Chapter 9 Problem 4 ideal mixture of toluene and hexane at 350K Pure vapor pressures: P t. *) Pa = Pa + x h. )P t.

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1 Chapter 9 Recommended Problems Chapter 9 Problem 4 ideal mixture of toluene and hexane at 350K Pure vapor pressures: P t * =3.57X104 Pa *=1.30X10 5 Pa a mole fraction hexane in the liquid mixture that has a normal boiling point of 350K. 1 atm = P = x t P t * + x h * = (1-x h P t * + x h * = P t * + x h ( *-P t * Pa = Pa + x h ( Pa = x h x h = x t = 1 - x h = b Raoult's law gives vapor pressures. =x h * = X 1.30X10 5 Pa = Pa P t = x t P t * = X 3.57X10 4 Pa = Pa Note: + P t = The difference from is round-off error. y h = /P = / = Chapter 9 Problem 7 molecular mass from osmotic pressure Π = 5.30X10 4 Pa at 298K. Density of solution is 997 kg/m 3. Concentration = 31.2 kg/m 3. Π = (n/v RT = [(m/v/m] RT where m/v is density and M is molecular mass M = (m/v RT / Π = 31.2 kg/m 3 X J/(mol K 298K / ( 5.30X10 4 Pa M = 1.46 kg/mol = 1460 g/mol (Engel and Reid say 1460 kg/mol Chapter 9 Problem 8 ideal mixture of benzene and hexane P b * = 120 torr. * = 189 torr x b =0.28. x h = 1-x b = = 0.72 P total = x b P b * + x h * = 0.28 X 120 torr X 189 torr = = 170 torr Chapter 9 Problem 9 ideal solution of A and B. * = 165 torr *=85.1 torr. P=110 torr 110 = 165 x A (1-x A = x A x A = ( /79.9 = 24.9 / 79.9 = 0.312

2 Chapter 9 Problem 11 liquid-vapor equilibrium and the lever rule At o C the vapor pressure of ethyl bromide is 10.0 torr and that of ethyl chloride is 40.0 torr. Assume the solution is ideal. Assume there is only a trace of liquid present and the mole fraction of ethyl chloride in the vapor is a What is the total pressure and what is 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 = P eb +P ec = x eb P* eb P* ec = (1-x ec P* eb P* ec = P* eb y ec = 0.80 = P ec /P = x ec P* ec / [ P* eb ] y ec [ P* eb ] = x ec P* ec y ec P* eb = x ec [ P* ec - y ec ] x ec = y ec P* eb / [ P* ec - y ec ] Substituting in numbers allows calculation of x ec. x ec = 0.80 X10 / [ ( ] = 8 / [ ] = 8 / 16 = 0.50 P = P* eb = ( = = 25 torr (b Apply the lever rule. Let z ec stand for the overall mole fraction ethyl chloride. n liquid ( z ec - x ec = n vapor ( y ec - z ec 5.00 mol ( z ec = 3.00 mol ( z ec 8.00 z ec = = 4.90 z ec = 4.90 / 8.00 = 0.61 (z eb = 0.39

3 Chapter 9 Problem 12 partial pressures of ideal solution of A and B *=84.3 torr *=41.2 torr x A =0.320 x B =1-x A = a calculate partial pressures in the gas phase = x A * = X 84.3 torr = 27.0 torr = x B * = X 41.2 torr = 28.0 torr Note that y A = /( + = 27/(27+28 = b A portion of the gas phase is removed and condensed. Its mole fraction x A = = x A * = X 84.3 torr = 41.4 torr = x B * = X 41.2 torr = 21.0 torr Chapter 9 Problem 14 freezing point depression constant K f of water. Δ fus H = 6008, T fus = K. From equation 9.33, K f = R M solvent T 2 fus / (Δ fus H K f = J/(mol K kg/mol ( K^2 /( 6008 J/mol = 1.86 K kg/mol K f = 1.86 K/molal Chapter 9 Problem 19 ideal mixture of A and B P=0.720 bar y A =0.510 x A =0.420 Calculate * and *. y A = /P so =y A P = X bar = bar * = /x A = bar / = bar P = + so = P-. = bar bar = bar x B = 1 - x A = = * = /x B = bar / = bar

