Calculate the vapor pressure at 25C of an aqueous solution that is 5.35% NaCl by mass. and it wants the final answer in torr Psoln = (Xsolvent)(Pvap of pure solvent) is the equation i believe im suppose to use i think the Pvap of water is 760 torr

Benzene, C6H6, has a boiling point of 80 degrees Celsius. The density of benzene vapor in equilibrium with liquid benzene at 50 degrees Celsius is 1.153 g/L. What is the vapor pressure of benzene in mmHg at 50 degrees Celsius?

the vapor pressure of a glucose (C6H12O6) solution is 17.01 mm Hg at 20 degrees Celcius, while that of pure H2O is 17.25 mm Hg at the same temperature. Calculate the molality of the solution.

Colligative Properties of Solutions

17.4) Vapor Pressure Lowering - Raoult’s Law17.5) Boiling Point Elevation and Freezing Point Depression17.6) Osmotic Pressure

We’ll start by focusing on nonvolatile nonelectrolitic solutes(e.g. sucrose in H2O)Later: volatile solutes, electrolytic solutesRaoult’s Law:Psolvent = Xsolvent P°solventX is the mole fraction

Figure 17.8, ZumdahlExample: Vapor Pressure LoweringProblem: Calculate the vapor pressure lowering (∆P) when 175g ofsucrose is dissolved into 350.00 ml of water at 75°C. The vapor pressure of pure water at 75°C is 289.1 mm Hg, and it’s density is 0.97489 g/ml.

Plan: Calculate the change in pressure from Raoult’s law using the vapor pressure of pure water at 75°C. We calculate the mole fraction of sugar in solution using the molecular formula of sucrose and density of water at 75°C.

Solution:molar mass of sucrose ( C12H22O11) = 342.30 g/mol 175g sucrose___342.30g sucrose/mol = 0.51125 mol sucrose

350.00 ml H2O x 0.97489g H2O = 341.21g H2O ml H2O

341.21 g H2O_18.02g H2O/mol = 18.935 mol H2OVapor Pressure Lowering (cont)

Xsucrose = mole sucrose________ moles of water + moles of sucrose

= 0.51125 mole sucrose = 0.2629 18.935 mol H2O + 0.51125 mol sucrose

∆P = Xsucrose * PoH2O = 0.2629 x 289.1 mm Hg = 7.600 mm Hg

Vapor Pressure lowering:∆P = Xsolute P°solvent

Boiling Point ElevationRecall: The normal boiling point (Tb) of a liquid is the temperature at which its vapor pressure equals atmospheric pressure. Nonvolatile solutes lower the vapor pressure of a liquid Æ greater temperature required to reach boiling point

Boiling Point Elevation:∆Tb = Kbmsolute

Kb = Molal boiling point constant for given liquidmsolute = molal solute concentrationKb = Molal boiling point constant for given liquid

Freezing Point Depression:∆Tf = KfmsoluteKf

= Molal freezing point constant for given liquid

Example: Boiling Point Elevation and Freezing Point Depression in an aqueous solution

Problem: We add 475g of sucrose (sugar) to 600g of water. What will be the Freezing and Boiling points of the solution?

Plan: Find the molality of the sucrose solution, and apply the equations for FP depression and BP elevation using the constants from table 17.5.

Solution: Sucrose (C12H22O11) has molar mass = 342.30 g/mol 475g sucrose___342.30gsucrose/mol = 1.388 mole sucrose

molality = 1.388 mole sucrose 0.600 kg H2O = 2.313 m

∆Tb = Kb . m = (2.313m)= 1.18oC

BP = 100.00oC + 1.18oC = 101.18oC0.512oC m∆Tf = Kf . m = (2.313 m) = 4.30oC FP = 0.00oC -4.30oC= -4.30oC1.86oC 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 nap 257g nap128.16g/molmolality = = = 4.01 mmoles napkg(CHCl3(2.0053 mol

