Chemistry B Colligative Properties

1. Chemistry B Colligative Properties © Mr. D. Scott; CHS

6. In optimal flight conditions, air moves smoothly over the wings of the plane. Deicing system not operating. Cells lie close to airfoil section. Ice is permitted to form. Ice on the wings disrupts airflow and adds weight, requiring more airspeed to maintain the same amount of lift. After deicer system has been put into operation, the center cell inflates cracking ice. Deicer boot on leading edge of wing When the center cell deflates, outer cells inflate. This raises the cracked ice causing it to be blown off by the air stream.

8. Colligative Properties of Solutions Colligative properties are the solvent properties that change when a solute is dissolved. They depend only on the number of solute particles in solution and not on the nature (identity) of the solute particles. Vapor-Pressure Lowering Boiling-Point Elevation Freezing-Point Depression Osmotic Pressure (p)

9. Vapor Pressure Lowering in a Solution The diagram below shows how a phase diagram is affected by dissolving a solute in a solvent. The black curve represents the pure liquid and the blue curve represents the solution. Notice the changes in the freezing & boiling points.

10. Vapor Pressure Lowering The presence of a non-volatilesolute results in reduced evaporation. The bonds between the solute-solvent particles cause more solvent particles to require more energy to change into the vapor phase.

11. Experimenting with Vapor Pressure Lowering The liquids in both beakers evaporate but not at the same rate. The vapor condenses back into both beakers equally. What is the long-term effect of this difference?

12. Example Problem 1 What is the vapor pressure of an aqueous solution containing 100. g of sucrose (C12H22O11) dissolved in 400. g of water at 25.0°C? (Po=23.8 mmHg) Raoult’s law

13. Mixtures of Volatile Liquids Since both liquids evaporate, they both contribute to the vapor pressure. These mixtures result in HIGHER total vapor pressures than would exist when they are separate.

14. Boiling-Point Elevation 0 DTb = Tb – T b 0 T b is the boiling point of the pure solvent 0 Tb > T b DTb = Kbm i T b is the boiling point of the solution DTb > 0 m is the molality of the solution Kbis the molal boiling-point elevation constant (°C/m) for a given solvent iis the van’t Hoff Factor for a given solute

15. Freezing-Point Depression 0 DTf = T f – Tf 0 T f is the freezing point of the pure solvent 0 T f > Tf DTf = Kfm i T f is the freezing point of the solution DTf > 0 m is the molality of the solution Kfis the molal freezing-point depression constant (°C/m) for a given solvent iis the van’t Hoff Factor for a given solute

17. Electrolyte Solutions & van’t Hoff Factors actual number of particles in soln after dissociation van’t Hoff factor (i) = number of formula units initially dissolved in soln Electrolytes divide into ions upon dissolving and result in yielding more particles in solution than their formulas indicate. Example: NaCl  Na+ + Cl- dissociation results in two ions Theoretically, the van’t Hoff Factor would be 2.0 The van’t Hoff Factor of 0.0500 M Electrolyte Solutions at 25°C

18. Example Problem 2 What is the freezing point of an aqueous solution containing 100. g of NaCl dissolved in 600. g of water? DTf = Kfm i Na+/Cl- Kf = 1.86 C°/m and i = 2.0 (two ions) Molal concentration must be calculated. Solution FP = Normal FP – DTf = 0.0 – 10.6 = - 10.6 °C

19. Example Problem 3 The Dead Sea is a hypersaline body of water. It has 33.7% salinity. It is 8.6 times more salty than the average ocean or sea. It has an average density of 1.24 Kg/L (374g salt/Kg H2O). If the salt concentration is 33.7 %w/v and the average molar mass of the salts is 85.4 g/mol, what is the boiling point of this water? m = mol/Kg = 4.38 m Solution BP = Normal BP + DTb 100 °C + 5.69 C° = 106 °C

20. Example Problem 4 moles of solute m= mass of solvent (kg) = 3.255 kg solvent 1 mol 62.0 g 1499 g x What is the freezing point of a solution containing 1499 g of ethylene glycol (antifreeze) in 3255 g of water? The molar mass of ethylene glycol is 62.0 g\mol. DTf = Kfm i Kf water = 1.86 0C/m = 7.43 m i = 1.0 for all non-electrolytes (molecules that don’t ionize) DTf = Kfm = 1.86 0C/m x 7.43 m = 13.8 0C Solution FP = Normal FP - DTf = 0.00 0C – 13.8 0C = -13.8 0C

21. Variationsof DT = K m i substitute

22. Molecular Weight by Freezing Point Depression Example Problem 5 2.00 g of an unknown molecular solute is dissolved in 10.0 g of solvent with Kf = 4.5 C°/m. The DTf is measured to be 5.00 C° (shown at left). What is the molar mass of the solute? Temperature DTf Molar mass = 180. g/mol Time

23. 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 p = MRTi where p = osmotic pressure i = van’t Hoff factor M = molarity R = ideal gas constant T = Kelvin temperature

24. Chemistry In Action: Desalination

25. Osmotic Pressure Example Problem 6 The Jubail desalination plant in Saudi Arabia is the largest in the world. The plant, located on the Persian Gulf, produces 800 million gallons per day (50% of Saudi Arabia’s fresh water supply). If the concentration of salt in the seawater is 0.60 M, what pressure in psi is required to start reverse osmosis (used in desalination)? Assume 25.0°C, i = 2.0, and 1atm=14.7psi For comparison, most car tires are inflated to between 45 and 50 psi.

26. Red blood cells (cell membrane only) Concentration: inside<outside inside=outside inside>outside Plant cells (cell membrane & cell wall)