4 Chapter 9 Problem 26 osmotic pressure of glucose in water moles glucose n g = 13.2g / g/mol n g = mol The volume of the final solution V = π r 2 h, where r=0.0125m and h is the height of the solution. The problem's hint suggests starting from the basics, chemical potentials. (A shorter solution of the problem would start with the van't Hoff equation, equation Let μ w * stand for the chemical potential of pure water at ambient pressure. At increased pressure Π (the osmotic pressure and diluted mole fraction x w, the chemical potential of water is μ w (P+ Π, x w. Assuming ideal-solution behavior of μ, μ w (P+Π, x w =μ w * (P+Π+RT ln(xw. The pressure increase from P to P+Π changes μ as follows: P+Π * * * μ (P+Π=μw w (P+ (V m, w dp. P μ w *(P+Π = μ w *(P + V * m,w (P+Π -P = μ w * + Π V* m,w where V * m,w is the molar volume of pure water. V* m,w =(M W /ρ* w, where ρ* w is the density of pure water, 997 kg/m 3, and M w = kg/mol. Overall, μ w (P+Π, x w =μ w * +Π V *m, w +RT ln(x w. At equilibrium the chemical potential of water is the same in the pure water as in the glucose solution, so μ w (P+ Π, x w = μ w *, and therefore, ΠV * m,w = - RT ln(x w The mole fraction water equals moles water over moles water plus moles glucose. x w = n w /(n w +n g = 1/[ 1 + (n g /n w ] ln(x w -(n g /n w, using the problem's hint

5 Chapter 9 Problem 26, continued Substituting that ln(x w into ΠV * m,w = - RT ln(x w yields Π V * m,w = (n g /n w RT The osmotic pressure Π equals the pressure exerted by the column of solution in the test tube. Π = ρ g h, where ρ is the solution's density. The density of the solution is not known so it will be approximated as equal to the density of water, ρ ρ w *. * As noted above, V m, w = M w * ρ, so Π V* m,w g h M w. w We have, so far, g h M w = (n g /n w RT The moles water in the solution, n w, is approximately the volume of the solution ( π r 2 h divided by the molar volume of pure water: Overall, n w ( π r 2 h / ( V * m,w = ( π r2 h (ρ* w /M w n g g h M w = (π r h( R M w n g RT T = 2 ρ * π r 2 * hρ w w M w Height may be isolated on the left-hand side. h 2 = n g RT π r 2 gρ w * Substituting in numbers: h mol J/(mol K298 K = π(0.0125m m/s 2 = 37.8 m kg/m h = 6.2 m One may use h to calculate osmotic pressure. Again, density is approximately water's density. Π = ρ g h = 997 kg/m m/s 2 X 6.2 m = 61 kpa

6 Chapter 9 Problem 28 vapor pressures of pure bromobutane and chlorobutane ln(p bromo / Pa= ln(p T / K chloro / Pa= T / K Let T=305K. Simple substitution into the above Antoine equations gives ln(p bromo = ln(p chloro = = 7113 Pa = Pa (The asterisks were inserted to emphasize that these pressures are of the pure liquids. The total pressure of the mixture is given: P = 9750 Pa. Because the solution is (stated to be ideal, P = x bromo 9750 Pa = x bromo 7113 Pa + (1-x bromo pa ( Pa = x bromo ( Pa x bromo = 8800 / x bromo = 0.77 y bromo follows from Raoult's and Dalton's laws (which apply to this ideal solution. y bromo = P bromo / P = x bromo /P = 0.77 X 7113 / 9750 y bromo = 0.56

7 Chapter 9 Problem 30 ideal mixture of bromobutane and chlorobutane Given: At 273K =3790 Pa and =1394 Pa. =0.75. b By Dalton's law, = P chloro /P = x chloro / [ (1-x chloro ] [ (1-x chloro ] = x chloro ( - = x chloro / [ - ( - ] = x chloro x chloro = 0.75X 1394 /[ ( ] x chloro = / 1993 = a Total pressure P = P bromo + P chloro = x bromo P = (1-x chloro = X X3790 P = 2650 Pa c Require that 4.86 mol liquid is in equilibrium with 3.21 mol gas at pressure 2650 Pa. Calculate z chloro. Consider the lever rule for coexisting liquid and gas phases. n gas ( - z chloro = n liquid (z chloro - x chloro Solve for z chloro. z chloro = ( n gas + n liquid x chloro / (n liquid + n gas z chloro = (3.21 X X / ( z chloro = ( / 8.07 z chloro = 0.61

8 Chapter 9 Problem 33 colligative properties of a benzene solution Benzene: K f =5.12 K kg/mol, K b =2.53 K kg/mol, density=876.6 kg/m 3, P*=103 torr. T=298K grams of some pure substance is dissolved in 825 g benzene. Boiling point elevation = K. n benzene = 825 g / = mol Calculate the freezing-point depression, P/P*, Π, and molecular mass of the substance. Assume the substance is nonvolatile. Freezing point depression ΔT f = ΔT b (K f /K b = 0.575K (5.12/2.53 = 1.16 K molality = n / 0.825kg where n is moles of the unknown substance Boiling point elevation = 0.575K = 2.53 K (n/0.825 kg so n = X / 2.53 = mol and molality = molal P/P* = x benzene = 10.56/(n = / ( = 10.56/ P / P* = Volume of benzene = kg / kg/m 3 = 9.41X10-4 m 3 Assume the volume of the solution is the same as the volume of benzene. Π = (n/v RT = (0.188 mol / 9.41X10-4 m J/(mol K 298K =4.95X10 5 Pa Molecular mass = 7.75 g / mol = 41.2 g/mol.