Problem Set 11 Solutions

Chemistry 360Dr. Jean M. StandardProblem Set 11 Solutions1. The vapor pressure of pure toluene is 400 torr and that of pure 1,2-dimethylbenzene is 150 torr at 90˚C.Determine the composition of the liquid and vapor phases if toluene and 1,2-dimethylbenzene are mixedat 90˚C and a total pressure of 0.5 atm. Assume ideal solution behavior.Assuming that the solution is ideal, it can be described by Raoult's Law,P i = x i P i * ,€where P i is the partial pressure of component i in the mixture, x i is the liquid phase mole fraction of*component i, and P i is the vapor pressure € of pure component i. The total pressure of the mixture is given bythe sum of the partial pressures (which are determined from Raoult's Law),€P = P€1 + P 2*= x 1 P 1 +* x2 P 2 .Using the relation€x 2 = 1− x 1 , the equation for total pressure (the bubble point line) becomes€*P = x 1 P 1 + 1 −*( x1 )P 2or P = P 1* − P2*( ) x 1 + P 2* .Solving this equation for the liquid phase mole fraction, we have€x 1 =P − P 2*P 1 * − P 2* .Let us assign component 1 as toluene and component 2 as 1,2-dimethylbenzene. Then, the equation abovebecomes an equation for the liquid phase € mole fraction of toluene,x toluene =*P − P dimeth*− P dimeth*P toluene.Substituting and using the total € pressure of 0.5 atm (= 380 torr), the liquid phase mole fraction of toluene is€x toluene =x toluene = 0.92.380 torr − 150 torr400 torr − 150 torrThen, the liquid phase mole fraction of 1,2-dimethylbenzene isx dimeth = 1 − x toluene= 1 − 0.92x dimeth = 0.08.€

43. The normal boiling points of propane and n-butane are –42.1˚C and –0.5˚C, respectively. The followingvapor pressure data have been measured.T (˚C) –31.2 –16.3P*, propane (kPa) 160.0 298.6P*, n-butane (kPa) 26.7 53.3Assume that the substances form ideal solutions.(a) Calculate the liquid phase mole fraction of propane in the solutions in which liquid and vapor are inequilibrium at 1.0 atm pressure at –31.2˚C and –16.3˚C.For each of these temperatures, we can use an approach similar to that taken in Problems 1 and 2 to get themole fraction of propane in the liquid phase. From Problem 1 (and the equation of the bubble point line),the liquid phase mole fraction isx 1 =P − P 2*P 1 * − P 2*.Assigning propane as component 1 and n-butane as component 2, at –31.2˚C we have€x propane ==x propane = 0.560.*P − P nbutane**P propane − P nbutane1.0 atm − 0.264 atm1.579 atm − 0.264 atmNote that the pressures in the table above are given in kPa, so the conversion factor 1 atm = 101.32 kPa wasused in the calculation of € the mole fraction.At –16.3˚C, the liquid phase mole fraction of propane isx propane ==x propane = 0.196.*P − P nbutane**P propane − P nbutane1.0 atm − 0.526 atm2.947 atm − 0.526 atm(b) Calculate the vapor phase € mole fraction of propane in each of the solutions.To obtain the vapor phase mole fraction of propane at –31.2˚C, the partial pressure is determined first fromRaoult's Law,*P propane = x propane P propane( )( 1.579 atm)= 0.560P propane = 0.884 atm.€

3 b.) Continued5Then, the mole fraction of propane in the vapor phase is determined from the definition,y propane = P propaneP0.884 atm=1.0 atmy propane = 0.884 .At –16.3˚C, the partial pressure of propane is€*P propane = x propane P propane( )( 2.947 atm)= 0.196P propane = 0.578 atm.The vapor phase mole fraction of propane is determined from the equation,€y propane = P propaneP0.578 atm=1.0 atmy propane = 0.578.€

4. A mixture of liquids A and B exhibits ideal behavior. At 84˚C, the total vapor pressure of a liquidsolution containing 1.2 mol A and 2.3 mol B is 331 torr. Upon the addition of 1 more mole of B to thesolution, the vapor pressure is 347 torr. Calculate the vapor pressures of pure A and pure B at 84˚C.6The ideal solution can be described by Raoult's Law,P i = x i P i * .The total pressure of the mixture is given by the sum of partial pressures,€P = P A + P Bor P = x A P A* + xB P B* .We are given the total pressure for two different liquid phase compositions, and are asked to determine the purevapor pressures. In the first case, € the mole fractions arex A =x A = 0.3431.2 mol1.2 mol + 2.3moland€x B = 1 − x A= 1 − 0.343x B = 0.657.In the second case, the mole fractions are€x A =x A = 0.2671.2 mol1.2 mol + 3.3moland€x B = 1 − x A= 1 − 0.267x B = 0.733.The total pressure, given by P = x A P * *A + x B P B , in the first case is€331torr = 0.343P * A + 0.657 P * B .€The total pressure in the second case is€347 torr = 0.267 P * A + 0.733P * B .€

