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Colligative Properties of Solutions

Colligative Properties of Solutions. are properties of solutions that depend solely on the number of particles of solute and NOT on their chemical identity. vapor pressure boiling point freezing point osmotic pressure. Vapor Pressure of a Solution.

jonah-contreras
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Colligative Properties of Solutions

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  1. Colligative Properties of Solutions • are properties of solutions that depend solely on the number of particles of solute and NOT on their chemical identity. • vapor pressure • boiling point • freezing point • osmotic pressure

  2. Vapor Pressure of a Solution A solute that is nonvolatile is one that has no measurable vapor pressure. We will study the effects of nonvolatile solutes on the properties of solutions. The presence of a nonvolatile solute causes the vapor pressure of the solution to be lower than the vapor pressure of the pure solvent.

  3. Vapor Pressure of a Solution – Raoult’s Law The extent to which the vapor pressure of a solvent is lowered by a nonvolatile solute is given, for an ideal solution, by Raoult’s Law. Raoult’s Law Psolvent over solution = XsolventP°solvent Xsolvent is the mole fraction of the solvent. P°solvent is the vapor pressure of the pure solvent at the solution temperature. Vapor pressure of the solution

  4. Raoult’s Law – Example: Nonvolatile Solute What is the vapor pressure at 25°C (room temperature) of a solution made by adding 226 g (1 cup) sugar (C12H22O11) to 118.5 mL (½ cup) of water? Pwater over sugar soln = XwaterP°water mol water = 118.5 mL x 0.99707 g x 1 mol mL 18.015 g mol sugar = 226 g x 1 mol 342.30 g Xwater = 6.559 = 0.9085 7.2192 = 6.559 mol = 0.660 mol

  5. Raoult’s Law – Example: Nonvolatile Solute What is the vapor pressure at 25°C (room temperature) of a solution made by adding 226 g (1 cup) sugar (C12H22O11) to ½ cup of water? Pwater over sugar solution = XwaterP°water Pwater over sugar solution = (0.9085) (23.76 torr) Adding sugar to the water lowered its vapor pressure. Adding more sugar would lower it still more. from Appendix B = 21.59 torr

  6. Raoult’s Law – Example: Two Volatile Components What is the vapor pressure at 25°C of 80-proof alcohol (40% alcohol by volume)? Alcohol: C2H5OH, MM = 46.07 g, v.p.(25°C) = 54.68 torr, density (25°C) = 0.786 g/mL Water: MM = 18.015 g, v.p.(25°C) = 23.76 torr, density (25°C) = 0.997 g/mL Apply Raoult’s Law to each volatile component. By convention, the liquid component present in larger volume is the “solvent.”

  7. Raoult’s Law – Example: Two Volatile Components What is the vapor pressure at 25°C of 80-proof alcohol (40% alcohol by volume)? Raoult’s Law for the water: Pwater over water/alcohol solution = XwaterP°water Find Xwater: We need a volume for the solution, don’t we? Any volume will do! 100 mL is convenient, though. 100 mL – 40 mL alcohol = 60 mL water

  8. Raoult’s Law – Example: Two Volatile Components Pwater over water/alcohol solution = XwaterP°water Find Xwater: 60 mL water x 0.997 g x 1 mol mL 18.015 g 40 mL alcohol x 0.786 g x 1 mol mL 46.07 g Xwater = 3.321 3.321+0.682 = 3.321 mol = 0.682 mol = 0.8296

  9. Raoult’s Law – Example: Two Volatile Components Pwater over water/alcohol solution = XwaterP°water Pwater over w/alcsoln = 0.8296 (23.76 torr) = 19.71 torr Palcohol over water/alc solution = XalcoholP°alcohol Xalcohol = 0.682 4.003 Palcohol over w/alcsoln = 0.170 (54.68 torr) =9.32 torr = 0.170

  10. Raoult’s Law – Example: Two Volatile Components Pwater over water/alcohol solution = 19.71 torr Palcohol over water/alcohol solution = 9.32 torr Now use Dalton’s Law of Partial Pressures to find the total vapor pressure of the solution: Ptot = 19.71 + 9.32 = 29.0 torr

  11. Boiling Point of a Solution • For water in an open container, the boiling point is the temperature at which the vapor pressure of water equals the prevailing atmospheric pressure. • Our sugar solution at 25°C has a lower vapor pressure than water at 25°C. • This means the temperature at which the sugar solution boils will be higher (102.8°C) than the temperature at which water boils (100.0°C). • This is called boiling point elevation.

