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 dissolved in water. 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. Freezing point elevation -  T f equals the freezing point of the pure solvent minus the freezing point of the solution.  T f = K f C m where K f = freezing point depression constant (depends on the solvent) C m = molal concentration (m = # mol solute/Kg solvent)