Chapter 9.4 – Colligative Properties CHM1111 Section 04 Instructor: Dr. Jules Carlson Class Time: M/W/F 1:30-2:20 Wednesday, November 23rd

Chapter Sections for the Exam Ch 1: (Entire Chapter) Ch 2: 2.1 to 2.6 Ch 4: 4.1 to 4.6 Ch 5: 5.1 to 5.4, 5.6 Ch 6: (Entire Chapter) Ch 7: 7.1 to 7.4 Ch 8: 8.1 to 8.4 Ch 9: 9.1 to 9.4 Ch 10: 10.1 to 10.2

Gas-Solution Equilibria   Chemistry, Canadian Edition ©2010 John Wiley & Sons Canada, Ltd.

Henry’s Law Example At 25⁰C, the equilibrium pressure of HCl vapour above a 0.100 M aqueous solution of HCl is 0.53 kPa. Calculate the Henry’s Law constant for HCl at 25⁰C, and determine the equilibrium pressure in bars of HCl vapour above a 6.00 M solution at the same temperature.

Vapour Pressure Reduction and Raoult’s Law A pure solvent in a closed system will reach a dynamic equilibrium between the liquid and vapour phases. The addition of a solute will decrease the vapour pressure because the solute molecules reduce the rate of escape of solvent molecules. This occurs because an increase in solute molecules decreases the number of solvent molecules at the air-liquid interface. Raoult’s Law quantifies this: Where A is the solvent, and B, C, … are volatile solutes. If the solutes are not volatile they are not included.  

Distillation Distillation is an important separation process. You can separate compounds by differences in their vapour pressures. Say you have a mixture of benzene and toluene. Suppose the mole fractions in the liquid mixture are Xbenzene = 0.400 and Xtoluene = 0.600, and vapour pressures are 12.7 kPa for Benzene and 3.79 kPa for Toluene at 25 C. Using Raoult’s Law we can get the partial pressures (equal to vapour pressures at equilibrium):        

Distillation  

Boiling P Elevation / Freezing P Depression Adding a solute decreases the escape from the liquid phase reducing the vapour pressure of the solution, so increases the boiling point. Adding a solute has the reverse effect on freezing point. Adding solute decreases the rate at which the solid captures solvent molecules from the solution.

Extent of FP Depression or BP Elevation      

Kf and Kb For Different Compounds Solvent Formula Melting Point (°C) Boiling Point (°C) Kf(°C/b) Kb(°C/b) Water H2O 0.00 100.000 1.858 0.512 Acetic acid HC2H3O2 16.60 118.5 3.59 3.08 Benzene C6H6 5.455 80.2 5.065 2.61 Camphor C10H16O 179.5 ... 40 Carbon disulfide CS2 46.3 2.40 Cyclohexane C6H12 6.55 80.74 20.0 2.79 Ethanol C2H5OH 78.3 1.07 Boiling point elevation and freezing point depression constants are unique for each molecule and account for intermolecular forces in the solvent. Greater intermolecular forces result in smaller Kf and Kb

Boiling Point Elevation Problem 1 Practice Exercise 9.8 from the text: Calculate the boiling point of an 8 % by mass aqueous solution of Na2SO4.

Boiling Point Elevation Problem 2 A solution was prepared by dissolving 18.00 g glucose in 150.0 g of water. The resulting solution was found to have a boiling point of 100.34 C. Calculate the molar mass of glucose.

Freezing Point Depression Problem What mass of ethylene glycol (C2H6O2, molar mass = 62.1 g/mol). The main component of antifreeze, must be added to 10.0 L water to produce a solution for use in a car’s radiator that freezes at -23.3 C. Assume the density of water is exactly 1 g/mL.

I clicker Question Which of the following statements are true: Henry’s Law constants are greater at higher temperatures. A distillation performed with mole fractions of 0.5 for methanol and ethanol in solution would have Xethanol > 0.5 in the gas phase. FeCl2 would produce a larger decrease in freezing point than would FeCl3 at the same concentration. The greater the molality of an aqueous NaCl solution over ice, the poorer the recapture of water molecules into the ice. None of the statements are true.

Osmosis Osmosis is the movement of solvent molecules through a semi-permeable membrane. Consider a membrane with water on one side and sugar in water on the other. Water will flow across to the side with water and sugar to try to equalize water concentrations (sugar cannot cross the membrane) reaching a dynamic equilibrium 55.5 M H2O < 55.5 M H2O

Osmotic Pressure  

Using Osmotic Pressure to Determine Molar Mass  

Osmotic Pressure Problem What concentration of sodium chloride in water is needed to produce a solution that is isotonic (has the same osmotic pressure) with blood (Π = 7.80 bar at 25 C)

Molar Mass From Osmotic Pressure Problem To determine the molar mass of a certain protein, 1.00 x 10-3 g of it was dissolved in enough water to make 1.00 mL of solution. The osmotic pressure of this solution was found to be 1.12 torr at 25 C. Calculate the molar mass of the protein.