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Chapter 9/ Ideal and Real Solutio Pter 9/ Ideal and Real So u 1 Pm =ZAP; 41—251); The minimum pressure results if ZA = yA. Using Equation 9.12, * =1: PAPB L
4 €3.03 miss his: 33:5 aez {1.925
* ~ 0.605 b 1— 0.5560 X33.
R4 : Plot!!! yBBma/ = ar X( ) = 1 bar 3 .
xA 0700 . Therefore, the Henry’s law constant in
P}; : PﬁxAyA —xA) : 0.311 barx(0.700><0.360—0.700) =1‘29 bar
(xA —1)yA (0.700—1)><0.360 N. Lewis and H. Storch, J. American Chemical Society 39 (1917), 2544]. Use these data
method to determine the Henry’s law constant for B and a graphical law constant for Br; in CCl4 at 25°C in terms of molallty.
r2 in CC14 at 25°C PBr2 (Torr) 5.43
9.57
9.83
10.27 218 219 4 " “ 2 l S mlSCible
liquids A and B are fOr x XB f01‘ < C and complete
9 9 TWO I .
L 23) C Sk tch the phase diagram showing as much information as you can from these
0° . e r , r 1,__T> 75. . _. . .. _. _. .. .. . . 2
mg“ (maul kg’i} The best ﬁt line in the plot is VIDA.”2 (Torr) = 61.9 111372 — 0.0243 . Therefore, the Henry’s law constant in terms of molality is 61.9 Torr (mol kg'l)'l. L “ ' ' ex' 0 er a
P9 22 l‘he d  ' format on 1ven. It may ist V
' . 7 _3 . ‘ 0 he hape of th 2 hase region cannot be determined from the m 1 g
. ) ensrtles of pure water and ethanol are 997 and 789 kg m , respectrvely, [he pamal m lat T S e p L mol—I, respectively. Calculate the change in A is
_ , . . , The vapor pressure of pure
 ' ‘ in 11 uids A and B at 298 K
volume relative to the pure components when 2.50 L of a solution with xemmwl = 0.35 is prepared. P9.24) An “1631 5011mm} IS formed by mlx g q . f A in the Vapor is 0,320, What is the mole
_  r 136 Torr and that of pure B is 93.7 Torr If the mole fraction 0
V : nHzO VH20 + "ethanol Vet/“mol _ ’
V . 7 _ w 3 I 3 1 _ fraction of A in the solution?
n = xH20VH20 + xethanolI/ethanol : x17‘8 X 10— L mol + X10“ Lmor y 13* Torr = total _ A B 2 . 1 x/l — Pa. +(P* _P:)yA 136 Torr + (93.7 Torr—136 Torr)><0 3
= 00309 L mor A B  ' f water. The vapor
"mm! P9 25) A solution is prepared by dissolving 89.6 g of a nonvolatrle solute 1n 175 g o T “h.
' ' 23.76 orr a IS
ntorgl = 0 03%;: L 1.1 2 80.9 mol 2 nHZO + "ethanol pressure above the solution is 20.62 Torr and the vapor pressure of pure water 1S
. mo
‘ ' lute?
ture. What 1s the mass of the SO
xethanol : "ethanol 2 0'35 "H20 : 52 6 m 01 naham), = mol , tempera
xHZO 17,120 0.650 PHZO 20.62 Torr _ 0 868
x = . = ﬂ—__ ’ '
Va, I = n h I Methanol +n MH20 H20 PHZO Torr
1 ea e1 ano H20
pe ano p 2 nxo ule .
