I. The Vapor Pressure of Solutions A.Solution vapor pressures differ from pure liquid vapor pressures 1.Solutes lower the vapor pressure by occupying surface positions 2.Raoult’s Law: P soln =  solvent P 0 solvent a.Solution vapor pressure is a fraction of the pure vapor pressure b.Linear equation (y = mx + b) c.Examples

Vapor Pressure of Ideal Solutions 1.Our Solutions so far have contained non-volatile (solid) solutes 2.Volatile (liquid) solutes will exert a Vapor Pressure of their own a.P T = P A + P B =  A P o A +  B P o B b.When the two liquids have very similar structures, Ideal behavior is observed c.When the two liquids interact strongly, P T is lower than expected d.When the two liquids interact very weakly, P T is higher than expected e.We will generally Assume Ideal Behavior

II. Colligative Properties A.Boiling Point Elevation 1.Normal Boiling Point occurs when vapor pressure = 1 atm 2.Solute molecules decrease the number of solvent molecules at the surface a.This lowers the vapor pressure (Rauolt’s Law) b.We have to heat the solution more to boil: Elevated the B.P.

3.The increase in b.p. depends only on the number of dissolved particles, not on what the dissolved particles are = Colligative Property 4.Boiling Point Elevation:  T b = K b m solute = ( o C kg/mol)(mol/kg) = o C a.K b = molal boiling point elevation constant for that solvent b.m solute = the molality of the solute c.Example

B.Freezing Point Depression 1.Dissolved solutes lower the f.p. of a liquid 2.The solute particles lower the rate at which liquid solvent enters the solid solvent. Solvent molecules are coming out of the solid faster, so the solid melts, or no solid forms. 3.As we cool the solution, the rate of solvent leaving the solid slows more than the reverse. So we eventually reach a temperature where the solid does form.

4.This is a useful property: we can use NaCl or CaCl 2 to help keep ice from forming on the sidewalk, or to help melt ice already there. 5.Freezing Point Depression:  T f = K f m solute = ( o C kg/mol)(mol/kg) = o C a.K f = molal freezing point depression constant for that solvent b.m solute = the molality of the solute c.Example C.Osmotic Pressure 1.Pure solvent will flow through a semipermeable membrane into a solution of the same solvent = Osmosis 2.Semipermeable membrane = one through which solvent, but not solute can pass 3.Dissolved solute interferes with solvent going from solution to pure side 4.Eventually, equilibrium is reached when the pressure from the solution side is greater than the pure solvent side. 5.Osmotic Pressure = minimum pressure needed to stop osmosis

OsmosisOsmotic Pressure

6.Osmotic Pressure =  = MRT a.M = molarity b.R = ideal gas constant = L atm/K mol c.T = temperature in Kelvins 7.We can determine M.W. from Osmotic Pressure: Example 8.Dialysis = osmosis of solvent plus small molecules Used to purify blood of people with kidney problem 9.Reverse Osmosis = external pressure causes flow from solution to solvent Used to produce drinking water from sea water

D.The colligative properties of Electrolytes 1.Nonelectrolytes (sugar) dissolve to give one particle per molecule 2.Electrolytes (NaCl) dissolve to give multiple particles per molecule a.van’t Hoff factor = i b.i = 2 for NaCl c.i = 4 for H 3 PO 4 d.i = 3 for K 2 SO 4 3.Ion-pairing reduces the total number of particles, depending on system a.The greater the concentration, the greater the ion-pairing b.0.1 M NaCl = 1.87 x particles of 0.1 M sugar c M NaCl = 1.97 x particles of M sugar 4.For electrolytes, use the following formulas:  T = imK  = iMRT

III. Colloids A.Colloid = nm particle that remains suspended but not dissolved in a liquid 1.Muddy water will eventually clear, except for tiny colloidal particles 2.Colloids stay suspended due to electrostatic repulsion for one another a.Each particle has same charge b.Particle is surrounded by opposite charges, which repel other particles 3.We Coagulate colloids by: a. Heating: particles collide and aggregate b. Add Electrolyte: disrupt charges