classes_fall08_110A-2ID25_110a_hw8_solns

classes_fall08_110A-2ID25_110a_hw8_solns - 11 If. P/torr...

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Unformatted text preview: 11 If. P/torr 215.0 220.0 230.0 CHEM 1 10A Fall 2007 8th HOMEWORK ASSIGNMENT Engel & Reid: Chapter 9: 4, 5, 7, 9, 13, 15, l6, 18, 21, 25, 28, 29. 1. The volume of two completely miscible liquids A and B can be represented by the expression V=a+bIA ~cgj +dxj, where a, b, c, and d are positive constants and [A is the mole fraction of component A. (Express all answers in terms of a, b, c, and d). a. What is the value of the molar volume of pure A ( V; ) and molar volume of pure B (V; )? b. Calculate the partial molar volume ofA (174) at [A = 0.1 (Assume n = HA + r23 = 1) 2. 0.00 0.00 Given the following data for the liquid vapor equilibrium of a two component (A,B) system at T= 400C: 0.06 0.15 0.30 0.44 0.56 0.64 0.73 0.80 0.89 0.94 0.97 1.00 0.032 0.081 0.18 0.31 0.42 0.52 0.63 0.73 0.84 0.93 0.96 1.00 249.0 277.0 287.0 301.2 316.0 327.0 343.0 353.0 355.0 356.0 Sketch and label the plot of PA and PB vs 1;. Include in the graph Raoult’s law behavior for component A and B. Calculate values of A in the Raoult’s law basis at I; = 0.18 and 0.84. Given this data and values for 7A , what can be said about the interaction between the constituent particles i.e., A—A, B—B, A-B interactions? 3: (see Chapter 10, Sections 3 and 4) a. Calculate the ionic strength of an aqueous solution containing 0.010 mol kg'1 of the strong electrolyte F eCl 3. Calculate the ionic strength of an aqueous solution containing 0.010 mol kg" A12(SO4)3 and 0.005 mol kg" NagSO4 at 2500. (Assume that the salts are strong electrolytes) Calculate the mean ionic molality mi for a 0.002 mol kg'] aqueous solution of the strong electrolyte LaClg. Given the experimental data below for HCl (in water and at 25°C): Mean Ionic Activity Coefficient m/m" 7: 0.001 0.965 0.01 0.904 0.1 0.796 1 0.809 10 10.44 i) Calculate the mean ionic activity at m/m0 = 10. ii) How does this compare with that predicte by the Debye— Huckel limiting law? Is this what you would expect? Explain. AZ‘ 3 (D V: ok*1o')l~~L)<h +459. «3 Eyreumk) ’X’t:l “JUL. cg gm EWL W4 WW Q) TLUK .5 q (634w; JJLva'cmen pron/z anmH‘s Lt.an Sb A~A a g—g {ARQAQXLQW‘ am 3(QA,+Q.(' mm M wwwmm. Sea (3 we Emwc QLK‘IOK Q) ¢’ V10 rev &x}al“13 ’ m : I I .. . 0..., .022 0 I I I T 0000000 I I I 0000900 I I I Mn.- 0 H H 000 0 000 0 000 000 0 H 0 H 0000000 0000000 000000.H 000000.H 00.000 0.0 00.00 00.0H 00.000 000 00.0 00.0 00.0 00.0 0H00000 0000000 000000.H 000000.H 0H.000 00.0H 0300 0H.H0 00.H00 000 00.0 00.0 00.0 00.0 00000H0 0000.0 000000.H 000000.H 00.000 0.00 00.000 00.00 00.000 000 0H0 00.0 HH.0 00.0 00H0000 H000000 0H000H.H 0H0000.H 00.0H0 00.00 00.000 0.00 0.H00 000 00.0 00.0 0.0 0.0 0000000 0000000 000000.H 000000H 00.000 00.00 00.000 00.00 00.000 0H0 00.0 00.0 00.0 00.0 0000000 000H00.0 000000.H 0H0H00.H 00.000 0.00H 0H.00H 000.00H 000.00H 0.80 00.0 00.0 00.0 00.0 0000000 H00H000 000H0.H 000000H 00.000 0.00H 00.00H 00.00H 00.00H 000 00.0 00.0 00.0 00.0 000H000 00000.0 000000.H 00000H.H H0000 00.00H 00.0: 0H.00H 00.H0H 000 00.0 H00 00.0 00.0 0000H00 H000000 0000000 0000H.H 00.000 0.00H 00.00 0.00H 0.00 000 00.0 0H0 0.0 0.0 0000000 H00000 0000000 H0000H.H H00.000 000.00H 000.00 0.00H 0.00 000 0H0.0 H000 00.0 0H0 00H00.0 0000000 0000000 00000H.H 0H0.0H0 0H.000 000.: 0.000 0.0H 000 000.0 000.0 00.0 00.0 H 0 H 0H0 0H0 0 0H0 0 0H0 H 0 H 0 0.002030 0.002030 0.080000 0005.03 0000.00 0000000 0000000 00 00 0 0.0x 0.0x >.000 flaw/«u 0 0. 