
*
*
55.03 Torr
PA
PA
, the
a A 0.0820
1.67
A
xA 0.0490
same calculation can be performed to get activity coefficients of water and ethanol at
different compositions. The results are shown below.
For example, XA=0.0490,
xA
0.0490
0.3120
0.4750
0.6535
0.7905
xA
0.0490
0.3120
0.4750
0.6535
0.7905
P (Torr)
257.9
211.3
184.4
156.0
125.7
yA
0.0175
0.1090
0.1710
0.2550
0.3565
yA
0.0175
0.1090
0.1710
0.2550
0.3565
P(Torr)
257.9
211.3
184.4
156.0
125.7
aA
0.082
0.419
0.573
0.723
0.814
aB
0.991
0.737
0.598
0.455
0.316
A
1.67
1.34
1.21
1.11
1.03
B
1.04
1.07
1.14
1.31
1.55
P8.16) The partial pressures of Br2 above a solution containing CCl4 as the solvent at
25C are found to have the values listed in the following table as a function of the mole
fraction of Br2 in the solution [G. N. Lewis and H. Storch, J. American Chemical Society
39 (1917), 2544]. Use these data and a graphical method to determine the Henrys law
constant for Br2 in CCl4 at 25C.
xBr
P (Torr)
2
xBr
P (Torr)
2
0.00394
1.52
0.0130
5.43
0.00420
1.60
0.0236
9.57
0.00599
2.39
0.0238
9.83
0.0102
4.27
0.0250
10.27
The best fit line in the plot is PBr2 Torr 413 xBr2 0.063. Therefore, the Henrys law
constant in terms of mole fraction is 413 Torr.
P9.1) Calculate Hreaction and Greaction for the reaction AgNO3(aq) + KCl(aq)
AgCl(s) + KNO3(aq).
H
reaction and G reaction are given by:
H
reaction H f AgCls H f K aq H f NO 3 aq
H K aq H Cl aq H Ag aq H NO 3 aq
f
f
f
f
1
1
1
H
reaction 127.0 kJ mol 252.4 kJ mol 207.4 kJ mol
252.4 kJ mol 1 167.2 kJ mol 1 105.6 kJ mol 1 207.4 kJ mol 1
65.4 kJ mol 1
G
reaction G f AgCls G f K aq G f NO 3 aq
G K aq G Cl aq G Ag aq G NO 3 aq
f
f
f
f
1
1
1
G
reaction 109.8 kJ mol 283.3 kJ mol 111.3 kJ mol
283.3 kJ mol1 131.2 kJ mol1 77.1 kJ mol1 111.3 kJ mol1
55.7 kJ mol1
P9.2) Calculate Hreaction and Greaction for the reaction Ba(NO3)2(aq) + 2KCl(aq)
BaCl2(s) + 2KNO3(aq).
Greaction = G f (BaCl2, s) + 2 G f (K+, aq) + 2 G f (NO3, aq) G f (Ba2+, aq)
2 G f (NO3, aq) 2 G f (K+, aq) 2 G f (Cl, aq)
Greaction = G f ( BaCl2, s) G f ( Ba2+, aq) 2 G f (Cl, aq)
Greaction = 806.7 kJ mol1 + 560.8 kJ mol1 + 2 131.2 kJ mol1 = 16.5 kJ mol1
H reaction = H f (BaCl2, s) H f ( Ba2+, aq) 2 H f (Cl, aq)
H reaction =855.0 kJ mol1 + 537.6 kJ mol1 + 2 167.2 kJ mol1 = 17.0 kJ mol1
P9.4) Calculate Sreaction for the reaction Ba(NO3)2(aq) + 2KCl(aq) BaCl2(s) +
2KNO3(aq).
Sreaction = S (BaCl2, s) S (Ba2+, aq) 2 S (Cl, aq)
Sreaction = 123.7 J K1 mol19.6 J K1 mol1 2 56.5 J K1 mol1 = 1.1 J K1 mol1
P9.5) Calculate Greaction in an aqueous solution for Cl(aq) using the Born model. The
radius of the Cl ion is 1.81 1010 m.
Gsolvation
Z 2 e2 N A 1
1
8 r r
r for water at 298 K is 78.5
For water at 25C
Gsolvation
z2
6.86 104 kJ mol-1
r/pm
Cl radius is 181 pm
1
Gsolvation
6.86 104 kJ mol1
181
379 kJ mol-1
P9.8) Express a in terms of a+ and a for (a) Li2CO3, (b) CaCl2, (c) Na3PO4, and (d)
K4Fe(CN)6. Assume complete dissociation.
1
1
v
v
2
a) Li2CO3 a a a a a 3
1
1
v
v
2
b) CaCl2 a a a a a 3
c) Na3PO4 a a a
v
1
v
a a
3
1
1
4
1
v
v
4
d) K4Fe(CN)6 a a a a a 5
P9.9) Express in terms of + and for (a) SrSO4, (b) MgBr2, (c) K3PO4, and (d)
Ca(NO3)2. Assume complete dissociation.
a) SrSO4
1
2
b) MgBr2
3
c) K3PO4
2
1
1
3
4
2
d) Ca(NO3)2
1
3
P9.14) Calculate the DebyeHckel screening length 1/ at 298 K in a 0.00100 m
solution of NaCl.
For water, the screening length at 298 K in m-1 can be calculated as:
2.91 1010
I/ mol kg -1 solvent
9.211 10 8
r
1
9.65 10 9 m 9.65 nm
0.00100.997
78.54
m 1 1.04 10 8 m 1