0.500 kg∆Tb = Kb . m = (4.01m) = 14.56oC normal BP = 61.7oC new BP = 76.3oC3.63oC m∆Tf = Kf . m = (4.01m) =18.85oC normal FP = - 63.5oC new FP = - 82.4oC4.70oC mOsmosis: The flow of solvent through asemipermeable membrane into a solutionThe semipermeable membrane allows solventmolecules to pass, but not solute moleculesSee Figures 17.15, 17.16Osmotic Pressure, Π = MRT(similar to ideal gas law!)R = gas constantM = molar concentration of soluteDetermining Molar Mass from Osmotic PressureProblem: A physician studying hemoglobin dissolves 21.5mg of theprotein in water at 5.0oC to make 1.5 ml of solution in order tomeasure its osmotic pressure. At equilibrium, the solution has anosmotic pressure of 3.61 torr. What is the molar mass (M) of thehemoglobin?Plan: 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 hemoglobin

from the known volume and use the known mass to find M.Solution: P = 3.61 torr . = 0.00475 atm 1 atm760 torr Temp = 5.00C + 273.15 = 278.15 KMolar Mass from Osmotic PressureM = = = 2.08 x10 - 4 MΠRT0.00475 atm0.082 L atm (278.2 K) mol KFinding # moles of solute:n = M . V = . 0.00150 L soln = 3.12 x10 - 7 mol2.08 x10 - 4 mol L solnCalculating molar mass of Hemoglobin (after changing mg to g):M = = 6.89 x104 g/mol0.0215 g3.12 x10-7 molConcentration from osmotic pressure:We started with nonvolatile nonelectrolytes (e.g. sucrose/H2O)Now let’s look at volatile solutes and electrolytesColligative Properties of Solutions17.4) Vapor Pressure Lowering - Raoult’s Law17.5) Boiling Point Elevation and Freezing Point Depression17.6) Osmotic PressureColligative Properties of Volatile Nonelectrolyte SolutionsFrom Raoult’s law: Psolvent = Xsolvent .P0solvent and Psolute = Xsolute

.P0soluteConsider a solution having equal molar quantities of acetone andchloroform, Xacetone = XCHCl3 = 0.500. At 350C, the vapor pressure ofpure acetone = 345 torr and pure chloroform = 293 torr.¾ Determine the vapor pressure of the solution and the partialpressure of each component. What are the mole fractions, X, of eachcomponent in the vapor phase?Pacetone = Xacetone . P0acetone = 0.500 . 345 torr = 172.5 torrPCHCl3 = XCHCl3 .P0CHCl3 = 0.500 . 293 torr = 146.5 torrFrom Dalton’s law of partial pressures we know that XA= PAPTotalXacetone = = = 0.541PacetonePTotal 172.5 torr172.5 + 146.5 torrXCHCl3 = = = 0.459PCHCl3PTotal 146.5 torr172.5 + 146.5 torrTotal Pressure = 319.0 torrVapor is enriched in acetone!Non-ideal solutions Ideal behavior is approached when solute and solvent are�involved in similar intermolecular interactions (∆Hsoln = 0)

When weaker solvent-solute interactions occur, heat is removed�upon dissolving (∆Hsoln > 0)Æ the observed vapor pressure is higher than ideal When stronger solute-solvent interactions occur, heat is released�upon dissolving (∆Hsoln < 0)Æ the observed vapor pressure is lower than idealFigure 17.11Colligative Properties of Ionic SolutionsFor ionic solutions we must take into account the number of ions present! i = van’t Hoff factor = “ionic strength”, or the number of ions presentFor Electrolyte Solutions:vapor pressure lowering: P = i XsoluteP 0solventboiling point elevation: Tb = i Kb mfreezing point depression: Tf = i Kf mosmotic pressure: π = i MRTNon-ideal Behaviorof ElectrolyteSolutions(0.05m aqueous)See Table 17.6