  12. Boiling Point Elevation The relationship between boiling point elevation and the number of particles of solute in the solution is given by ΔTb = Kbm where ΔTb = Tbp(solution) - Tbp(pure solvent) Kb is the molal boiling-point-elevation constant and is for the solvent. m is the molality of particles from the solute.

  13. Boiling Point Elevation Now you can calculate the boiling point of our sugar solution yourself (Kb of water is 0.51°C/m): ΔTb = Kbm The molality of solute particles in our sugar solution is the same as the molality of the sugar itself. m = mol sugar = 0.660 mol sugar kg water 0.11815 kg water ΔTb = (0.51°C/m) (5.586m) ΔTb = Tbp(solution) - Tbp(pure solvent) = 2.8°C Tbp(solution) = 100.00°C + 2.8°C = 102.8°C = 5.586 m = 2.8°C

  14. Boiling Point Elevation - Electrolytes Electrolytes dissolve in water to form ions. Each ion is a solute particle. ΔTb = Kbm If we made our solution up with 0.660 mol of NaCl instead of sugar, the boiling point elevation would be different from that of sugar. m = 0.660 mol salt = 1.32 mol ions 0.11815 kg water 0.11815 kg water = 11.18 m

  15. Boiling Point Elevation - Electrolytes Electrolytes dissolve in water to form ions. Each ion is a solute particle. ΔTb = Kbm and m = 11.18 ΔTb = (0.51°C/m) (11.18 m) ΔTb = Tbp(solution) - Tbp(pure solvent) = 5.7°C Tbp(solution) = 100.00°C + 5.7°C = 105.7°C = 5.7°C

  16. Boiling Point Elevation - Electrolytes ΔTb = Kbm If we made our solution up with 0.660 mol CaCl2 instead of sugar or salt, the boiling point elevation would be even more, because CaCl2 dissolves in water to release 3 ions per mol.

  17. Freezing Point Depression The addition of a nonvolatile solute to a solution lowers the freezing point of the solution relative to that of the pure solvent.

  18. Freezing Point Depression The relationship betweenfreezing point depression and the number of particles of solute in the solution is given by ΔTf = Kfm where ΔTf = Tfp(pure solvent) - Tfp(solution) Kf is the molal freezing-point-depression constant and is for the solvent. m is the molality of particles from the solute. note the difference!!

  19. Freezing Point Depression We will now calculate the freezing point of our sugar solution (Kf of water is 1.86°C/m): ΔTf = Kfm m = mol sugar = 0.660 mol sugar kg water 0.11815 kg water ΔTf = (1.86°C/m) (5.586m) ΔTf = Tfp(pure solvent) - Tfp(solution) = 10.4°C Tfp(solution) = 0.00°C – 10.4°C = -10.4°C = 5.586 m = 10.4°C

  20. Freezing Point Depression We will now calculate the freezing point of our salt solution: ΔTf = Kfm m = mol ions = 1.32 mol ions kg water 0.11815 kg water ΔTf = (1.86°C/m) (11.17m) ΔTf = Tfp(pure solvent) - Tfp(solution) = 20.8°C Tfp(solution) = 0.00°C – 20.8°C = - 20.8°C If we had used CaCl2, Tfp(solution) would be even lower. That’s why CaCl2 is sometimes used to salt icy sidewalks. = 11.17 m = 20.8°C

  21. Boiling Point Elevation and Freezing Point Depression Adding a nonvolatile solute to a solvent raises its boiling point and lowers its freezing point.

  22. Boiling Point Elevation and Freezing Point Depression Another way to look at things: Adding a nonvolatile solute to a solvent expands its liquid range.

  23. Finding the Molar Mass of a Solute from Boiling Point Elevation or Freezing Point Depression Measurement • Using either ΔTb = Kbm or ΔTf = Kfm • If you know the mass of solute that is not an electrolyte and the mass of solvent used to make a solution, and • you can measure the freezing point depression or boiling point elevation of the solution, • you can calculate the molar mass of the solute.

