1. Solutions

2. Occur in all phases u The solvent does the dissolving. u The solute is dissolved. u There are examples of all types of solvents dissolving all types of solvent. u We will focus on aqueous solutions.

3. Energy of Making Solutions  Heat of solution (  H soln ) is the energy change for making a solution. u Most easily understood if broken into steps. u 1.Break apart solvent u 2.Break apart solute u 3. Mixing solvent and solute

4. 1. Break apart Solvent  Have to overcome attractive forces.  H 1 >0 2. Break apart Solute.  Have to overcome attractive forces.  H 2 >0

5. 3. Mixing solvent and solute   H 3 depends on what you are mixing.  If molecules can attract each other  H 3 is large and negative.  Molecules can’t attract-  H 3 is small and negative. u This explains the rule “Like dissolves Like”

6. EnergyEnergy Reactants Solution H1H1 H2H2 H3H3 Solvent Solute and Solvent  Size of  H 3 helps determine whether a solution will form H3H3 Solution

7. Ways of Measuring u Molarity = moles of solute Liters of solution u % mass = Mass of solute x 100 Mass of solution  Mole fraction of component A  A = n A n A + n B

8. u Molality = moles of solute Kilograms of solvent  Molality is abbreviated m u Normality - read but don’t focus on it. u It is molarity x number of active pieces Ways of Measuring

10. Vapor Pressure of Solutions u A nonvolatile solute lowers the vapor pressure of the solution. u The molecules of the solvent must overcome the force of both the other solvent molecules and the solute molecules.

11. Raoult’s Law:  P soln =  solvent x P solvent u Vapor pressure of the solution = mole fraction of solvent x vapor pressure of the pure solvent u Applies only to an ideal solution where the solute doesn’t contribute to the vapor pressure.

12. Aqueous Solution Pure water u Water has a higher vapor pressure than a solution

13. Aqueous Solution Pure water u Water evaporates faster from for water than solution

14. u The water condenses faster in the solution so it should all end up there. Aqueous Solution Pure water

15. Practice Problem u A solution of cyclopentane with a nonvolatile compound has vapor pressure of 211 torr. If vapor pressure of the pure liquid is 313 torr, what is the mole fraction of the cyclopentane? u P soln = X cp P cp u 211 torr = X cp (313 torr) u.674

16. Try one on your own u Determine the vapor pressure of a solution at 25  C that has 45 grams of C 6 H 12 O 6, glucose, dissolved in 72 grams of H 2 O. The vapor pressure of pure water at 25  C is 23.8 torr. u P solution = X solvent P solvent u P solution =.941(23.8 torr) u P solution = 22.4 torr

17. u Liquid-liquid solutions where both are volatile. u Modify Raoult’s Law to  P total = P A + P B =  A P A 0 +  B P B 0 u P total = vapor pressure of mixture u P A 0 = vapor pressure of pure A u If this equation works then the solution is ideal. Ideal solutions

18. χbχb χAχA Vapor Pressure P of pure A P of pure B Vapor Pressure of solution

19. Deviations u If solvent has a strong affinity for solute (H bonding). u Lowers solvent’s ability to escape. u Lower vapor pressure than expected. u Negative deviation from Raoult’s law.   H soln is large and negative exothermic.  Endothermic  H soln indicates positive deviation.

20. χbχb χAχA Vapor Pressure Positive deviations- Weak attraction between solute and solvent Positive ΔH soln

21. χbχb χAχA Vapor Pressure Negative deviations- Strong attraction between solute and solvent Negative ΔH soln

22. Problem #1 u The vapor pressure of a solution containing 53.6g of glycerin C 3 H 8 O 3 in133.7g ethanol C 2 H 5 OH is 113 torr at 40C. Calculate the vapor pressure of pure ethanol at 40C assuming that the glycerin is a non volatile, nonelectrolyte solute in ethanol.

23. Answer to #1 P soln = X eth P eth 113torr = 2.90mol/3.48mol (P eth ) 135.6 torr = P eth

24. Problem #2 u At a certain temperature, the vapor pressure of pure benzene C 6 H 6 is 0.930atm. A solution was prepared by dissolving 10.0g of a nondissociating, nonvolatile solute in 78.11g of benzene at that temperature. The vapor pressure was found to be 0.900atm. Assuming the solution behave ideally, determine the molar mass of the solute.