46017110“3 k [110‘1 18 02 103 k r1 ' :0'1322 n in 0’
YO u 3 H2
= 28.3 molx ‘ X am" + 52.6 molx—'~X—~§_TIEL = 2.60 L ' ”
789 kgm 998 kg m  mm 0 132 175 g
X _
AV=V_V;deaI L xsolute MHZO —— ‘ grnol1 mol
"solute : T  M = = 60.6 gmol‘l
1.48 mol 221
220 ' _ R ISolutions
Chapter 9/ Ideal and Real Solutions: pter 9/ Ideal and ea 2688.1
.612 —
_T_  55 .725
beaker of water at 298 K so that the bottom of the test tube is only slightly below the level of the water K
t290 0 K and a total pressure of 7325 Pa.
in the beaker The densny of water at this temperature IS 997 kg m‘3. After equilibrium is reached how ‘ , g i d6 all solution behavior, calculate Xbromo and ybmm" a '
sumln
high is the level of the water 1n the tube above that in the beaker? What IS the value of the osmotic 1 pressure? You may ﬁnd the approximation ln(1/(1+ X» z —x useful. Using Equation 9.39 and expressing the number of = 14877 Pa. chloro 300.0K, mm = 5719 Pa and P° , a _ Pf’
P l : xbromofhroma + xbromo) thoro
moles of solvent in terms of the densrty and height of « "0 the column of water, _ Bola] — chﬂom : : xlmmo = W 3564 Pa ~9301 Pa
L bromo C 0”"
751/”: + xsolmt : + = 0 y : Xbromopijmo : = solvent + nsucrose bmmO P 1 l ‘ i h Cous
10“ . .  4.255 mol are in t 6 gas
IDA/1 L 9 29) I an ideal solution of A and B, 2.75 mol are 1n the liquld Phase and ~~ 1
* * 1 ' n . _ = Calcu ate yA
pgh V’” + RTln pAh M : pgh V’" +RTln n M 2 0 phase The overall composition of the system IS ZA — 0.420 and M
+ n, , . 
M .SMCI‘OAB _
Expandlng the argument of the logarithmic term in a Taylor series lnm z —x A ( Z 8 4 x8 ) = 1112;,” (My _ ZB) ( 80 0 690) + 4 255 m0 1X 0 580
to! 1x 0.5 *  ‘ '
to: Z _x +nva orZB _ m0 _ "sue/223M = 0 yB : mgSL—J— — p Z vapor
h __ nsucroxeM RTnxucrme = — 2 * I : — = 0 r
V p Ang V Ma y —10509 ‘ _ . = 3790.
A d 1ehlorobutane form an ideal solutlon. At 273 K, Pchloro
8 314 Jm 1‘, K_1X298 KX 4.55><10‘3 kg p930) Assume that lbromobutane an I 0 630
. O ' . .  2  '
sent at 273 K, ychloro
= 0180216 kg “‘01 :131 m Pa and Pblmm z, 1394 pa. When only a trace of hquld IS Pre
997kgm'3><3.14x(2.50><102 m) x981 m s“2 1 tion
3 2 4 a_ Calculate the total pl‘CSSUI‘B abOVe the SO u ' .
7: = pg]? = 997kgm x981 ms x1.81 m = 1.77><10 Pa b calculate the mole fraction of 1_Ch10r0butane in the solutlon
P9 27) A volume of 13.75 L of air IS bubbled through liquid toluene at 298 K, thus reducmg the mass of
toluene in the beaker by 5.95 g Assuming that the C 070 61)
.1 _ BLoraRo/al ‘12::laro‘Pb:amo
P=M= :0 bar ych/WO— P (13* ~B)*m())
V total chloro r0 Pa
' 3790. Paxpm —3790. Fax
Pgm8) The vapor pressures of I'brOmObUtane and 1chlor0butane can be expressed 1n the form 0.630 = P ><(3790 Pat—1394 Pa)
total '
In Biro/no : _‘ 8
Pa 3790. Pa><1394 Pa = 2.32X103 Pa
1) ________________________________ total ‘— P3—1394 Pa) 223
222 Chapter 9/ Ideal and Real Solutions
b)
P ~ x P“ * total “ chloro chlom + — xchlaro )Bn‘omo utions The Scatchard plot of the data is:
x _ [Total .— 1317:0mo _ Pa Pa chlom * a: —' :3 11km — 11mm 3790. Pa — 1394 Pa * P P" P"
_ chloro total _ clzluro bromo ychloro __ :1: a:
P ( Cir/um _ Batohm ) total _3790 Pax23l7Pa—3790 Pa><1394 Pa 0630
~W: o
2317 Pax(3790 Pa—l394 Pa) y a «516800); + 114m 1‘ x __ ychluro [Jbromo
ch/aro _' =1: * *
chloro + (Pbromo _ [Jab/am chloro 0.630xl394 Pa
=M=0385
3790 Pa + (1394 Pa ~3790 Pa)><0.630 a ' ' .02 ' 7' arm ass ’V 1 X _ n p
I]; ( C 7/0r0 Chloro ) Vzpor (yr/1,0”)  hIOI'D ) lot to!
n _‘ vaporychloro + n/[q xchlnro _ + chloro 1 2.8 base triplets per binding site.