4.) Continued7There are two equations and two unknowns. One way to solve this is to solve the first equation for*substitute into the second equation. So, solving the first equation for P A yieldsP A* and0.343P A* = 331 − 0.657 PB**or P A = 965 − € * 1.915PB .€Substituting this result into the second equation leads to the vapor pressure of pure B,€347 = 0.267 P * *A + 0.733P B( ) + 0.733P B **347 = ( 0.267) 965 − 1.915P B347 = 257.7 − 0.511P B * + 0.733P B*89.3 = 0.222 P B*P B * = 402 torr .From the first equation, we had P **A = 965 − 1.915P B . Substituting the vapor pressure of pure B allows us todetermine the vapor pressure € of pure A,€P A* = 965 − 1.915PB*P A *= 965 − 1.915( 402 torr)P A* = 195torr .€

85. Determine the mole fraction of each component in the vapor phase in equilibrium with a liquid phase inwhich there is a 1:1 molar ratio of n-hexane and cyclohexane. The vapor pressures of pure n-hexaneand cyclohexane are 151.4 torr and 97.6 torr, respectively.The partial pressures can be determined from Raoult's Law,P i = x i P i * .Since the molar ratio is 1:1 in the liquid phase, the liquid phase mole fractions are both 0.5. For n-hexane, thepartial pressure is€*P nhexane = x nhexane P nhexane( )( 151.4 torr)= 0.5P nhexane = 75.7 torr .For cyclohexane, the partial pressure is€*P cyclohex = x cyclohex P cyclohex( )( 97.6 torr)= 0.5P cyclohex = 48.8 torr .The total pressure is€P = P nhexane + P cyclohex= 75.7 torr + 48.8 torrP = 124.5torr .Then, the mole fraction of n-hexane in the vapor phase isThe vapor phase mole fraction of cyclohexane is€€y nhexane = P nhexaneP75.7 torr=124.5torry nhexane = 0.608.y cyclohex = P cyclohexP48.8 torr=124.5torry cyclohex = 0.392.€

6. The Henry's Law constant for carbon dioxide in water is 1.25×10 6 torr at 25˚C. Calculate the solubilityof carbon dioxide in water at 25˚C when its partial pressure in air is (a) 4.0 kPa and (b) 100 kPa.9Henry's Law isp i = K H x i . Solving for the mole fraction yields€x i =p iK H.(a) For a partial pressure of 4.0 kPa or 30.0 torr (1 torr = 133 Pa), the mole fraction of carbon dioxide dissolvedin water is€30.0 torrx i =1.25×10 6 torrx i = 2.4 ×10 −5 .(b) For a partial pressure of 100 kPa € or 750 torr, the mole fraction of carbon dioxide dissolved in water is750 torrx i =1.25×10 6 torrx i = 6.0 ×10 −4 .€7. At 25˚C, the mole fraction of air dissolved in water is 1.388×10 –5 .(a) Determine the molarity of the solution.The molarity is defined as moles of air divided by liters of solution,[ air] =n airL solution.Since the mole fraction is so small, we can assume that the liters of solution are equal to liters of water,€[ air] =n airL H 2 O.In order to calculate the molarity, we need to determine the moles of air. The definition of mole fraction is€x air =n air.n air + n H 2 O€

7 a.) Continued10The moles of water in 1 L can be calculated using the density and molecular weight,n H 2 O ==D H 2 OMW H 2 O1 g/mL18.015g/mol ⋅ # 1000 mL &% ($ 1L 'n H 2 O = 55.5mol/L .From the definition of mole fraction, we can solve for the moles of air,Substituting, we can determine the moles of air,€€x air =n airn air + n H 2 Oor n air = x air n H 2 O1 − x air.n air = x air n H 2 O1 − x air( 1.388 ×10 −5) 55.5mol=1 − 1.388 ×10 −5n air = 7.70 ×10 −4 mol.( )Finally the molarity of air can be calculated,€[ air] =≈n airL solutionn airL H 2 O= 7.70 ×10−4 mol1L[ air] = 7.70 ×10 −4 mol/L .€

7.) Continued11(b) Calculate the Henry's Law constant for air (in water). Compare your results with the literaturevalues of 6.80×10 7 and 3.27×10 7 torr, respectively, for the Henry's Law constants of N 2 and O 2dissolved in water at 25ºC.Henry's Law isp i = K H x i . Solving for the Henry's Law constant,€K H = p ix i.The partial pressure of air is 1 atm. Substituting,€K H = p ix i1atm=1.388 ×10 −5K H = 72050 atm.The literature values of the Henry's Law constants of N 2 and O 2 in water are 6.80×10 7 and 3.27×10 7 torr,respectively. Using 1 atm = 760 € torr, the Henry's Law constants of N 2 and O 2 are 89500 and 43000 atm,respectively. The Henry's Law constant calculated for air lies in between these values as expected, since airis a mixture primarily of N 2 and O 2 (and it lies much closer to the value of N 2 since air is about 78% N 2 ).(c) Would you expect the solubility of air to increase or decrease with an increase in temperature?The solubility of a gas in a liquid decreases as the temperature increases. This may be observedqualitatively for carbonated beverages. As the beverage warms up, you can see the bubbles of CO 2 comingout of solution. For example, shown below are measured Henry's Law constants of CO 2 in H 2 O at varioustemperatures [from J. J. Carroll, J. D. Slupsky, and A. E. Mather, J. Phys. Chem. Ref. Data 1991, 20, 1201-1209]. Since the Henry's Law constant is inversely proportional to solubility, as the temperature goes up,the solubility of CO 2 in H 2 O goes down.