50,0 50.0 V‘shfl3 fed“ Xiwc’g’l’d (aw, AW.) javfy ~ ~ L VB HWLJ E t‘Hu : _!.H’31"‘\F 'n - F(0W\ LK‘W‘xA-«C/vvé' (max = ~0me .— I‘ts cwpeo+bi, M$L‘\+L;ULQ,{ ngmjgw‘o) law 40¢; vLo‘r wDrL wle im‘l‘US case Menu/~91 «A,» (ow/L Shel/gm =(V). flex/‘1)» m («5+ 14Amo(avaqrox(L5 In LS is Law (I a)\ Work Sic,be [OH 99 33A w? KQCA (3M Chapter 9: Ideal and Real Solutions Problem numbers in italics indicate that the solution is included in the Student’s Solutions Manual. Questions Q9.1) Using the differential form of G, dG = VdP — SdT, show that if AGng = nRTZ x, In xi, then AH,”in = AVW = 0. AGmmg = nRTZxI. ln xi aAG . . ASWI." = — fl = ~nRZ xi ln xi and aAGmixin AVmixing : —_§'_ 2 0 BP T,n,,n2 AHmimg = AGmmng + MSW = nRTZ xi In x]. — T [nRZ x1. In L.) = 0 Q92) For a pure substance, the liquid and gaseous phases can only coexist for a single value of the pressure at a given temperature. Is this also the case for an ideal solution of two volatile liquids? No, an ideal solution of two volatile liquids can exist over a range of pressures that are limited by the pressure for which only a trace of liquid remains, and the pressure for which only a trace of gas remains. Q93) Fractional distillation of a particular binary liquid mixture leaves behind a liquid consisting of both components in which the composition does not change as the liquid is boiled off. Is this behavior characteristic of a maximum or a minimum boiling point azeotrope? This behavior is characteristic of a maximum-boiling azeotrope. After initially giving off the more volatile component, the liquid remaining tends to the composition of the maximum boiling point at intermediate composition. After the more volatile component has boiled away, the azeotrope evaporates at constant composition. Chapter 9/Ideal and Real Solutions (29.4) Why is the magnitude of the boiling point elevation less than that of the freezing point depression? The boiling point elevation is less than the freezing point depression because the chemical potential of the vapor is a much more steeply decreasing function of temperature than the solid, as seen in Figure 9.1121. This is due to the relation d,u = —SdT at constant P and the fact that the molar entropy of a vapor is much larger than that of a solid. When the Miun curve is displaced down by the addition of the solute (see Figure 9.11a), the intersection of the ,uliquid curve with the ,usolid curve and the fig”, curve dertermine the shift in the freezing and boiling temperatures. Because the magnitude of the slope of the lug,” curve is greater than that of the [use/id curve, T 1, moves up less than the T m moves down. Q95) Why is the preferred standard state for the solvent in an ideal dilute solution the Roualt's law standard state set? Why is the preferred standard state for the solute in an ideal dilute solution the Henry's law standard state? Is there a preferred standard state for the solution in which xmlm, = xmlm = 0.5? Raoult's law defines the appropriate standard state of the solvent in a dilute solution because in this limit