  24. Finding the Molar Mass of a Solute from Boiling Point Elevation or Freezing Point Depression Measurement ΔTf = Kfm = Kfmol solute kg solvent = Kf (mass solute) (molar mass of solute)(kg solvent) Rearranging the equation gives: molar mass of solute = Kf (mass solute) ΔTf (kg solvent)

  25. Molar Mass from Boiling Point Elevation Data – Example A solution of 0.64 g of adrenaline in 36.0 g of CCl4 has a b.p. of 77.03°C. a) What is the molar mass of adrenaline? b) What is the f.p. of the solution? CCl4: b.p. (760 torr) = 76.54°C Kb = 5.02°C/m m.p. (760 torr) = -22.3°C Kf = 29.8°C/m ΔTb = Kbm ΔTb = 77.03 – 76.54 = 0.49°C m = ΔTb / Kb = 0.49°C = 0.0976 m 5.02°C/m

  26. Molar Mass from Boiling Point Elevation Data – Example A solution of 0.64 g of adrenaline in 36.0 g of CCl4 has a b.p. of 77.03°C. a) What is the molar mass of adrenaline? b) What is the f.p. of the solution? CCl4: b.p. (760 torr) = 76.54°C Kb = 5.02°C/m m.p. (760 torr) = -22.3°C Kf = 29.8°C/m Molar mass of adrenaline = (5.02°C/m) (0.64 g) 0.49°C (0.0360 kg) = 182 g/mol (really 180)

  27. Molar Mass from Boiling Point Elevation Data – Example A solution of 0.64 g of adrenaline in 36.0 g of CCl4 has a b.p. of 77.03°C. a) What is the molar mass of adrenaline? b) What is the f.p. of the solution? CCl4: b.p. (760 torr) = 76.54°C Kb = 5.02°C/m m.p. (760 torr) = -22.3°C Kf = 29.8°C/m ΔTf = Kfm ΔTf = 29.8°C (0.0976 m) m = 2.908°C

  28. Molar Mass from Boiling Point Elevation Data – Example A solution of 0.64 g of adrenaline in 36.0 g of CCl4 has a b.p. of 77.03°C. a) What is the molar mass of adrenaline? b) What is the f.p. of the solution? ΔTf = 29.8°C (0.0976 m) = 2.908°C m ΔTf =Tf (CCl4) - Tf(soln) = 2.908°C Tf(soln) = Tf(CCl4) - 2.908 = -22.3 - 2.908 = -25.2°C

  29. Osmotic Pressure • The last colligative property we will study is osmotic pressure. • It is based on the tendency of solvent molecules to move toward an area of lesser concentration. • This movement causes osmotic pressure when the areas of differing solvent concentration are separated by a semipermeable membrane.

  30. Osmotic Pressure Osmotic pressure is the pressure that must be applied to the solution in order to just stop the movement of solvent molecules into the solution.

  31. Osmotic Pressure The equation relating osmotic pressure (π) to concentration is very similar to the ideal gas law π = MRT M = molarity particles in the solution R = gas constant T = temperature in K

  32. Hypertonic Solutions Osmotic pressure plays an important role in living systems. For example, the membranes of red blood cells are semipermeable. When we eat too much salt, the high concentration of saltin our plasma makes ithypertonicrelative to the inside of the red blood cell and causes water to diffuse out of the red blood cells. A red blood cell in a hypertonic solution shrinks.

  33. Hypotonic Solutions When we perspire heavily and then drink a lot of water (not gatorade), the low concentration of saltin our plasma makes it hypotonic relative to the inside of the red blood cell and causes water to diffuse into the red blood cells. A red blood cell in a hypotonic solution expands and may burst.

  34. Isotonic Solutions When we lose a lot of fluids and have to replace them, the ideal situation is to receive fluids that are neither hypertonic nor hypotonic. Fluids that have the same osmotic pressure are said to be isotonic. The osmotic pressure of blood is 7.7 atm at 37°C. What concentration of saline solution (NaCl in sterile water) is isotonic with blood at human body temperature? Using M = π /RT, M = molarity of solute particles = 7.7 atm. 0.08206 L-atm (310. K) mol-K Molarity of NaCl for isotonic saline = 0.15 M In mass percent, an isotonic saline solution is 0.9% NaCl. = 0.30 M

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