1. 1 28-2-2011

2. 2 Colligative Properties of solutions A colligative property is that which depends on the relative number of solute and solvent molecules. The number of solute molecules relative to solvent, would reflect a change in some physical properties. Physical properties which will be studied include vapor pressure, boiling point, melting point, as well as osmotic pressure.

3. 3 Vapor pressure lowering The presence of solutes in liquid solutions has great effects on the physical properties of the resulting solution. The vapor pressure of the solution will vary depending on the amount and nature of dissolved solute. Raoult's law discusses the effects of solutes on the vapor pressure of solutions provided that these solutes are non dissociable (non ionics) and nonvolatile.

4. 4 Raoult’s law states that the vapor pressure of solution (P soln ) is directly proportional to the mole fraction of the solvent. This can be represented by the relation: P soln = x solvent * P o solvent Where x solvent is the mole fraction of the solvent and P o solvent is its vapor pressure. Therefore, it is obvious that:

5. 5 1. Dissolving a nonvolatile solute will result in a decrease in the vapor pressure of the solution. 2. When the vapor pressure of the solute equals the vapor pressure of the solvent, no change in the vapor pressure of the solution would be observed. 3. When the vapor pressure of the solute is greater than the vapor pressure of the solvent, the vapor pressure of the solution would be increased.

6. 6 Example 10.0 g of a paraffin (FW = 282 g/mol) oil, a nonvolatile solute, was dissolved in 50.0 g of benzene (C 6 H 6, FW = 78.1 g/mol). At 53 o C, the vapor pressure of pure benzene is 300 torr. What is the vapor pressure of the solution at that temperature? P soln = x benzene * P o benzene

7. 7 Solution n solvent = 50.0 g/(78.1 g/mol) = 0.640 n solute = 10.0 g/(282 g/mol) = 0.035 x solvent = 0.640/(0.640 + 0.035) = 0.948 P soln = x solvent * P o solvent P soln = 0.948 * 300 torr = 284 torr

8. 8 Example At 25 o C, the vapor pressure of pure water is 23.76 mmHg and that of a urea (FW = 60 g/mol) solution is 22.98 mmHg. Find the molality of solution. We only have nonvolatile urea and water. Therefore: P soln = x water * P o water

9. 9 X urea = 1 – x water P soln = (1 – x urea ) P o water P o water - P soln = x urea P o water  P = x urea P o water X urea = (23.76 – 22.98)/23.76 = 0.033 X urea = n urea /(n urea + n water )

10. 10 Number of moles of water in 1 kg = 1000/18 = 55.6, therefore: 0.033 = n urea /(55.6 + n urea ) n urea = 1.9 Therefore, the molality of the urea soln is about 1.9 m

11. 11 Solutions with more than One Volatile Component When solutions contain more than one volatile component, the vapor pressure of the solution is the sum of the partial vapor pressures of all components. Assume a situation where a solution is composed from three volatile components A, B, and C, the vapor pressure of the solution is: P soln = x A * P o A + x B * P o B + x C * P o C

12. 12 Example A solution containing 50.0 g of CCl 4 (FW = 153.8 g/mol) and 50.0 g of CHCl 3 (FW = 119.4 g/mol). At 50 o C, if the vapor pressure of pure CCl 4 and CHCl 3 is 317 and 526 torr, respectively, find the partial pressure of each substance and the vapor pressure of the solution. Solution n CCl4 = 50.0 g /(153.8 g/mol) = 0.325 mol n CHCl3 = 50.0 g /(119.4 g/mol) = 0.419 mol

13. 13 x CCl4 = 0.325/(0.325 + 0.419) = 0.437 x CHCl3 = 1 – 0.437 = 0.563 P CCl4 = x CCl4 * P o CCl4 P CCl4 = 0.437 * 317 torr = 139 torr P CHCl3 = x CHCl3 * P o CHCl3 P CHCl3 = 0.563 * 526 torr = 296 torr P soln = P CCl4 + P CHCl3 P soln = 139 torr + 296 torr = 435 torr

14. 14 The P soln can also be calculated directly form the relation below without calculation of partial pressures in separate steps. P soln = x A * P o A + x B * P o B P soln = 0.437 * 317 torr + 0.563 * 526 torr = 435 torr

15. 15 It should be emphasized here that Raoult's law is valid for dilute solutions only while concentrated solutions usually show deviations from that law. Solutions obeying Raoult's law are referred to as ideal solutions while non ideal solutions show either negative or positive deviations from Raoult's law. Three situations can be stated:

16. 16 1. The situation where the intermolecular forces between solute-solute and solvent solvent are of the same strength (  H soln = 0). No deviations from Raoult's law as this is an ideal solution (like benzene-carbon tetrachloride solution).

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18. 18 2. The situation where the intermolecular forces between solute- solvent molecules are larger than solute-solute and solvent- solvent intermolecular forces (  H soln = -ve, exothermic process). Negative deviations from Raoult's law is observed (like acetone-water solution).

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20. 20 3. The situation where the intermolecular forces between solute-solvent molecules are less than solute-solute and solvent- solvent intermolecular forces (  H soln = +ve, endothermic process). Positive deviations from Raoult's law is observed (like ethanol-hexane solution).

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22. 22 Boiling point elevation

23. 23 Boiling-Point Elevation  T b = T b – T b 0 T b > T b 0  T b > 0 T b is the boiling point of the pure solvent 0 T b is the boiling point of the solution  T b = K b m m is the molality of the solution K b is the molal boiling-point elevation constant ( 0 C/m)

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