 Download Presentation Solutions part 2

# Solutions part 2

Download Presentation ## Solutions part 2

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript

1. Solutionspart 2 • Colligative Properties of Solutions Dr. Hisham E Abdellatef Professor of pharmaceutical analytical chemistry http://www.staff.zu.edu.eg/ezzat_hisham/browseMyFiles.asp?path=./userdownloads/physical%20chemistry%20for%20clinical%20pharmacy/

2. Colligative Properties of Solutions • There are four common types of colligative properties: • Vapor pressure lowering • Freezing point depression • Boiling point elevation • Osmotic pressure • Vapor pressure lowering is the key to all four of the colligative properties.

3. Lowering of Vapor Pressure and Raoult’s Law • Addition of a nonvolatile solute to a solution lowers the vapor pressure of the solution. • The effect is simply due to fewer solvent molecules at the solution’s surface. • The solute molecules occupy some of the spaces that would normally be occupied by solvent. • Raoult’s Law models this effect in ideal solutions.

4. Lowering of Vapor Pressure and Raoult’s Law • Derivation of Raoult’s Law.

5. Lowering of Vapor Pressure and Raoult’s Law • Lowering of vapor pressure, Psolvent, is defined as:

6. Lowering of Vapor Pressure and Raoult’s Law • Remember that the sum of the mole fractions must equal 1. • Thus Xsolvent + Xsolute = 1, which we can substitute into our expression.

7. Lowering of Vapor Pressure and Raoult’s Law • This graph shows how the solution’s vapor pressure is changed by the mole fraction of the solute, which is Raoult’s law.

8. Examples The vapor pressure of water is 17.5 torr at 20°C. Imagine holding the temperature constant while adding glucose, C6H12O6, to the water so that the resulting solution has XH2O = 0.80 and XGlu = 0.20. What is , the vapor pressure of water over the solution = 14 torr

9. Glycerin, C3H8O3, is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25°C. Calculate the vapor pressure at 25°C of a solution made by adding 50.0 mL of glycerin to 500.0 mL of water. The vapor pressure of pure water at 25°C is 23.8 torr

10. The vapor pressure of pure water at 110°C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.00 atm at 110°C. Assuming that Raoult's law is obeyed, what is the mole fraction of ethylene glycol in the solution? Answer: 0.290 P°H2O =1070 torr PH2O = 1 Atm = 760 torr PH2O P°H2O 760 torr 1070 torr XH2O = --------- = --------- = XH2O + XEG = 1 0.7103 + XEG = 1 1- 0.7103 = XEG = 0.290 XEG =

11. Many solutions do not obey Raoult's law exactly: They are not ideal solutions. If the intermolecular forces between solvent and solute are weaker than those between solvent and solvent and between solute and solute, then the solvent vapor pressure tends to be greater than predicted by Raoult's law. Conversely, when the interactions between solute and solvent are exceptionally strong, as might be the case when hydrogen bonding exists, the solvent vapor pressure is lower than Raoult's law predicts. Although you should be aware that these departures from ideal solution occur, we will ignore them for the remainder of this chapter.

12. More Examples Sucrose is a nonvolatile, nonionizing solute in water. Determine the vapor pressure lowering, at 27°C, of a solution of 75.0 grams of sucrose, C12H22O11, dissolved in 180. g of water. The vapor pressure of pure water at 27°C is 26.7 torr. Assume the solution is ideal. Vapor Pressure Lowered = 26.7-26.1= 0.6

13. solution is made by mixing 52.1 g of propyl chloride, C3H8Cl, and 38.4 g of propyl bromide, C3H8Br. What is the vapor pressure of propyl chloride in the solution at 25°C? The vapor pressure of pure propyl chloride is 347 torr at 25°C and that of pure propyl bromide is 133 torr at 25°C. Assume that the solution is an ideal solution.

14. . At 25°C a solution consists of 0.450 mole of pentane, C5H12, and 0.250 mole of cyclopentane, C5H10. What is the mole fraction of cyclopentane in the vapor that is in equilibrium with this solution? The vapor pressure of the pure liquids at 25°C are 451 torr for pentane and 321 torr for cyclopentane. Assume that the solution is an ideal solution.

15. Fractional Distillation • Distillation is a technique used to separate solutions that have two or more volatile components with differing boiling points. • A simple distillation has a single distilling column. • Simple distillations give reasonable separations. • A fractional distillation gives increased separations because of the increased surface area. • Commonly, glass beads or steel wool are inserted into the distilling column.

16. Boiling Point Elevation • Addition of a nonvolatile solute to a solution raises the boiling point of the solution above that of the pure solvent. • This effect is because the solution’s vapor pressure is lowered as described by Raoult’s law. • The solution’s temperature must be raised to make the solution’s vapor pressure equal to the atmospheric pressure. • The amount that the temperature is elevated is determined by the number of moles of solute dissolved in the solution.

17. Boiling Point Elevation • Boiling point elevation relationship is:

18. Boiling Point Elevation • Example 14-4: What is the normal boiling point of a 2.50 m glucose, C6H12O6, solution?

19. Boiling-Point Elevation The addition of a nonvolatile solute lowers the vapor pressure of the solution. At any given temperature, the vapor pressure of the solution is lower than that of the pure liquid

20. The increase in boiling point relative to that of the pure solvent, DTb, is directly proportional to the number of solute particles per mole of solvent molecules. Molality expresses the number of moles of solute per 1000 g of solvent, which represents a fixed number of moles of solvent

21. Automotive antifreeze consists of ethylene glycol, C2H6O2, a nonvolatile nonelectrolyte. Calculate the boiling point of a 25.0 mass percent solution of ethylene glycol in water.

