1. Colligative Properties Consider three beakers:  50.0 g of ice  50.0 g of ice + 0.15 moles NaCl  50.0 g of ice + 0.015 moles sugar (sucrose) What will the temperature of each beaker be?  T f = K f  m  Beaker 1: 0C  Beaker 2: -5.58 C  Beaker 3: -.558 C

2. Colligative Properties The reduction of the freezing point of a substance is an example of a colligative property:  A property of a solvent that depends on the total number of solute particles present There are four colligative properties to consider:  Vapor pressure lowering (Raoult’s Law)  Freezing point depression  Boiling point elevation  Osmotic pressure

3. Colligative Properties – Vapor Pressure A solvent in a closed container reaches a state of dynamic equilibrium. The pressure exerted by the vapor in the headspace is referred to as the vapor pressure of the solvent. The addition of any nonvolatile solute (one with no measurable vapor pressure) to any solvent reduces the vapor pressure of the solvent.

4. Colligative Properties – Vapor Pressure Nonvolatile solutes reduce the ability of the surface solvent molecules to escape the liquid.  Vapor pressure is lowered. The extent of vapor pressure lowering depends on the amount of solute.  Raoult’s Law quantifies the amount of vapor pressure lowering observed.

5. Colligative Properties – Vapor Pressure Raoult’s Law: P A = X A P O A where P A = partial pressure of the solvent vapor above the solution (ie with the solute) X A = mole fraction of the solvent P o A = vapor pressure of the pure solvent

6. Colligative Properties – Vapor Pressure Example: The vapor pressure of water at 20 o C is 17.5 torr. Calculate the vapor pressure of an aqueous solution prepared by adding 36.0 g of glucose (C 6 H 12 O 6 ) to 14.4 g of water. Given: P o H2O = 17.5 torr mass solute = 36.0 g of glucose=.2 mol mass solvent = 14.4 g of water=.8 mol Find: P H2O Raoult’s Law: P A = X A P O A

7. Colligative Properties – Vapor Pressure Solution: Raoult’s Law: P A = X A P O A P o H2O = 17.5 torr mass solute = 36.0 g of glucose=.2 mol mass solvent = 14.4 g of water=.8 mol So,.8/(.2 +.8) x 17.5 torr = Answer: 14.0 torr

8. Colligative Properties – Vapor Pressure Example: The vapor pressure of pure water at 110 o C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.10 atm at the same temperature. What is the mole fraction of ethylene glycol in the solution? Both ethylene glycol and water are liquids. How do you know which one is the solvent and which one is the solute?

9. Colligative Properties – Vapor Pressure Given: P o H2O = 1070 torr P soln = 1.10 atm Find: X EG Solution: Answer: X EG = 0.219 Raoult’s Law: P A = X A P O A

10. Colligative Properties – Vapor Pressure Ideal solutions are those that obey Raoult’s Law. Real solutions show approximately ideal behavior when:  The solution concentration is low  The solute and solvent have similarly sized molecules  The solute and solvent have similar types of intermolecular forces.

11. Colligative Properties – Vapor Pressure Raoult’s Law breaks down when solvent- solvent and solute-solute intermolecular forces of attraction are much stronger or weaker than solute-solvent intermolecular forces.

12. Raoult’s Law: Mixing Two Volatile Liquids Since BOTH liquids are volatile and contribute to the vapor, the total vapor pressure can be represented using Dalton’s Law: P T = P A + P B The vapor pressure from each component follows Raoult’s Law: P T =  A P° A +  B P° B Also,  A +  B = 1 (since there are 2 components)

13. Boiling Point Elevation

14. Colligative Properties – BP Elevation The addition of a nonvolatile solute causes solutions to have higher boiling points than the pure solvent.  Vapor pressure decreases with addition of non- volatile solute.  Higher temperature is needed in order for vapor pressure to equal 1 atm.

15. Colligative Properties- BP Elevation The change in boiling point is proportional to the number of solute particles present and can be related to the molality of the solution:  T b = K b. m where  T b = boiling point elevation K b = molal boiling point elevation constant m = molality of solution The value of K b depends only on the identity of the solvent (see Table 11.5).

16. Colligative Properties - BP Elevation Example: Calculate the boiling point of an aqueous solution that contains 20.0 mass % ethylene glycol (C 2 H 6 O 2, a nonvolatile liquid). Solute = 20g Solvent =80g K b (solvent) =.51 C/m  T b = K b  m

17. Colligative Properties - BP Elevation Molality of solute: Solute = 20g Solvent =80g K b (solvent) =.51 C/m  T b = BP = 102.1 o C  T b = K b  m

18. Colligative Properties - BP Elevation Example: The boiling point of an aqueous solution that is 1.0 m in NaCl is 101.02 o C whereas the boiling point of an aqueous solution that is 1.0 m in glucose (C 6 H 12 O 6 ) is 100.51 o C. Explain why. Answer: Van’t Hoff Factor

19. Van’t Hoff Factor (i) NaCl is a strong electrolyte and should dissolve into two distinct ions Glucose is also dissolvable, but will only form one molecule. Therefore NaCl should lower the vapor pressure by twice as much as glucose does. i= moles of particles in solution moles of solute dissolved

20. van’t Hoff Factor However, One mole of NaCl in water does not really give rise to two moles of ions.