the solute has only a weak effect on the solvent. Henry’s law defines the appropriate standard state for the solute at a low solute concentration. However, neither standard state adequately describes the standard state for xsalvem = xsolm = 0.5. In this case, either standard state can be used, but neither has a firm physical basis. Q9.6) Is a whale likely to get the bends when it dives deep into the ocean and resurfaces? Answer this question by considering the likelihood of a diver getting the bends if he or she dives and resurfaces on one lung full of air as opposed to breathing air for a long time at the deepest point of the dive. The whale is unlikely to get the bends on one lungful of air. There is not enough N2 present in a single lungful to yield a saturated solution of N2 in the blood. Breathing air for a long time at depth, however, will result in saturated blood for the diver. For this reason, the diver is more susceptible to getting the bends than the whale. Q9.7) The statement “The boiling point of a typical liquid mixture can be reduced by approximately 100°C if the pressure is reduced from 760 to 20 Torr” is found in Section 9.4. What figure(s) in Chapter 8 can you identify to support this statement in a qualitative sense? Figure 9.11(b) shows the P—T phase diagram of a pure substance (and a corresponding solution). By looking at the liquid—vapor coexistence curves, the dependence of boiling point on pressure can be traced. Qualitatively, it is seen that a decrease in P leads to a decrease in T b. Chapter 9/Ideal and Real Solutions Q9.8) Explain why chemists doing quantitative work using liquid solutions prefer to express concentration in terms of molality rather than molarity. The molality of a solution is the preferred unit because it is independent of P and T. A kilogram of a substance is a conserved quantity, independent of temperature and pressure. The volume, however, changes as T or P are varied because the thermal expansion coefficient and the isothermal compressibility are not zero. Moles per kilogram is thus a more useful concentration unit than moles per liter, which changes with the thermodynamic state. Q9.9) Explain the usefulness of a tie line on a P-Z phase diagram such as that of Figure 9.4. The tie line allows the compositions of the liquid and vapor phases to be determined geometrically for a given total composition and pressure. Specifically, the ratio of moles in liquid and vapor phase is inversely proportional to the ratio of tie line distances. Q9.10) Explain why colligative properties depend only on the concentration, and not on the identity of the molecule. The origin of the colligative properties is the dilution of the solvent, which lowers its vapor pressure. This is the case because solute molecules at the surface of the liquid affects the relative rates of evaporation and condensation. This statistical likelihood of being near the interface doesn't depend on the identity of the solute—only the number in solution. Therefore, colligative properties (ideally) depend only on the number of species in solution. Problems P9.I) At 303 K, the vapor pressure of benzene is 118 Torr and that of cyclohexane is 122 Torr. Calculate the vapor pressure of a solution for which xbemne = 0.25 assuming ideal behavior. P P total _ xbenzene benzene it * +1 'cyclohexene cyc/ohexene = 0.25X118 Torr+(l—0.25)><122 Torr = 121 Torr P9.2) A volume of 5.50 L of air is bubbled through liquid toluene at 298 K, thus reducing the mass of toluene in the beaker by 2.38 g. Assuming that the air emerging from the beaker is saturated with toluene, determine the vapor pressure of toluene at this temperature. 