0’ = 0.501
"vapor + "liq m01+ mol ‘ ' — ' d t in Table 9.3
P9 32) Calculate the activity and activity coefﬁc1ent for CS; at x“2 — 0.573 usmg the a a P931) DNA is capable of forming complex helical structures. An unusual triplehelix structure of for both a Raoult’s law and a Henry’s law standard state.
poly(dA).2poly(dT) DNA was studied by P. V. Scaria and R. H. Shafer (Journal of Biological an : Pcsz Z M = 0.820
. . . . . . . . C51 Pi 512.3 Torr
Chemlstry, 266 (1991) 5417) where the intercalation of ethidium brom1de was studied usrng UV ('52
' R
absorption and circular dichroism spectroscopy. The following representative data were obtained using yR = 61052 = = 1.43
(332 2 0.573
the results of this study: XL“
arr = PCSz =W=0209
('52 kHSCSZ 2010Torr
H
751 = “€32 = 99.03 : 0.365
  . . . . LS2 xm 0.573
Usmg the above data, determme K and N for the binding of ethidium bromide to the DNA triple—helical " 2 ’ ' th boilin point by
t t P9 33) The dissolution of 5.25 g of a substance 1n 565 g of benzene at 298 K raises e g
3 rue ure. . ‘ ' g * =103 Torr at 298 K.
the pure solvent the osmotic pressure, and the “mar mass Of the SOlute' Pbmzm 225
224 Chapter 9/ Ideal and Real Solutions AT .
A]; : Klimsolule; msolute : b 2 O K K 2 53 K kg mol“ 7 0247 “‘01 kg].
17 o M: 5.25 g E = 37.6 gmol'1
0.247 mol kg x0565 kg A7} = «me = —5. 12 K kg mol‘1x0.247 mol kg‘1 = —1.26 K benzene _
h —
* benzene benzene "benzene + nsolule 565 g _ 78.11 g mol“ *
_ W —— 0.981 78 11 11 + 0.247 mol kg‘1x0.565 kg
. g mo _ "benzene 3
52510 kg _1 x8314 Jmor‘ K“ x298 K
_ anRT 37.6x10 kgmol 7“ =\6=537x105pa
V 565x10 kg 876.6 kgm'3 Si in Figure 9.26 from 300°C to 1300°C. The transition follows a path along a vertical line beginning at 40% Si composition and 300. °C. There would be no change in the solid until 363°C at which it would melt to give a liquid at the eutectic composition and sOlid Si. As the temperature increases beyond 363 °C, the liquid becomes enriched in — liquid + solid curve extending from the eutectic point to the 100% ' ' . about 820°C, all the Si melts and there is no further change in composition of the solution as ’ '
to 1300°C. Si, following the liquid We obtain the equations by replacing P; by k: . 226 lapter 9/ Ideal and Real Solutions rzewn—nb. ;[email protected]—aﬂ P9 36) Describe the changes you would observe as the temperature of a mixture of phenol and water at . . , . t of
point a in Figure 9.21 is increased until the system is at p01nt a. How does the relative amoun . . . . 9
separate phases of phenolrich liquid L2 and water rich liquid L1 change along this path.