22. Calculate the boiling point of a solution of 2.0 molal of NaCl. Kb, water= 0.52 °C /mola. Dt = Kbm NaCl(aq)  Na+ + Cl- 2.0 m 2.0 m 2.0 m 2.0 m + 2.0 m = 4.0m Dt = (0.52 °C/molal)(4.0 molal) =2.08 °C BP = NBP +Dt = 100.00°C +2.08 °C = 102.08° C

23. Freezing Point Depression • Addition of a nonvolatile solute to a solution lowers the freezing point of the solution relative to the pure solvent. • See table 14-2 for a compilation of boiling point and freezing point elevation constants.

25. Fundamentally, freezing point depression and boiling point elevation are the same phenomenon. The only differences are the size of the effect which is reflected in the sizes of the constants, Kf & Kb. This is easily seen on a phase diagram for a solution. Freezing Point Depression • Notice the similarity of the two relationships for freezing point depression and boiling point elevation.

26. Freezing Point Depression

27. Freezing Point Depression • Example 14-5: Calculate the freezing point of a 2.50 m aqueous glucose solution.

28. Freezing Point Depression • Example 14-6: Calculate the freezing point of a solution that contains 8.50 g of benzoic acid (C6H5COOH, MW = 122) in 75.0 g of benzene, C6H6. You do it!

29. Freezing Point Depression

30. Determination of Molecular Weight by Freezing Point Depression • The size of the freezing point depression depends on two things: • The size of the Kf for a given solvent, which are well known. • And the molal concentration of the solution which depends on the number of moles of solute and the kg of solvent. • If Kf and kg of solvent are known, as is often the case in an experiment, then we can determine # of moles of solute and use it to determine the molecular weight.

31. Determination of Molecular Weight by Freezing Point Depression • Example 14-7: A 37.0 g sample of a new covalent compound, a nonelectrolyte, was dissolved in 2.00 x 102 g of water. The resulting solution froze at -5.58oC. What is the molecular weight of the compound?

32. Colligative Properties and Dissociation of Electrolytes • Electrolytes have larger effects on boiling point elevation and freezing point depression than nonelectrolytes. • This is because the number of particles released in solution is greater for electrolytes • One mole of sugar dissolves in water to produce one mole of aqueous sugar molecules. • One mole of NaCl dissolves in water to produce two moles of aqueous ions: • 1 mole of Na+ and 1 mole of Cl- ions

33. Colligative Properties and Dissociation of Electrolytes • Remember colligative properties depend on the number of dissolved particles. • Since NaCl has twice the number of particles we can expect twice the effect for NaCl than for sugar. • The table of observed freezing point depressions in the lecture outline shows this effect.

34. Colligative Properties and Dissociation of Electrolytes • Ion pairingor association of ions prevents the effect from being exactly equal to the number of dissociated ions

35. Colligative Properties and Dissociation of Electrolytes • The van’t Hoff factor, symbol i, is used to introduce this effect into the calculations. • i is a measure of the extent of ionization or dissociation of the electrolyte in the solution.

36. i has an ideal value of 2 for 1:1 electrolytes like NaCl, KI, LiBr, etc. i has an ideal value of 3 for 2:1 electrolytes like K2SO4, CaCl2, SrI2, etc. Colligative Properties and Dissociation of Electrolytes

37. Colligative Properties and Dissociation of Electrolytes • Example 14-8: The freezing point of 0.0100 m NaCl solution is -0.0360oC. Calculate the van’t Hoff factor and apparent percent dissociation of NaCl in this aqueous solution. • meffective = total number of moles of solute particles/kg solvent • First let’s calculate the i factor.

38. Colligative Properties and Dissociation of Electrolytes

39. Colligative Properties and Dissociation of Electrolytes • Next, we will calculate the apparent percent dissociation. • Let x = mNaCl that is apparently dissociated.

40. Colligative Properties and Dissociation of Electrolytes

41. Colligative Properties and Dissociation of Electrolytes

42. Colligative Properties and Dissociation of Electrolytes

43. Colligative Properties and Dissociation of Electrolytes • Example 14-9: A 0.0500 m acetic acid solution freezes at -0.0948oC. Calculate the percent ionization of CH3COOH in this solution. You do it!

44. Colligative Properties and Dissociation of Electrolytes

45. Osmotic Pressure • Osmosis is the net flow of a solvent between two solutions separated by a semipermeable membrane. • The solvent passes from the lower concentration solution into the higher concentration solution. • Examples of semipermeable membranes include: • cellophane and saran wrap • skin • cell membranes

46. Osmotic Pressure semipermeable membrane H2O 2O H2O H2O sugar dissolved in water H2O H2O net solvent flow H2O H2O

47. Osmotic Pressure

48. Osmotic Pressure • Osmosis is a rate controlled phenomenon. • The solvent is passing from the dilute solution into the concentrated solution at a faster rate than in opposite direction, i.e. establishing an equilibrium. • The osmotic pressure is the pressure exerted by a column of the solvent in an osmosis experiment.

49. Osmotic Pressure • For very dilute aqueous solutions, molarity and molality are nearly equal. • M m