9-3 Chapter 9/Ideal and Real Solutions x0.08314 L bar K51 X298 K = 0.116 bar 2.38 g><—lmo~1—«:T _ nRT _ 78.11 gmol V 5.50 L P P93) 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? x _ yAP; 0.450x82.1 Torr A —.——.—.——=————————— =0.272 PA +(P, —P, )yA 180 Torr+ (82.1 Torr—180 Torr)x0.450 P9.4) A and B form an ideal solution. At a total pressure of 0.900 bar, yA = 0.450 and M = 0.650. Using this information, calculate the vapor pressure of pure A and of pure B. Bola] : xAPz: + yBBalaI . — 0.900 b 1—0.550 Pa : 13mm] yBPtolaI : arx ( ) : bar x, 0.450 x _ MB? A _ i w :0! P. + (PR —P,. )y. .4 * 0.650 = 0 SOPB P; +0.450(P; —P;) P; _ 0.650><(1—0.450) _ 0.450><(1—0.650) . = 2.27 P P; = 1.414 bar A P9.5) A and B form an ideal solution at 298 K, with xA = 0.600, Pg: 105 Torr and P; = 63.5 Torr. 3) 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 = xAP; = 0.600X105 Torr = 63.0 Torr P5 = (1 — xA)P; = 0.400x63.5 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 9-4 Chapter 9/Ideal and Real Solutions _ PA _ 63.0 Torr PA + PB 88.4 Torr yA 20.71; yB 20.287 For the portion removed, the new xA and x 8 values are the previous y A and y8 values. P = xAP; = 0.713x105 Torr = 74.9 Torr A P8 = (1 — xA)P,: = 0.287x63.5 Torr: 18.2 Torr P9. 6) The vapor pressures of l-bromobutane and 1-chlorobutane can be expressed in the Hiram!) : _ and In [imam : Assuming Pa «T-—111.88 Pa Z—55.725 K K ideal solution behavior, calculate xbmmo and ybmmo at 300.0 K and a total pressure of 8741 Pa. form ln = 5719 Pa and P = 14877 Pa. chlaro At 300.0K, P° bromo Plota/ : xbmmofzmma + — xbmmu ) Pch/(m) Ram, — BLOW 8741 Pa — 14877 Pa xbramo : —°___o— = —— = Pbmma — Pchmm 5719 Pa — 14877 Pa P° , ybromo : Xbmma mama = W = P 8741 Pa total P9.7) Assume that l-bromobutane and l-chlorobutane form an ideal solution. At 273 K, P‘ = 3790 Torr and P* = 1394 Torr. When only a trace of liquid is present at 273 ch [am hmm 0 K9 ychlaro = a) Calculate the total pressure above the solution. b) Calculate the mole fraction of l—chlorobutane in the solution. c) What value would Zchlom 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)? [N0te.' This composition is different from that of part (a).] a) Calculate the total pressure above the solution. P‘ P — P“ P‘ y M : chlora Iota: ail/on: bromo C m Plalul (Poll/am _ Emma) 0.75 2 3790 PaxPW — 3790 Pa><1394 Pa PW, ><(3790 Pa—1394 Pa) 3790 Pa><1394 Pa Rota] : Pa. 3790 Pa — 0.75><(379O Pa —1394 Pa) b) Calculate the mole fraction of l—chlorobutane in the solution. 9-5 Chapter 9/Ideal and Real Solutions a: Rota] : xch/oroBIl/Dro + _ xchlum ) thmo P — * P — x 2 W, Pbm _ 2651 a 1394 Pa 2 0.525 chloro f)m _ Pl. — Pa — Pa chloro brumu c) What value would Zch/m 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).) P‘ P — P” P“ ch/am total chlara bromo I! It Prom] (Pch/oro — BJromo ) _ 3790 Pa><2651 Pa — 3790 PaX1394 Pa 2651 Pa><(3790 Pa—1394 Pa) 5. ._ ychlarUImema Chlara _ :c t P: < bramo — chlom ) ych/oru ychloro ‘— = 0.750 L'h/aro 0.750X1394 Pa = ————————————— = 0.525 3790 Pa + (1394 Pa — 3790 Pa)><0.750 Ir)! _ m! nliq (Zchlam — xchloro) _ nvapur (yak/um Zchlum ) I0! (I)! nvaporychloro + nliqxchmrg _ + n"” + 11,7: 3.21 mol+ 4.86 mol vapor chlom 20.614 P9. 