1. Unit # 4 Colligative Properties.

2. Colligative Properties - Properties that depend on the concentration of solute molecules or ions in solution, but do not depend on the chemical identity of the solute. (For example, addition of Ethylene glycol or Urea to water will have the same effect.)

3. Examples of Colligative Properties: 3.Vapor pressure lowering 1.Boiling point elevation 2.Freezing point depression

4. The above colligative properties depend on the mole fraction of the solvent. Mole fraction - The mole fraction of substance “A” is represented as X A X A = # of moles of “A” Total moles of solution (solute + solvent)

5. Example #1: A solution is 1 mol Ethylene glycol and 9 mol water. Calculate the mole fraction of both the solute and the solvent.

6. X Ethylene glycol = 1 mol Ethylene glycol 10 mol solution x 100 X Ethylene glycol = 10 %

7. X Water = 9 mol water 10 mol solution x 100 X Water = 90 %

8. Example #2: What are the mole fractions of glucose and water in a solution containing 5.67g of glucose dissolved in 25.2g of water ?

9. Clues: 1.Determine the chart mass of the solute and the solvent. 2.Determine the number of moles of the solute and solvent 3.Use the mole fraction formula to solve.

10. Glucose = C 6 H 12 O 6 = 180.1572 g/mol Use “DIMO” to calculate moles. 5.67g glucose =0.0315 mol glucose

11. Water = H 2 O = 18.0152 g/mol Use “DIMO” to calculate moles. 25.2g water =1.40 mol water

12. Mole fraction glucose 0.0315 mol glucose 1.4315 mol solution = 0.0220 x 100 = 2.20 % Mole fraction water 1.40 mol water 1.4315 mol solution = 0.978 x 100 = 97.8 %

13. Example #3: A bleaching solution contains Sodium hypochlorite, NaClO, dissolved in water. The bleach is 0.750m NaClO. What is the mole fraction of Sodium hypochlorite ?

14. Clues: 1.Recall the definition of molality. 2.Convert Kg of solvent to moles 3.Use the mole fraction formula to solve

15. 0.750m = 0.750 mol NaClO 1 Kg solvent 1 Kg solvent (water) = 55.5 mol water X NaClO = 0.750 mol NaClO 56.25 mol solution = 0.013 %

16. Example #4: Vinegar is 0.763M Acetic acid, CH 3 COOH. The density is 1.004 g/mL. What is the mole fraction of the acetic acid ?

17. Vapor pressure lowering - Vapor pressure lowering of a solvent is equal to the vapor pressure of the pure solvent minus the vapor pressure of the solution. There are three formulas associated with Vapor Pressure Lowering.

18. 1.  P lowering = Vapor pressure solvent - VP solution 2.  P lowering = P Å x X B P Å = vapor pressure of pure solvent X B = mole fraction of solute 3.  P lowering = P Å - P A P A = partial pressure of solvent

19. V.P. lowering example #1: V.P. water = 17.54 mmHg V.P. 0.0100m antifreeze sol’n = 17.36 mmHg V.P. lowering = 0.18 mmHg Lowering vapor pressure, increases boiling point.

20. V.P. lowering example #2: How much will the vapor pressure of water drop when 5.67g of glucose are dissolved in 25.2g of water at 25  C and what will the new vapor pressure be ? V.P. water 25  = 23.8 mmHg Clue: Use:  P lowering = P Å x X B

21.  P lowering = P Å x X B 5.67g glucose = 0.0315 mol 25.2g water = 1.40 mol Total mol = 1.4315 mol Mol fraction glucose = 0.0220 = 23.8 mmHg x 0.0220 = 0.524 mmHg

22. New V.P. = V.P. pure solvent - drop in V.P. = 23.8 mmHg - 0.524 mmHg = 23.3 mmHg

23. V.P. lowering example #3: Calculate the vapor pressure at 35  C of a solution made by dissolving 20.2g of sucrose, C 12 H 22 O 11, in 60.5g of water. V.P. water 35  C = 42.2 mmHg

24. V.P. lowering example #4: Naphthalene, C 10 H 8, is used to make mothballs. Suppose 0.515g of Naphthalene are dissolved into 60.8g of chloroform, CHCl 3 ; calculate the new vapor pressure of the solvent. V.P. chloroform 20  C = 156 mmHg

25. Boiling point elevation -  T b equals the boiling point of the solution minus the boiling point of the pure solvent.  T b = i K b C m where i = The number of ions if solvent is polar. K b = boiling point elevation constant (depends on the solvent) C m = molal concentration (m = # mol solute/Kg solvent)

26. What is the “Normal Boiling Point” of a substance? The temperature at which the vapor pressure equals 1 atmosphere. Therefore if: Vapor Pressure increases, Boiling Point decreases. Vapor Pressure decreases, Boiling Point increases.

