1. CHAPTER 4
First year
Solutions C By
Dr. Hisham Ezzat

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

6. 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 atm and the vapor pressure of octane is atm.

7. 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
= x x 0.04
= atm atm.
= atm.

8. 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 atm.

9. Solution :
From example 2 the mole fraction of water in 1.0m solution is
PH2O = XH2O PH2O = x = atm.

10. Problem:
At 140°C, the V.P of C6H5CI is torr and that of C6H5Br is 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?

11. Non- ideal solutions Negative deviation Positive deviation
1- The force of attraction
increase by mixing
A - A, B-B < A-B
The force of attraction decrease by mixing A-A , B-B > A-B
2- The vapor pressure will be lower than that given by Roault's law
The vapor pressure will be higher than that given by Raoult's law.
3- H solution :- Ve (exothermic)
 H solution: + Ve (endothermic)
4- Temperature change when solution is formed: increase
Temperature change when solution is formed: decrease.
5- Example: Acetone-water
Ethanol-hexane

12. Fig.2: Vapour pressure of non-ideal solution (-ve deviation)

13. 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

14. 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

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

16. 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.

17. 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.

18. 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 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.

20. Colligative Properties of Solutions

21. 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.

22. 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.

23. Lowering of Vapor Pressure and Raoult’s Law
Derivation of Raoult’s Law.

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

25. 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.

26. 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.

27. 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 = What is , the vapor pressure of water over the solution
= 14 torr

28. Glycerin, C3H8O3, is a nonvolatile nonelectrolyte with a density of 1
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 mL of water. The vapor pressure of pure water at 25°C is 23.8 torr

29. The vapor pressure of pure water at 110°C is 1070 torr
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
XEG = 1
= XEG
= 0.290
XEG =

30. 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 = = 0.6

31. solution is made by mixing 52. 1 g of propyl chloride, C3H8Cl, and 38
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.

32. At 25°C a solution consists of 0. 450 mole of pentane, C5H12, and 0
. At 25°C a solution consists of mole of pentane, C5H12, and 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.

33. 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.

34. Boiling Point Elevation
Boiling point elevation relationship is:

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

36. 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

37. 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
Solvent
B.Point (°C)
Kb (°C/m)
Freezing P. (°C)
Kf (°C/m)
Water, H2O
100.0
0.52
0.00
1.86
Benzen, C6H6
80.1
2.53
5.5
5.12
Ethanol, C2H6O
78.4
1.22
-114.0
1.99
Carbon tetrachloride, CCl4
76.8
5.02
-22
29.8
Chloroform, CHCl3
61.2
3.63
-63.5
4.68

38. 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.

39. 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.

40. Freezing Point Depression
Relationship for freezing point depression is:

41. Freezing Point Depression
Notice the similarity of the two relationships for freezing point depression and boiling point elevation.
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.

42. Freezing Point Depression

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

44. 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!

45. Freezing Point Depression

46. 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.

47. 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?

48. Determination of Molecular Weight by Freezing Point Depression

49. 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

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

51. 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.

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

53. Osmotic Pressure Osmotic pressures can be very large.
For example, a 1 M sugar solution has an osmotic pressure of 22.4 atm or 330 p.s.i.
Since this is a large effect, the osmotic pressure measurements can be used to determine the molar masses of very large molecules such as:
Polymers
Biomolecules like
proteins
ribonucleotides

54. Osmotic Pressure
Example : A 1.00 g sample of a biological material was dissolved in enough water to give 1.00 x 102 mL of solution. The osmotic pressure of the solution was 2.80 torr at 25oC. Calculate the molarity and approximate molecular weight of the material.
You do it!

55. Osmotic Pressure

56. Osmotic Pressure

57. Application of Osmotic Pressure
1. Water Purification by Reverse Osmosis
If we apply enough external pressure to an osmotic system to overcome the osmotic pressure, the semipermeable membrane becomes an efficient filter for salt and other dissolved solutes.
Ft. Myers, FL gets it drinking water from the Gulf of Mexico using reverse osmosis.
Dialysis is another example of this phenomenon.

58. 2) Isotonic solution: In the living cells, the osmotic pressure of solution is equal to the osmotic pressure of the cell.
e.g: NaCI (0.9%) has the same osmotic pressure as blood.
3) Hypertonic solution: A solution of higher osmotic pressure. In this solution red blood cells shrink. The cells are called plasmolysed.
4) Hypotonic solution: A solution of lower osmotic pressure. In this solution red blood cells swells up and burst. The cell is said to be haemolysed

59. End of Chapter 2
Human Beings are solution chemistry in action!