8) An ideal solution at 298 K is made up of the volatile liquids A and B, for whichP; =125 Torr and P; = 46.3 Torr. As the pressure is reduced from 450 Torr, the first vapor is observed at a pressure of 70.0 Torr. Calculate am. The first vapor is observed at a pressure of * =xAP/l +(l_xA)PB Plum, —P1; 70.0 Torr —46.3 Torr x = f = —‘———“‘“ A PA ~ P8 125 Torr — 46.3 Torr P tutu] = 0.301 P93) At —47°C, the vapor pressure of ethyl bromide is 10.0 Torr and that of ethyl chloride is 40.0 Torr. Assume that the solution is ideal. Assume there is only a trace of liquid present and the mole fraction of ethyl chloride in the vapor is 0.80 and answer these questions: a) What is the total pressure and the mole fraction of ethyl chloride in the liquid? b) If there are 5.00 mol of liquid and 3.00 mol of vapor present at the same pressure as in part (a), what is the overall composition of the system? a) P _ 1358—ch tataI_ >a= a: PEC+ FEB—PEC)yEC 10.0 Torr — 40 Torr : 40.0 Torr+ (10 Torr~40 Torr)x0.80 = 25.0 Torr Rota] : XECPEC + (1" xEC)PL:B 130m, —P;B _ 25.0 Torr—10.0 Torr ch — 13;; 40.0 Torr ~10.0 Torr = 0.50 xEC : b) We use the lever rule. (ZB _x3):ni:tp (ya ‘ZB) ZB—xB =(l—ZA)—(l—xA)=xA—ZA yB-ZB =(1-yA)-(1—ZA)=ZA-yi Therefore, we know that xEC = 0.50 and yEC = 0.80 (oso——zflqzzg(zfli—050) ZEC 20.613 ZEB = (1 — ZEC) = 0.387 Chapter 9/Ideal and Real Solutions P9.10) At —3l.2°C, pure propane and n-butane have a vapor pressure of 1200 and 200 Torr, respectively. a) Calculate the mole fraction of propane in the liquid mixture that boils at —3 12°C 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). 08W=n$+0=nfli 760 Torr =1200xp + 200(1— xp) xp = 0.560 9—7 Chapter 9/Ideal and Real Solutions b) _ xPPI: yB’P;+<I::—P;)x,. _ 0.56><1200 Torr 200 Torr + 0.56X (1200 Torr— 200 Torr) = 0.884 P9.II) In an ideal solution of A and B, 3.50 mol are in the liquid phase and 4.75 mol are in the gaseous phase. The overall composition of the system is ZA = 0.300 and xA = 0.250. Calculate yA. "1,2(23 _ x3) = nézizur (ye _ ZB) _ (Z3 — xB)+ nm' Z3 _ 3.50 mol><(0.300— 0.250) + 4.75 mol><0.300 vapor B 11’” 4.75 mol vapor : 0.337 P9.12) Given the vapor pressures of the pure liquids and the overall composition of the system, what are the upper and lower limits of pressure between which liquid and vapor coexist in an ideal solution? Referring to Figure 9.4, it is seen that the maximum pressure results if Z, = xA. * P =ZAP;+(1—ZA)PB l'l’laX The minimum pressure results if ZA = y. Using Equation (9.12), RIP; P- ZT—f PA+(PB—PA)ZA mm P9.I3) At 399°C, a solution of ethanol (x1 = 0.9006, P;=130.4 Torr) and isooctane (P; = 43.9 Torr) forms a vapor phase with y1 = 0.6667 at a total pressure of 185.9 Torr. a) Calculate the activity and activity coefficient of each component. b) Calculate the total pressure that the solution would have if it were ideal. a) The activity and activity coefficient for ethanol are given by a 2 Mle = W = 0,9504 1 B‘ 130.4 Torr _ “1 0.9504 7/1 _. —— 2 : x1 0.9006 Similarly, the activity and activity coefficient for isooctane are given by 1.055 