128. Why do people who live at high altitudes add salt to water when boiling foods like pasta? Determine themolality of NaCl required to raise the boiling point of water by 3ºC. Assume that K b for water is0.51ºC/molal. Does adding about 1 teaspoon of NaCl to four quarts of water substantially change theboiling point?The equation for boiling point elevation isΔT b = K b m solute .In order to calculate the molality of NaCl required to raise the boiling point of water by 3ºC, we must solve forthe molality and substitute €m solute = ΔT bK bNote that this is the total molality of the solute particles required. Since for each molecule of NaCl, twoparticles are released in solution (an € Na + ion and a Cl – ion), the molality of NaCl required is really half theamount calculated, or=3 ! C0.51 ! C/mm solute = 5.88 molal.m NaCl = 2.94 molal.How much NaCl is this for four quarts of water? Four quarts of water can be converted to liters, usingconversion factors from the CRC, €"4 Quarts 0.946 L %$ ' = 3.784 L .# 1Quart &Using the density of water as 1.0 g/mL, 3.784 L of water weighs about 3784 g or 3.784 kg. The molality ofNaCl desired to raise the boiling € point of water by 3ºC is 2.94 molal = 2.94 moles of NaCl per kg of water, sofor 3.784 kg of water, this would be 11.125 moles of NaCl. The amount of NaCl in grams required to raise theboiling point of water by 3ºC is"11.125moles 58.44 g %$ ' = 650 g NaCl .# mol &650 grams is much more than 1 teaspoon; 650 g is about 1.5 pounds of NaCl! Thus, the teaspoon of salt addedto boiling water is not really € added to raise the boiling point; it is just for taste.

9. Estimate the freezing and boiling points of seawater, assuming its composition is approximately 1.08molal in NaCl. Assume that K f for water is 1.86ºC/molal and K b for water is 0.51ºC/molal.13The equation for freezing point depression isΔT f = K f m solute .Substituting the molality of NaCl in seawater, we have€ΔT f = K f m solute( ) 2.16 m= 1.86 ! C/m( )= 4.02 ! C .Note that since for each NaCl molecule, we get two particles in solution, we have doubled the molality of theparticles in solution. So, since the € freezing point of pure water is 0ºC, the freezing point of seawater is loweredto –4.0ºC.The equation for boiling point elevation isΔT b = K b m solute .Substituting the molality of NaCl in seawater, we have€ΔT b = K b m solute( ) 2.16 m= 0.51 ! C/m( )= 1.10 ! C .Once again we have doubled the molality of the particles in solution since for each NaCl molecule we get twoparticles in solution. Since the boiling € point of pure water is 100ºC, the boiling point of seawater is increased to101.1ºC.

1410. A solution containing 1.470 g of an unknown compound in 50.000 g of benzene boils at 80.60ºC at a totalpressure of 1 bar. The boiling point of pure benzene is 80.09ºC and the molar enthalpy of vaporization ofbenzene is 32.0 kJ/mol. Determine the molar mass of the unknown compound.In order to determine the molality of the unknown compound, we must first calculate the boiling point elevationconstant K b of the solvent, benzene, since it was not given in the problem. The boiling point elevationconstant K b is defined as€€K b =( ) 2*M solvent R T vap,( 1000 g/kg) ΔH vap,mwhere M solvent is the molar mass of the solvent (in this case, benzene), is the boiling point of the puresolvent, ΔH vap,m is the enthalpy € of vaporization of the solvent, and the factor of 1000 carries units of g/kg toconvert mass units.€€€ Substituting the values for benzene yields the boiling point elevation constant,K b ==( ) 2*M solvent R T vap( 1000 g/kg) ΔH vap,m( 78.69 g/mol) 8.314 J/molKK b = 2.55 K/molal .€Now that the boiling point elevation constant has been determined, the molality of the unknown compound maybe calculated. The equation for boiling point elevation is*T vap( )( 353.24 K) 2( 1000 g/kg) 32.0 ×10 3 J/mol( )Note that here, the unit 'm' indicates molality in moles of solute per kg of solvent.ΔT b = K b m solute .Solving for the molality of the solute and substituting gives€€m solute =m solute = ΔT bK b353.75 K − 353.24 K=2.55 K/mm solute = 0.200 molal.The moles of the unknown compound may be calculated from the molality,ormol solutekg solventmol solute = m solute ⋅ kg solvent,( )( 0.0500 kg solvent)= 0.200 mol solute/kg solventmol solute = 0.0100 moles.€

10.) Continued15Finally, the molar mass of the unknown compound may be calculated,€g soluteM solute =mol solute1.470 g=0.0100 molM solute = 147 g/mol.As an aside, the unknown compound corresponds to dichlorobenzene.