21. van’t Hoff Factor Some Na + and Cl − reassociate for a short time, so the true concentration of particles is somewhat less than two times the concentration of NaCl.

22. The van’t Hoff Factor Reassociation is more likely at higher concentration. Therefore, the number of particles present is concentration dependent, and (i) can only be determined experimentally.

23. The van’t Hoff Factor We modify the previous equations by multiplying by the van’t Hoff factor, i  T f = i  K f  m

24. Colligative Properties - BP Elevation Example: A solution containing 4.5 g of glycerol, C 3 H 5 (OH) 3 a nonvolatile nonelectrolyte, in 100.0 g of ethanol, C 2 H 6 O, has a boiling point of 79.0 o C. If the normal BP of ethanol is 78.4 o C, calculate the molar mass of glycerol. Given:  T b = 79.0 o C - 78.4 o C = 0.6 o C mass solute = 4.5 g mass solvent = 100.0 g = 0.100 kg K b = 1.22 o C/m (Table 11.5) Find:molar mass (g/mol)  T b = i  K b  m

25. Freezing Point Depression

26. Colligative Properties - Freezing Pt Depression  Freezing point of the solution is lower than that of the pure solvent. The addition of a nonvolatile solute causes solutions to have lower freezing points than the pure solvent. Solid-liquid equilibrium line rises ~ vertically from the triple point, which is lower than that of pure solvent.

27. Colligative Properties - Freezing Pt Depression The magnitude of the freezing point depression is proportional to the number of solute particles and can be related to the molality of the solution.  T f =i  K f  m where  T f = freezing point depression K b = molal freezing point depression constant m = molality of solution The value of K f depends only on the identity of the solvent (see Table 11.5).

28. Colligative Properties - Freezing Pt Depression Example: Calculate the freezing point depression of a solution that contains 5.15 g of benzene (C 6 H 6 ) dissolved in 50.0 g of CCl 4. Given: mass solute = mass solvent = K f solvent = Find:  T f =39.3  T f = i  K f  m

29. Colligative Properties - Freezing Pt Depression Example: Which of the following will give the lowest freezing point when added to 1 kg of water: 1 mol of Co(C 2 H 3 O 2 ) 2, 2 mol KCl, or 3 mol of ethylene glycol (C 2 H 6 O 2 )? Explain why. KCl don’t forget the van’t Hoff Factor applies here as well we end up with 4 mol of ions the most out of this group.

30. Osmosis

31. Colligative Properties - Osmosis Some substances form semipermeable membranes, allowing some smaller particles to pass through, but blocking other larger particles. In biological systems, most semipermeable membranes allow water to pass through, but solutes are not free to do so. If two solutions with identical concentration (isotonic solutions) are separated by a semipermeable membrane, no net movement of solvent occurs.

32. Colligative Properties - Osmosis Osmosis: the net movement of a solvent through a semipermeable membrane toward the solution with greater solute concentration. In osmosis, there is net movement of solvent from the area of lower solute concentration to the area of higher solute concentration.  Movement of solvent from high solvent concentration to low solvent concentration

33. Colligative Properties - Osmosis Osmosis plays an important role in living systems:  Membranes of red blood cells are semipermeable. Placing a red blood cell in a hypertonic solution (solute concentration outside the cell is greater than inside the cell) causes water to flow out of the cell in a process called crenation.

34. Colligative Properties Placing a red blood cell in a hypotonic solution (solute concentration outside the cell is less than that inside the cell) causes water to flow into the cell.  The cell ruptures in a process called hemolysis.

35. Colligative Properties - Osmosis Other everyday examples of osmosis:  A cucumber placed in brine solution loses water and becomes a pickle.  A limp carrot placed in water becomes firm because water enters by osmosis.  Eating large quantities of salty food causes retention of water and swelling of tissues (edema).

36. Osmotic pressure Osmosis is the spontaneous movement of water across a semi- permeable membrane from an area of low solute concentration to an area of high solute concentration Osmotic Pressure - The Pressure that must be applied to stop osmosis  = iMRT where  – osmotic pressure (atm) i = van’t Hoff factor M = molarity R = 0.08206 L∙atm/mol∙K T = Kelvin temperature

37. Colloids: Suspensions of particles larger than individual ions or molecules, but too small to be settled out by gravity.

38. Tyndall Effect Colloidal suspensions can scatter rays of light. This phenomenon is known as the Tyndall effect.

39. Colloids in Biological Systems Some molecules have a polar, hydrophilic (water- loving) end and a nonpolar, hydrophobic (water- hating) end.

40. Colloids in Biological Systems Sodium stearate is one example of such a molecule.

41. Colloids in Biological Systems These molecules can aid in the emulsification of fats and oils in aqueous solutions.

42. Surfactants Change the surface properties so that two things that would not normally mix do  Emulsifying agent  Soap  Detergent

43. Hydrophobic – water-fearing (nonpolar) Hydrophilic – water-loving (polar)

44. Action of soap on oil