As T increases the number of moles in separate phases decreases until the system consrsts of a Single hase at point a’ The number of moles of each of the two immiscible solutions can be determined at any
P  . temperature by applying the lever rule to the appropriate tie line. For example at pomt a, molsL1><(x2 —xl)=mols L2 ><(x3 —x2). P9 37) Describe the changes you would observe as the temperature of a mixture of triethylamme and ‘ ' ‘ " ' mount
3 water at point a in Figure 9.22 is increased until the system 1S at pomt a . How does the relative a ‘ of separate phases of triethylamine and water change along this path? . . . . int
At point a the two components are completely misc1ble, and phase separation does not occur until p0 ' ' ' arate hases increases
a’ is reached. As the temperature increases further, the amount of material in sep p continuously The relative amounts of L1 and L2 can be determined at each temperature usmg the lever ’ — = — x .
rule and the appropriate tie line. For example at pornt a, mols L1 x (xa. x“) mols L2 x (x,2 a ) P938) Describe the changes in a beaker containing water and butanol that you would observe along the
path a —> b —> c in Figure 19.24b. How would you calculate the relative amounts of different phases present along the path? ' . . Initially, the temperature of L2 is increased with some change in its composition as indicated by the left
boundary of the L2 area. At T b, boiling begins. If the vapor is collected, it has the composrtion of pomt b.
If this vapor is cooled, it separates into L1 and L2 phases at 940C. The mole fraction .of butanol in these
phases can be determined from the Z values corresponding to points g and h, respectively. As the
mixture is cooled further to point 0, liquid L1 becomes more concentrated in water and L2 becomes more ' ture usin
Concentrated in butanol. The relative amounts of L1 and L2 can be determined at each tempera g 227 Chapter 10: Electrolyte Solutions mols Ll x(xc —xd)=mols L2 ><(x —xc). e i ' ' 'n he
1 numbers in italics indicate t at t e solutlon is included 1 t
19 em h h
em ’5 Solutions Manual. istionS 0“ Concepts
present along the path?  9 ( )
i _ becomes less concentrated in water and L2 y i e incre ed.
i cons . I ll! ()l an elaCUOl t6 ()1 1
Electric 9 ut ()II a] as ' reads 0
‘ T increases the screening length. The random thermal motion sp ut the cloud of e potential attracting the counter ion j curve as the temperature increases. At point j, nearly the Whole ‘ weaker, causing it to spread out more.
System is in the vapor phase, with a trace of liquid having the composition k. At point __ _ ) Increasing gr (0) Increasing ionic strength decreases ) + Cl_(aq), which can be As dis H+
lated to the Gibbs energy for the reaction 1/2H2(g) + l/2C12(g) —> (aq
__ re  ° + 1272 kJ mol“.
. 0 ﬂ 0 > 9a + AGso va ion 7 aq) +
 determined experimentally by th€ equatlon AGR “‘ AGsoIvauun( 9') 1 I
present along the path? 3 C ( ) e~Hiickel limiting law always less than
As the temperature is increased from point f, liquid L1 becomes less concentrated in water and L2 Q10 3) Why are activity coefﬁcients calculated using the Deby
becomes less concentrated in butanol. At 94° C, the two liquids present having the composition given by one?
h and g coexist with the vapor phase, which has the composition 1’. As T increases further to point j, e—Hﬁckel limiting law are always less than one because
the Activity coefﬁcients calculated using the Deby  u . . . l . .1 1 l . . . 5 C l 5 rather than repulsive. . 1 P9.41) Describe the system at points a and c in Figure 19.25b. How would you calculate the relative __ [um/me v+lu+ + V4“.
amounts Of diffﬂem phases present at these points? [Hi V v m ' ' ’ it ?
Q10 5) How is the chemical potentlal of a solute related to its activ y
Point a corresponds to essentially pure solid A in equilibrium with a solution whose c0 position is  — 0 RT In a
. . . . . . ., : ° RTlna+ and ,LL—IL+
given by pomt b. Pomt c corresponds to essentially pure SOlld B in equilibrium with a solution whose ’u+ ’u+ + ’ ' ' chemical potential gives
. i I ' i i i i ' ‘ s in terms of the mean ionic
Composmon IS given by pomt c. The relative amounts of pure B and the llquld can be determined at each Expressmg the relation temperature using the lever rule and the appropriate tie line. For example at point c, #i = y; + RTlna:
mols A x (x6, — O) 2 mols Liquid x (xd — x6). 229
228 ...
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 Vapor pressure, Vapor, Water vapor

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