HW9_Answer
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HW9_Answer

Course Number: CHEM 342, Spring 2011

College/University: Ill. Chicago

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Chem340 Homework 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...

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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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2010/05/09**z*z0z*(,vR+R+**E100f 10080y= f (d )um31604402203.30204060l 100801004*0G2u(t)TspTCTmG1, T1T00E@Xu:4*C1G1 (T T1 ) = G24Kp =:T1=G2 C1G1 G110G2u(t)T1Kp ==G2 C1G1 G1TspTCTmG1, T1T
Zhejiang University - EE - math 402
20100525****DMMX=CMPMMMMMMMX=CMMM()*1mM2 mM3 m*(7R*(7R(7R*(7RH ( s) MG (s) MD ( s) M*5C*5CGDttHH0HH2HH1HtH1H (s)K1K2=+D( s)s T2 s + 1H ( s ) K 0 s=eG (s)stH (s)K1K2=+D( s)s T2 s
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