27. B.P. Elevation example #1: Naphthalene, C 10 H 8, is dissolved in Benzene until a 0.100m solution is attained. Benzene’s normal boiling point is 80.2  C. It’s boiling point elevation constant is 2.61  C /m. What is the new boiling point ?

28. Clues: 1.Write the formula that you will use. 2.Plug the given values into the formula; or, set up the proportion needed to calculate the increase in boiling point. 3.Add the calculated increase in temperature to the Normal Boiling Point to determine the new boiling point.

29.  T b = i K b C m = 1 x K b x C m = 1 x 2.61  C /m x 0.100m = 0.261  C increase New B.P. = 80.2  C + 0.261  C = 80.461  C

30. B.P. Elevation example #2: 89.75g of Sodium chloride were added to 250mL water. K b H20 = 0.512  C /m. What is the new boiling point ?

31.  T b = i K b C m First calculate the molality. 89.75g NaCl = mol“X”1.54 250mL H 2 O = g H 2 O = “X” 250 Kg H 2 O“X” 0.250 Therefore: 1.54 mol 0.250 Kg =6.14m

32.  T b = i K b C m = 2 x K b x C m = 2 x 0.512  C /m x 6.14m = 6.29  C increase New B.P. = 100  C + 6.29  C = 106.29  C

33. B.P. Elevation example #3: An aqueous solution is 0.0222m Glucose. Water’s boiling point elevation constant is 0.512  C /m. What is the new boiling point ?

34.  T b = i K b C m = 1 x K b x C m = 1 x 0.512  C /m x 0.0222m = 0.0114  C increase New B.P. = 100  C + 0.0114  C = 100.0114  C

35. B.P. Elevation example #4: What is the boiling point of a solution of 0.152g of Glycerol, C 3 H 8 O 3, in 20g of water ?

36. Clues: 1.Determine the molality of the solution. 2.Solve using the formulas as before.

37. m = mol solute Kg solvent C 3 H 8 O 3 = 92.0944 g/mol = 0.152g =0.00165 mol C 3 H 8 O 3 20g H 2 O =XKg0.020 = 0.00165 mol C 3 H 8 O 3 0.020 Kg = 0.0825m

38.  T b = i K b C m = 1 x K b x C m = 1 x 0.512  C /m x 0.0825m = 0.0422  C increase New B.P. = 100  C + 0.0422  C = 100.0422  C

39. B.P. Elevation example #5: A solution was prepared by dissolving 0.915g of Sulfur, S 8, in 100.0g of Acetic acid. What is the new boiling point ? B.P. of Acetic acid = 118.5  C K b of Acetic acid = 3.08  C /m

40. Determination of molar mass from B.P. elevation. Example #1: A solution of 9.75g of a solute in 29.38g of water boils at 100.78  C at 1 atm of pressure. What is the molar mass of the solute ?

41. Clues: 1.Determine the molality of the solution. 2.From the molality, find the number of molesof the solute in the mass of solvent you areworking with. 3.From the given mass of solute and determined number of moles in the problem; determinethe number of grams of solute in one mole.

42.  T b = i K b C m = 1 x K b x C m = 1 x 0.512  C /m x “X”m = “X”m 0.78  C 1.52m

43. 1.52 mol 1 Kg = 0.02938 Kg mol “X” 0.0447 And if… 9.75g 0.0447 mol = 1 mol g “X” 218.33

44. Freezing point elevation -  T f equals the freezing point of the pure solvent minus the freezing point of the solution.  T f = i K f C m where i = The number of ions if solvent is polar. K f = freezing point depression constant (depends on the solvent) C m = molal concentration (m = # mol solute/Kg solvent)

45. Freezing point depression example #1: How many grams of Ethylene glycol, C 2 H 6 O 2, must be added to 37.8g of water to give a freezing point of -0.150  C ?  T f H2O = 1.858  C /m

46. Determination of molar mass from F.P. depression. Example #2: Safrole is contained in oil of sassafras and was once used to flavor root beer. A 2.39mg sample of safrole was dissolved in 103.0mg of Diphenyl ether. The solution had a melting point of 25.70  C. Calculate the molecular weight of Safrole. F.P. pure Diphenyl ether = 26.84  C Kf = 8.00  C /m

47. Clues: 1.Write the formula that you will use. 2.Plug the given values into the formula; or, set up the proportion needed to calculate the increase in boiling point. 3.Add the calculated increase in temperature to the Normal Boiling Point to determine the new boiling point.

48.  T f = i K f C m C m = 0.1425m  T f = 1.14  C You must calculate the following to solve: Final Answer = 162.83 g/mol

49. Determination of molar mass from F.P. depression. Example #3: An aqueous solution was prepared by dissolving 0.131g of a substance in 25.5g of water. The molality of the solution was determined by Freezing Point Depression to be 0.056m. What is the molecular weight of the solute ?