html5-img
1 / 59

CHAPTER 4

CHAPTER 4. First year. Solutions C By Dr. Hisham Ezzat 2011- 2012. Completely miscible liquids Ideal solution Non - ideal solution Completely immiscible liquids H 2 O and aniline, H 2 O and chlorobenzene Partially immiscible liquids, H 2 O and phenol, H 2 O and ether.

athena-snyder
Download Presentation

CHAPTER 4

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CHAPTER 4 First year • Solutions C By Dr. HishamEzzat 2011- 2012

  2. Completely miscible liquids • Ideal solution • Non - ideal solution • Completely immiscible liquids H2O and aniline, H2O and chlorobenzene • Partially immiscible liquids, H2O and phenol, H2O and ether

  3. Ideal Solution • The force of attraction between all molecules are identical i.e. the attraction force is not affected by addition of other components A - A = B-B = A - B. • No heat is evolved or absorbed during mixing i.e.  H soln. = Zero • The volume of solution is the sum of volumes of the two liquids. • The solution obeys Raoult's law.

  4. Figure (1): Vapor pressure of ideal solutions

  5. Example 6: • Heptane (C7H16) and octane (C8H18) form ideal solutions What is the vapor pressure at 40°C of a solution that contains 3.0 mol of heptane and 5 mol of octane? At 40°C, the vapor pressure of heptane is 0.121 atm and the vapor pressure of octane is 0.041 atm.

  6. Solution: The total number of moles is 8.0. therefore X heptane = 3.0/8.0 = 0.375 X octane = 5.0/8.0 = 0.625 Total = X heptane . Po heptane + X octane. Po octane = 0.375 x 0.12 +0.625 x 0.04 = 0.045 atm + 0.026 atm. = 0.071 atm.

  7. Example 7: • Assuming ideality, calculate the vapor pressure of 1.0 m solution of a non - volatile, on dissociating solute in water at 50°C. The vapor pressure of water 50°C is 0.122 atm.

  8. Solution : • From example 2 the mole fraction of water in 1.0m solution is 0.982. • PH2O = XH2O PH2O = 0.982 x 0.122 = 0.120 atm.

  9. Problem: • At 140°C, the V.P of C6H5CI is 939.4 torr and that of C6H5Br is 495.8 torr. Assuming that these two liquids from an ideal solution. Find the composition of a mixture of two liquids which boils at 140°C under 1 atm pressure?

  10. Non- ideal solutions

  11. Fig.2: Vapour pressure of non-ideal solution (-ve deviation) Fig.3: Vapour pressure of non-ideal solution (+ve deviation)

  12. Fractional Distillation of Binary Miscible liquids • The separation of mixture of volatile liquids into their components is called fractional distillation, • the distillate containing the more volatile component and the residue the less volatile one

  13. a) Ideal solutions • If a mixture of 2 liquids (A and B) form a completely miscible ideal solution and PA > PB result in B.P. of A < B.P of B thus on boiling:- • 1) The Liquid A boils at lower B.P than that of liquid B. • 2) The liquid A which is more volatile will be passed from the fractionating column and the liquid B which is less volatile returned again to the distallating flask. • A solution of intermediate B.p. between 2 pure liquid -called azeotropic solution

  14. b) Non - ideal solutions (solutions that exhibit deviations from Raoults law)

  15. Non - ideal solutions with minimum boiling point: If a solution having any other compositions is distilled, the azeotropic mixture will distill first and the excess of (A) or (B) will remains in the flask e.g 95 % ethanol and 5 % H2O.

  16. 2) Non - ideal solutions with maximum boiling point: • If a solution having any other composition is distilled, the execs of acetone or CHCI3 will distill first leaving the azeotropic mixture in the flask.

  17. Example 8: • A solution is prepared by mixing 5.81 g acetone C3H6O, (M. wt = 58.1 g/mole) 11.9 g chloroform (CHCI3 M.wt 119.4 g/mole). At 35°C this solution has a total vapor pressure of 260 torr. Is this an ideal solution? Comment? The vapor pressure of pure acetone and pure CHCI3 at 35°C are 345 and 293 torr, respectively.

  18. Colligative Properties of Solutions

  19. 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.

  20. 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.

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

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

  23. 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.

  24. 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.

  25. 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

  26. 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

  27. 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 =

  28. 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

  29. 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.

  30. . 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.

  31. 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.

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

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

  34. 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

  35. 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

  36. 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.

  37. 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 for a compilation of boiling point and freezing point elevation constants.

  38. Freezing Point Depression • Relationship for freezing point depression is:

  39. 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.

  40. Freezing Point Depression

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

  42. Freezing Point Depression • Example : 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!

  43. Freezing Point Depression

  44. 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.

  45. Determination of Molecular Weight by Freezing Point Depression • Example : 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?

  46. Determination of Molecular Weight by Freezing Point Depression

  47. 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

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

More Related