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Chapter 23:Answers to Concept Review Questions1. In response to the U.S. senators description as an electronic pyramid scheme, it can be saidthat derivative securities can be very risky, and they can be used for speculative purposes.However, derivativ

Park - FI - 360DLB

Chapter 1:Answers to Concept Review Questions1. Many companies have connections between other functional areas and finance. For example,any company with international dealings must look at the impact of foreign exchange on itsbusiness. Does the firm g

Park - FI - 360DLB

Mid-Term (WK5)1. The ultimate owner(s) of an ongoingcorporation are (Points: 2) (CH1, P18)the federal government.the debt holders.the equity holders.the executive staff of the corporation.2. Which of the following is a valid criticismconcerning th

Park - FI - 360DLB

FI360 Final Exam Study Guide1. Maximizing profit as the managers goal has several flaws, including:a. Earnings figures focus on past performance rather than current or future performanceb. The timing of the profits may be ignoredc. Focusing on earning

Park - FI - 360DLB

FI360 Final Review Study GuideThe final is 42 multiple choice questions of which you will need to answer 33. Youget to choose which 33 you answer. The subject matter listed below will help youtremendously to prepare for the final. NOTE: I will only gra

Park - FI - 360DLB

GlossaryGlossaryChapter 1agency costsCosts that arise due to conflicts of interest between shareholders and managers.agencyproblemsThe conflict between the goals of a firm's owners and its managers.boards ofdirectorsElected by shareholders to be

DeVry Sacramento - FINANCE - FI560

Disney Analysis 1Analysis of the Walt Disney Company (DIS)Dalma GomezFI560August 16, 2011Disney Analysis 2AbstractThe Walt Disney Company has been able to gain recognition and is known across theworld through its Disney movies like The Little Merm

JFK - 2 - 2

Zhejiang University - EE - math 402

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Zhejiang University - EE - math 402

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Zhejiang University - EE - math 402

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Zhejiang University - EE - math 402

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Zhejiang University - EE - math 402

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Zhejiang University - EDUCATION - 212

Zhejiang University - EDUCATION - 212

Zhejiang University - EDUCATION - 212

Zhejiang University - EDUCATION - 212

Zhejiang University - EDUCATION - 212

Zhejiang University - EDUCATION - 212

Zhejiang University - EDUCATION - 212

Zhejiang University - EDUCATION - 212

Zhejiang University - EDUCATION - 212

Zhejiang University - EDUCATION - 212

Zhejiang University - EDUCATION - 212

Zhejiang University - EDUCATION - 212

Zhejiang University - EDUCATION - 212

Zhejiang University - EDUCATION - 212

Zhejiang University - EDUCATION - 212

Zhejiang University - EDUCATION - 212

Zhejiang University - EDUCATION - 212

Zhejiang University - EDUCATION - 212

Zhejiang University - EDUCATION - 212