9-8 Chapter 9/Idea1 and Real Solutions a _ (1—y1)Pmm, _ 0.3333X185.9 Torr 2 P2 43.9 Torr y2=fl=fl-=i4.2o x2 1—0.9006 21.411 b) If the solution were ideal, Raoult’s law would apply. PTotaI : x11): + x21); = 0.9006 x 130.4 Torr + (1— 0.9006) x 43.9 Torr = 121.8 Torr P9.14) Ratcliffe and Chao [Canadian Journal of Chemical Engineering 47, (1969), 148] obtained the following tabulated results for the variation of the total pressure above a solution of isopropanol ( Pf= 1008 Torr) and n-decane (P; = 48.3 Torr) as a function of the mole fraction of the n-decane in the solution and vapor phases. Using these data, calculate the activity coefficients for both components using a Raoult’s law standard state. a) for isoproponal Pf 21008 Torr for n-decane P; = 48.3 Torr Using the relations = yiBml, a]. = / P: , and y]. = fl , the calculated activities and x. activity coefficients are shown below. P(TOI’I‘) X2 yz a1 a2 Y1 Y2 942.6 0.1312 0.0243 0.912 0.474 1.05 3.61 909.6 0.2040 0.0300 0.875 0.565 1.10 2.77 883.3 0.2714 0.0342 0.846 0.625 1.16 2.30 868.4 0.3360 0.0362 0.831 0.651 1.25 1.94 830.2 0.4425 0.0411 0.790 0.706 1.42 1.60 786.8 0.5578 0.0451 0.745 0.735 1.69 1.32 758.7 0.6036 0.0489 0.716 0.768 1.81 l.27+ 9—9 Chapter 9/Idea1 and Real Solutions P9.]5) At 399°C, the vapor pressure of water is 55.03 Torr (component A) and that of methanol (component B) is 255.6 Torr. Using data from the following table, calculate the activity coefficients for both components using a Raoult’s law standard state. See the solution to Problem P9.15 for the method used. The results are shown below. m- — . — P9.16) The partial pressures of Brz above a solution containing CC14 as the solvent at 25°C are found to have the values listed in the following table as a function of the mole fraction of Brz 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 Henry’s law constant for Brz in CCl4 at 25°C. 0 00394 0 00420 9-10 Chapter 9/Ideal and Real Solutions 12 ll) A g: 8 O C p: 6 Ga 9" 4 2 0.005 0.01 0.015 0.02 0.025 0.03 XBr2 The best fit line in the plot is PBrz (Torr) = 413 xBrz — 0.063. Therefore, the Henry’s law constant in terms of mole fraction is 413 Torr. P9.17) The data from Problem P9.l6 can be expressed in terms of the molality rather than the mole fraction of Br2. Use the data from the following table and a graphical mgr2 The best fit line in the plot is PB,2 (Torr) = 61.9 mgr2 — 0.0243 . Therefore, the Henry’s law constant in terms of molality is 61.9 Torr. 9-11 Chapter 9/ldeal and Real Solutions P9.18) The partial molar volumes of ethanol in a solution with xHZO = 0.60 at 25°C are 17 and 57 cm3 mol—1, respectively. Calculate the volume change upon 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 cm_3, respectively, at this temperature. V ‘3 nHZOVfllo +I’ZEIVE, VHZO = 17.0 cm3 mol‘1 and V5; = 57.0 cm3 mol‘] n nHZO = 2.00 and xHZO = —”2—0—~ = 0.600 "H20 + nEt 2m” 20.600; 171921.333 2 mol+nE, The total mixed volume is given by Vmixea’ : "moi/120 + "Etr/E’ = 2.00 molx17.0 cm3 mol‘1+1.333 molx57.0 cm3 mol”l = 109.98 cm3 M Vimmixea’ : "H20 H20 + "5! ME! pHZO p51 —3 = 2.00 molx18.02 g molx 10‘“ 0.997 g 1 —3 +1333 molx46.07 g mol—1 xfl— 0.7873 g = 36.15 cm3 + 78.00 cm3 2114.15 cm3 AV = V —V " 2109.98 cm3 —114.15 cm3 = —4.2 cm3 mixed unmixed P9.I9) A solution is prepared by dissolving 