25. Answer #2 P soln = X benzene P benzene.900atm = X benzene (.930atm) X benzene =.9677 X solute =.0323 MM = 10.0g/.0323mol = 310g/mol

26. Problem #3 A solution of NaCl in water has a vapor pressure of 19.6 torr at 25C. What is the mol fraction of solute particle in this solution if the vapor pressure of water is 23.8 torr at 25C?

27. Answer #3 P soln = X water P water 19.6torr = X water (23.8torr).824 = X water therefore X solute =.176

28. Problem #3 For the same problem as #3: What is the vapor pressure of the solution at 45C if the vapor pressure of water is 71.9 torr at 45C?

29. Answer #3 P soln =.824(71.9torr) P soln = 59.2 torr

30. Problem #4 A solution is made from 0.0300mol CH 2 Cl 2 and 0.0500mol CH 2 Br 2 at 25C. Assuming that the solution is ideal, calculate the % composition of the vapor at 25C. P CH2Cl2 = 133 torr P CH2Br2 = 11.4 torr

31. Answer #4 P soln = X CH2Cl2 P + X CH2Br2 P P soln = (.03/.08)(133torr) + (.05/.08)(11.4 torr) P soln = 57.0 torr X CH2Cl2 = 49.9 / 57 =.875 = 87.5% X CH2Br2 = 7.13 / 57 =.125 = 12.5%

33. Colligative Properties u Because dissolved particles affect vapor pressure - they affect phase changes. u Colligative properties depend only on the number - not the kind of solute particles present u Useful for determining molar mass

34. Water

35. 1 atm Vapor Pressure of solution Vapor Pressure of pure water

36. 1 atm Freezing and boiling points of solvent

37. 1 atm Freezing and boiling points of solution

38. 1 atm TfTf TbTb

39. Boiling point Elevation u Because a non-volatile solute lowers the vapor pressure it raises the boiling point.  The equation is:  T = K b m solute   T is the change in the boiling point u K b is a constant determined by the solvent.  m solute is the molality of the solute

40. Freezing point Depression u Because a non-volatile solute lowers the vapor pressure of the solution it lowers the freezing point. Freezing occurs when Pv (l) = Pv(s). Because the Pv(soln) is reduced, the temp must be lowered to form a solid.  The equation is:  T = -K f m solute   T is the change in the freezing point u K f is a constant determined by the solvent  m solute is the molality of the solute

41. Electrolytes in solution u Since colligative properties only depend on the number of molecules. u Ionic compounds should have a bigger effect. u When they dissolve they dissociate. u Individual Na and Cl ions fall apart. u 1 mole of NaCl makes 2 moles of ions. u 1mole Al(NO 3 ) 3 makes 4 moles ions.

42. u Electrolytes have a bigger impact on on melting and freezing points per mole because they make more pieces.  Relationship is expressed using the van’t Hoff factor i i = Moles of particles in solution Moles of solute dissolved u The expected value can be determined from the formula of the compound.

43. u The actual value is usually less because u At any given instant some of the ions in solution will be paired up. u Ion pairing increases with concentration.  i decreases with increasing concentration. u We can change our formulas to  = iKm

44. Problem #1 A 2.00 gram sample of a large biomolecule was dissolved in 15.0 grams of CCl 4. The boiling point of this solution was found to be 77.85C. Calculate the molar mass of the biomolecule. K b = 5.03C kg / mol BP CCl4 = 76.5C

45. Answer #1  T = K b m (77.85 – 76.5C) = 5.03C Kg/mol x m.268mol/Kg = m.268 mol/Kg = Xmol /.015kg =.00402 mol 2.00g /.00402mol = 498g/mol

46. Problem #2 The freezing point of t-butanol is 25.5C and K f = 9.1C Kg/mol. Usually t-butanol absorbs water on exposure to air. If the freezing point of a 10.0 gram sample of t- butanol is 24.59C, how many grams of water are present in the sample?

47. Answer #2  T = -K f m (24.59 – 25.5C) = -9.1 C Kg/mol x m.1 mol/Kg = m.1 m = X mol H 2 O /.01kg =.001mol H 2 O.018g H 2 O

48. Problem #3 Calculate the freezing point and the boiling point for each:.050m MgCl 2.050m FeCl 3 K f = 1.86C Kg/mol K b = 0.51C Kg/mol Predict which will have higher BP and which will have the lower FP

49. Answer #3 Dissociate each: MgCl 2 Mg 2+ + 2Cl - i= 2.7 FeCl 3 Fe 3+ + 3Cl - i= 3.4 MgCl 2 T f = -.25C T b = 100.069C FeCl 3 T f = -.32C T b = 100.087C