1. 1 2-3-2011

2. 2 Change in Freezing Point Common Applications of Freezing Point Depression Propylene glycol Ethylene glycol – deadly to small animals

3. 3 Freezing point depression

4. 4 Freezing-Point Depression  T f = T f – T f 0 T f > T f 0  T f > 0 T f is the freezing point of the pure solvent 0 T f is the freezing point of the solution  T f = K f m m is the molality of the solution K f is the molal freezing-point depression constant ( 0 C/m)

5. 5 What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62.01 g.  T f = K f m m =m = moles of solute mass of solvent (kg) = 2.41 m = 3.202 kg solvent 478 g x 1 mol 62.01 g K f (water) = 1.86 0 C/m  T f = K f m = 1.86 0 C/m x 2.41 m = 4.48 0 C  T f = T f – T f 0 T f = T f –  T f 0 = 0.00 0 C – 4.48 0 C = -4.48 0 C

6. 6 Osmotic Pressure (  ) Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one. A semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules. Osmotic pressure (  ) is the pressure required to stop osmosis.

7. 7 dilute more concentrated

8. 8 Higher PLower P Osmotic Pressure (  )  = MRT M is the molarity of the solution R is the gas constant T is the temperature (in K)

9. 9 hypertonic solution A cell in an: isotonic solution hypotonic solution Dilute Soln Conc. Soln Conc Dilute Soln

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11. 11 Colligative Properties of Nonelectrolyte Solutions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. Vapor-Pressure Lowering P 1 = X 1 P 1 0 Boiling-Point Elevation  T b = K b m Freezing-Point Depression  T f = K f m Osmotic Pressure (  )  = MRT

12. 12 Molecular mass from colligative properties From MolarityFrom Molality 1.Get molarity 2.Get no. of moles Mol = M * V L 3. FW = (g solute/no. mol) 1.Get molality 2.Get no. of moles Mol = m * kg solvent 3. FW = (g solute/no. mol)

13. 13 Molecular Weight of Unknown What is the molecular mass of a sample if 250 grams of the sample is placed into 1000 grams of water and the temperature rose by 3.5°C?  T b = K b m 3.5 o C = (0.52 o C/m)* m m = 6.73

14. 14 Find the number of moles in 1000 g No. mol = 6.73 mol No. mol = mass/FW FW = 250/6.73 = 37 g/mol

15. 15 A 5.50 g of a newly synthesized compound was dissolved in 250 g of benzene )k f = 5.12 o C/m) and the freezing point depression was found to be 1.2 o C. Find the molecular mass of the compound.  T f = k f m 1.2 o C 5.12 = o C/m * m Molality = 0.199 m

16. 16 Mol = m * kg solvent Mol solute = (0.199 mol/kg) * (0.250 kg) mol solute = 0.0498 moles mol solute = Wt solute/FW FW = 5.50 g/0.0498 mol = 111 g/mol It should be realized that a solvent with high k f is an advantage for such experimental calculations of molecular masses.

17. 17 Example A solution of 0.85 g of a compound in 100 g of benzene has a freezing point of 5.16 o C. What are the molality and the molar mass of the solute? The normal freezing point of benzene is 5.5 o C, k f = 5.12 o C/m.  T f = k f m m = (5.5 – 5.16)/5.12 = 0.066 mol/kg benzene

18. 18 Mol = m * kg solvent Number of moles of compound = (0.066 mol/kg benzene)* 0.100 kg benzene = 6.6*10 -3 FW = g compound/mol FW = 0.85 g/(6.6*10 -3 mol) = 128 g/mol

19. 19 Molecular mass from osmotic pressure A solution is prepared by dissolving 2.47 g of an organic polymer in 202 mL of benzene. The solution has an osmotic pressure of 8.63 mmHg at 21 0 C. Find the molar mass of the polymer. Can calculate the molarity, where:  = MRT

20. 20 Get the number of moles of polymer where 1 L of polymer solution contains 4.53*10 -4 mol Mol polymer = (4.53*10 -4 mol/1L soln) * 0.202 L soln = 9.15*10 -5 mol Molar mass = g polymer/no. mol Molar mass = (2.47 g / 9.15*10 -5 mol) = 26,992 g/mol

21. 21 Remember that osmotic pressure can be easily and accurately measured in mmHg. However, it is not as easy to measure freezing point depression when the concentration of solute is very small. Therefore, osmotic pressure is better suited for calculation of molar mass of solutes dissolved in solutions.

22. 22 Colligative Properties of Electrolyte Solutions 0.1 m NaCl solution 0.1 m Na + ions & 0.1 m Cl - ions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. 0.1 m NaCl solution0.2 m ions in solution van’t Hoff factor (i) = actual number of particles in soln after dissociation number of formula units initially dissolved in soln nonelectrolytes NaCl CaCl 2 i should be 1 2 3

23. 23 Boiling-Point Elevation  T b = i K b m Freezing-Point Depression  T f = i K f m Osmotic Pressure (  )  = iMRT Colligative Properties of Electrolyte Solutions

24. 24 At what temperature will a 5.4 molal solution of NaCl freeze? Assume i = 2 Solution ∆T f = K f m i ∆T f = (1.86 o C/m) 5.4 m 2 ∆T f = (1.86 o C/m) 5.4 m 2 ∆T f = 20.1 o C ∆T f = 20.1 o C FP = 0 – 20.1 = -20.1 o C FP = 0 – 20.1 = -20.1 o C Freezing Point Depression

25. 25 What is the freezing point depression of a 0.15 m aqueous solution of Al 2 (SO 4 ) 3 ? (k f = 1.86 o C/m)? Solution Al 2 (SO 4 ) 3 = 2 Al 3+ + 3 SO 4 2- Therefore, the overall effective molality of the solution is: m = 0.15 * 5 = 0.75, neglecting interionic attractions  T f = {1.86 o C/m}*0.75 m = 1.4 o C

26. 26 Vapor pressure lowering A solution of CaCl 2 (FW = 111 g/mol) was prepared by dissolving 25.0 g of CaCl 2 in exactly 500 g of H 2 O. What is the expected vapor pressure at 80 o C (P o water at 80 o C = 355 torr)? What would the vapor pressure of the solution be if CaCl 2 were not an electrolyte? P soln = x solvent P o solvent Therefore, calculate the number of moles of water and CaCl 2 n water = 500g/{18.0 g/mol} = 27.8

27. 27 n CaCl2 = 25.0 g/{111 g/mol} = 0.225 n species = 3 * 0.225 = 0.675 x water = 27.8/{27.8 + 0.675} = 0.975 P soln = 0.975 * 355 torr = 346 torr If CaCl 2 were not an electrolyte, CaCl 2 will not dissociate and the number of moles of all species of solute will be just 0.225 X water = 27.8/{27.8 + 0.225} = 0.993 P soln = 0.993 * 355 torr = 352 torr

28. 28 Boiling point elevation Find the boiling point elevation of a 0.100 m MgSO 4 aqueous solution (k b = 0.51 o C/m, i = 1.3).  T b = ik b * m  T b = 1.3 * 0.51 o C/m * 0.100m = 0.066 o C If MgSO 4 was not an electrolyte  T b = 0.51 o C/m * 0.100 m = 0.051 o C.

29. 29 Selected Problems: 6, 7 – 13, 15 - 17, 20 - 24, 27, 29, 31, 35, 39 – 44, 49, 50 - 52, 54 – 56, 60 - 64, 70, 71, 73, 75 -77, 79, 80.