32.5 g of a nonvolatile solute in 200 g of water. The vapor pressure above the solution is 21.85 Torr and the vapor pressure of pure water is 23.76 Torr at this temperature. What is the molecular weight of the solute? PH20 21.85 Torr X = * : ____— = H20 P1120 23.76 Torr xxolute : = nso/ule + "H20 0 0804 n 0.0804><1_§(%)9z‘%_1_1 Z = __-_gm°_ = 0.970 mol 0.9617 091% Mzfizgj gmolil 0.970 mol 9—12 Chapter 9/ldeal and Real Solutions P920) The heat of fusion of water is 6.008 X 103 J mol’1 at its normal melting point of 273.15 K. Calculate the freezing point depression constant Kf. _ RM T2 8.314 Jmor‘K-1 x18.02><10'3kgm01‘1 ><(273.15 K)2 K solvent fusion f 3 Al AH 6.008X10 Jmol filsion K]. = 1.86 Kkg mol‘1 P921) The dissolution of 5.25 g of a substance in 565 g of benzene at 298 K raises the boiling point by 0.625°C. Note that Kf: 5.12 K kg molTI, Kb = 2.53 K kg mol—1, and the density of benzene is 876.6 kg m_3. Calculate the freezing point depression, the ratio of the vapor pressure above the solution to that of the pure solvent, the osmotic pressure, :0: and the molecular weight of the solute. Pbmne = 103 Torr at 298 K. AT 0. K A]; : Kbmmlute; mmlute : b : if] = m01 kg-1 ’ ‘ Kb 2.53 Kkgmol M = ——5'Eig—-—= 37.6 gmor‘ 0.247 molkg x0565 kg AT, = —Kfmm,m = —5.12 Kkg morl x0247 molkg’l = —1.26 K P bfnzene : xbenzene : "benzene anzene "benzene + nsolule 565 g 71 2 565 78.11 gmol : 0.981 ___£_j + 0.247 molkg“ x0565 kg 78.11 g mol 73 RT fill-ELF]me Jmor‘IC1 x298 K it: "in/"’9 = ' g m A3 = 5.37 x 105Pa V 565x10 kg 876.6 kg m’3 P922) A sample of glucose (C6H1206) 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 m'3. 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? 1 1+x You may find the approximation 1n z —x useful. 9—13 Chapter 9/Ideal and Real Solutions flfl+RTmKWW=flfl+RTm—~flflfl——=O nso/vem + "sucrose pAh hV*+RT1-——4M4——=- hVV+RTm———i————0 pg m n _ m nmcroreM _ _—— + nYllL'WH‘e 1 + I i M ‘ ‘ pAh . . . 1 Expanding the argument of the logarithmic term 1n a Taylor series, 1n1-+— z ——x x _ nsucmseM ___ 0 pAh h _ R TnsucroseM _ R Tnxucrose fogK; pAg 1.25x10'3kg 0.18016 kg mol‘1 997 kgm’3 ><3.14x104m2 ><9.81ms’2 7r: pgh = 997 kg m‘3 ><9.81ms’2 X237 m = 2.32X104Pa 8.314 Jmol“K’1 x298 K>< 22.37 m P923) The osmotic pressure of an unknown substance is measured at 298 K. Determine the molecular weight if the concentration of this substance is 25.5 kg m4, and the osmotic pressure is 4.50 X 104 Pa. The density of the solution is 997 kg m'3. nmlmeR T : csolulepxolulionR T _ psalu’i‘m Gimme/kg R T fl: Msolute "' V Mrs/we fl. ‘3 2 . “3 . “K’1 Mmlmez997 kgm X 55kgm X8314Jmol ><298K:1400kgmol_1 ‘ 4.50X 10 Pa P924) An ideal dilute solution is formed by dissolving the solute A in the solvent B. Write expressions equivalent to Equations (9.9) through (9.13) for this case. We obtain the equations by replacing P; by kj' . Plot :PA +PB :xaklj +(1—XA)P B * y _ PA _ xAklj A——_-—*——*— Pm! P3(k1/11_PB)XA _ n5 xA- A * A kH+(PB_kH)yA m3 (at :k;+(PBK_k;)yA ;m%%%-%§VHMm’§fl 9-14 Chapter 9/Ideal and Real Solutions P925) A solution is made up of 184.2 g of ethanol and 108.1 g of H20. If the volume of the solution is 333.4 cm3 and the partial molar volume of H20 is 17.0 cm}, what is the partial molar volume of ethanol under these conditions? V : "H20 VHZO + nethanaIVemanol V : nHZOVHZO + nethanol I/ethan0/ — 333.4 cm3 ——1—()81g—4Xl7.0 cm3mol‘x _ — ——-—~V — nHZoVHZO — —————-————-—-—18'02 gm01 - 57 8 cm3mol’1 Ethan”, — nethanol — _ . 