Solutions 71. For an ideal solution,µA - µA* = RT lnxA = (8.3145 J K-1 mol-1)×(353.25 K) ln(0.30) = -3.536 kJ mol-1aA = pA / pA*γA xA = pA / pA* pA = γA xA pA*At the normal boiling point, pA* = 1 atm = 760 TorrpA = 0.93 × 0.30 × 760 Torr = 212 Torr2. € ΔmixS = −nR xAln xA+ xBln xB( )€ xB= 1− xA€ ΔmixS = −nR xAln xA+ 1− xA( )ln 1− xA( )( )€ dΔmixSdxA= −nR ln xA+1− ln 1− xA( )−1( )= −nR lnxA1− xAThe maximum ΔmixS is reached when € dΔmixSdxA= 0, i.e. at € xA1− xA= 1Therefore, xA = xB = 0.5(a) hexane and heptane should be mixed each with the mole fraction of 0.5.(b) to have the same chemical amounts of hexane and heptane, their mass ratio should beequal to the ratio of their molar masses:m(hexane) / m(heptane) = M(C6H14) / M(C7H16) = (86.17 g mol-1) / (100.2 g mol-1) = 0.863. For ideal mixing,€ ΔmixG = nRT xAln xA+ xBln xB( )xA = 0.25 (MCH) xB = 0.75 (THF)ΔmixG (ideal) = (4.0 mol)×(8.3145 J K-1 mol-1)×(303.15 K)×{0.25ln(0.25) +0.75ln(0.75)} = -5.67 kJGE = RTx(1 – x){0.4857 – 0.1077(2x – 1) + 0.0191(2x – 1)2} x = 0.25GE = 0.1021RT = 0.257 kJ mol-1 (note that GE is a molar quantity)ΔmixG (real) = ΔmixG (ideal) + nGE = (-5.67 kJ) + (4.0 mol)×( 0.257 kJ mol-1) = -4.6 kJ4. Using the mole fraction in the vapor phase, we can calculate partial vaporpressure of each component:yA = pA/(pA + pB) = pA/101.3 kPa = 0.314 pA = 0.314×101.3 kPa = 31.8 kPapB = p – pA = 101.3 kPa – 31.8 kPa = 69.5 kPaaA = pA/pA* = 31.8 kPa / 73.0 kPa = 0.436aB = pB/pB* = 69.5 kPa / 92.1 kPa = 0.755γA = aA/xA = 0.436 / 0.220 = 1.98γB = aA/xB = 0.755 / (1 – 0.220) = 0.968https://www.coursehero.com/file/11467744/Homework-7-Solution-on-Physical-Chemistry/This study resource wasshared via CourseHero.com5. € I =12bibΘ      i∑zi2For an MpXq salt, € b+/bΘ= pb/bΘ, € b−/bΘ= qb /bΘ, so€ I =12pz+2+ qz−2( )bibΘI = I(K3[Fe(CN)6]) + I(KCl) + I(NaBr) = (1/2)(3 + 32)b(K3[Fe(CN)6])/bΘ +b(KCl)/bΘ + b(NaBr)/bΘ = 6×0.040 + 0.03 + 0.05 = 0.3206.7. Denote the mole fraction of methylbenzene as xM and mole fraction ofdimethylbenzene as xD. xD = 1 –xMAccording to Raoult’s law, pM = xMpM*pD = xDpD*When the mixture the mixture bolis at 0.5 atm, the total vapor pressure should be0.50 atm (380 Torr):p = pM + pD = xMpM* + xDpD* = xMpM* + (1 – xM)pD*xM = (p – pD*)/(pM* – pD*) = (380 Torr – 150 Torr)/(400 Torr – 150 Torr) = 0.92xD = 1 –xM = 0.08Vapor composition (according to Dalton’s law):https://www.coursehero.com/file/11467744/Homework-7-Solution-on-Physical-Chemistry/This study resource wasshared via CourseHero.comyM = pM/p = xMpM*/p = 0.92×(400 Torr)/(380 Torr) = 0.968yD = 1 – yM = 0.0328. At the boiling point the total vapor pressure of solution should be 1 atm = 760Torr. For an ideal solution, according to Raoult’s law,p = pA + pB = xApA* + xBpB* = 0.6589×(957 Torr) + (1 – 0.6589)×(379.5Torr) = 760.01 Torr ≈ 760 TorrTherefore, the solution is very close to ideal.yA = pA/p = xApA*/p = 0.6589×(957 Torr)/(760 Torr) = 0.830yB = 1 – yA = 0.1709. Retrace the argument given in the derivation of van’t Hoff’s equation forosmotic pressure in the text. At equilibrium,€ µA*p( )=µAxA, p + Π( )For a real solution, € µ=µ*+ RT ln a€ µA*p( )=µA*p + Π( )+ RT ln aA=µA*p( )+ Vmdppp+Π∫+ RT ln aATherefore, € Vmdppp+Π∫= −RT ln aAFor an incompressible solution, € Vmdppp+Π∫= ΠVm, so € ΠVm= −RT ln aAThe osmotic coefficient is defined as € φ= −xAxBln aA€ ΠVm= RTφxBxA≈ RTφnBnA(because nB << nA)€ ΠVmnA≈ RTφnB€ VmnA≈ V€ ΠV ≈ RTφnB€ nBV= [B] - molar concentration of solute. Therefore, we obtain€ Π = RTφB[ ]https://www.coursehero.com/file/11467744/Homework-7-Solution-on-Physical-Chemistry/This study resource wasshared via CourseHero.comPowered by TCPDF