Answers

2014-10-08T13:01:31+08:00
Colligative Properties of Solutions17.4) Vapor Pressure Lowering - Raoult’s Law17.5) Boiling Point Elevation and Freezing Point DepressionWe’ll start by focusing on 17.6) Osmotic Pressurenonvolatile nonelectrolitic solutes(e.g. sucrose in H2O)Later: volatile solutes, electrolytic solutesSee Figure 17.9PRaoult’s Law:solvent = Xsolvent . P°solventX is the mole fractionFigure 17.8, ZumdahlExample: Vapor Pressure LoweringProblem: Calculate the vapor pressure lowering (∆P) when 175g ofsucrose is dissolved into 350.00 ml of water at 75C. The vapor 0 pressure of pure water at 75C is 289.1 mm Hg, and it’s 0Plan: density is 0.97489 g/ml. Calculate the change in pressure from Raoult’s law using the vapor pressure of pure water at 75C. We calculate the mole 0 sucrose and density of water at 75 fraction of sugar in solution using the molecular formula of C. 0Solution:molar mass of sucrose ( C12H22O11) = 342.30 g/mol342.30g sucrose/mol 175g sucrose= 0.51125 mol sucrose350.00 ml H2O x 0.97489g H2O = 341.21g H2O ml H2O 341.21 g H2O18.02g H2O/mol= 18.935 mol H2OVapor Pressure Lowering (cont)Xsucrose = moles of water + moles of sucrose mole sucrose= = 0.2629 18.935 mol H 0.51125 mole sucrose2O + 0.51125 mol sucrose∆P = Xsucrose * PoH2O = 0.2629 x 289.1 mm Hg = 7.600 mm Hg∆Vapor Pressure lowering:P = Xsolute . P°solventBoiling Point ElevationRecall: The normal boiling point (Tb) of a liquid is theNonvolatile solutes lower the pressure.temperature at which its vapor pressure equals atmosphericvapor pressure of a liquidÆ greater temperature required to reach boiling point∆Boiling Point Elevation:Tb = KbmsoluteKb = Molal boiling point constant for given liquidmsolute = molal solute concentrationKb = Molal boiling point constant for given liquidSee Table 17.5See Figure 17.3∆Depression:Freezing PointTf = KfmsoluteKf = Molal freezing point constant for given liquidSee Table 17.5See Figure 17.12Example: Boiling Point Elevation and Freezing PointDepression in an aqueous solutionProblem: We add 475g of sucrose (sugar) to 600g of water. What willbe the Freezing and Boiling points of the solution?Plan: Find the molality of the sucrose solution, and apply the equationsSolution:for FP depression and BP elevation using the constants from table 17.5. Sucrose (C12H22O11) has molar mass = 342.30 g/mol342.30gsucrose/mol 475g sucrose= 1.388 mole sucrosemolality =  = 2.313 m 0.600 kg H1.388 mole sucrose2O∆Tb = Kb . m = (2.313m)= 1.18oC BP = 100.00C + 1.18 o Co = 101.18Co0.512oC m∆Tf = Kf . m = (2.313 m) = 4.30oC  FP = 0.00C -4.30 o C= -4.30 o Co1.86oC mExample: Boiling Point Elevation and Freezing PointDepression in a Non-Aqueous SolutionProblem: Calculate the Boiling Point and Freezing Point of asolution having 257g of napthalene (C10H8) dissolved into 500.00g ofchloroform (CHCl3).Plan: Just like the first example.Solution: napthalene = 128.16g/mol chloroform = 119.37g/molmolesnap = =2.0053 mol nap128.16g/mol 257g napmolality = = = 4.01 mkg(CHClmoles nap3) 0.500 kg2.0053 mol ∆Tb = Kb . m = (4.01m) = 14.56oC normal BP = 61.7oC new BP = 76.3Co3.63oC m∆Tf = Kf . m = (4.01m) =18.85oC normal FP = - 63.5oC new FP = - 82.4Co4.70oC msemipermeable membraneOsmosis: The flow of solvent through a into a solutionThe semipermeable membrane allows solventmolecules to pass, but not solute molecules17.15, 17.16See Figures Osmotic Pressure, Π = MRTM = molar concentration of soluteR = gas constant(similar to ideal gas law!) Osmotic PressureDetermining Molar Mass from Problem: A physician studying hemoglobin dissolves 21.5mg of theprotein in water at 5.0oC to make 1.5 ml of solution in order toosmotic pressure of 3.61 torr. What is the measure its osmotic pressure. At equilibrium, the solution has anmolar mass (M) of thePlan:hemoglobin? We know Π, R, and T. We convert Π from torr to atm, and Tfrom oC to K, and then use the osmotic pressure equation to solve formolarity (M). Then we calculate the number of moles of hemoglobinfrom the known volume and use the known mass to find M.Solution: P = 3.61 torr .  = 0.00475 atm760 torr  1 atmTemp = 5.0C + 273.15 = 0 278.15 KM = = = 2.08 x10Molar Mass from Osmotic Pressure - 4 MRT Π mol K0.082 L atm (278.2 K)0.00475 atmn = MFinding # moles of solute: .
0