46.04 g mol”1 P9.2 6) Calculate the solubility of H2S in l L of water if its pressure above the solution is 3.25 bar. The density of water at this temperature is 997 kg rrf3 . "H15 nHZS P1125 _ bar =—'—: = H ——-—=5.72><10'3 ’1st + "H20 nHlo kHZS bar n = pHZOV : 10‘3m3 x997 kg m’3 “3" M 18.02 x10"3kg mor1 H20 nHlS = 59.25an = 5.72 ><10"3 x 55.4 = 0.327 mol tzs =55.4 P927) The densities of pure water and ethanol are 997 and 789 kg ms, respectively. The partial molar volumes of ethanol and water in a solution with xethan01= 0.20 are 55.2 and 17.8 X 10’3 L mol‘l, respectively. Calculate the change in volume relative to the pure components when 1.00 L of a solution with xemano; = 0.20 is prepared. V : nHZOVHZO + "ethanol I/e'tl1am)/ V _ _ : waOVHWO + xe,hm,I/;,ham,, = 0.80 x 17.8 ><10'3 Lmor‘ + 0.20x 55.2 x10‘3L mol" ntuta/ - g V = 0.0253 Lmol" "total 1.00 L n =————— =39.6 mol=n +n 10ml L molil HZO ethanol 1 x“"”’“” = —n"’h”""/ = —; nfljo = 31.7 mol nwham, = 7.90 mol x1110 "H20 4 _ M V7dea/ : nelhanol Me’ham/ + nH20 H20 pethanol pHZO 46.07 X10‘3k l" 18.02 x10‘3k l“ = 7.90 mol><————g—UL + 31.7 mom—fl = 1.034 L 789 kg nf3 998 kg nf3 AV = V — Vim, = —0.034 L 9—15 Chapter 9/Ideal and Real Solutions P9.28) At a given temperature, a nonideal solution of the volatile components A and B has a vapor pressure of 832 Torr. For this solution, M = 0.404. In addition, xA = 0.285, P; = 591 Torr, and P; = 503 Torr. Calculate the activity and activity coefficient of A and B. PA = yAPW, = 0.404>< 832 Torr = 336 Torr PB = 832 Torr — 336 Torr = 496 Torr _ PA _ 336 Torr aA — , — = 0.569 PA 591 Torr yA = 3A— : 9&9— = 2.00 xA 0.285 a8 : P6 = 496 Torr : 0986 P8 503 Torr yB 261—3: 0.986 =1-38 x8 0.715 P929) Calculate the activity and activity coefficient for CS; at xCSZ = 0.7220 using the data in Table 9.3 for both a Raoult’s law and a Henry’s law standard state. R Pcsz 446.9 Torr aCS = . =———:0.8723 3 PCS2 512.3 Torr YR : c1532 2 0.8723 21208 “1 xCS? 0.7220 ' P ag; = C52 =M=02223 ‘ kH‘CS2 2010Torr H a yg’s = C52 = 02223 =0.3079 2 xCSZ 0.7220 P930) At high altitudes, mountain climbers are unable to absorb a sufficient amount of 02 into their bloodstreams to maintain a high activity level. At a pressure of 1 bar, blood is typically 95% saturated with 02, but near 18,000 feet where the pressure is 0.50 bar, the corresponding degree of saturation is 71%. Assuming that the Henry’s law constant for blood is the same as for water, calculate the amount of 02 dissolved in 1.00 L of blood for pressures of 1 bar and 0.500 bar. Air contains 20.99% 0; by volume. Assume that the density of blood is 998 kg m_3. By Dalton's law, P 2 = 0.2099P. At sea level, 9—16 Chapter 9/Idea1 and Real Solutions P02 pHZOVPOZ flat H0 0 0 2 kHz kHzMHZO _ 998 kg m_3 X 1013m3 X 0.2099 bar X 0.95 4.95 x104bar x 18.02 x10‘3kg mol“ mass = nM = 2.31X104molx 32.0 g mol“1 2 7.14x10’3g n = 17,120ch2 2 n = 2.31x104m01 At 18,000 feet, P02 : szoVPOme: k? 1932 MHZO n _ 998 kg m’3 ><10'3m3 x 0.2099 x0500 bar X 0.71 4.95 bar X 18.02 X10'3kg mol“‘ mass = nM = 8.34X10’5molx320 g mol'1 2 2.67X10’3g ’7 : nhwxo2 : "H20 = 8.34X10’5m01 9—17 ...
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