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2-3-2011

2-3-2011. Change in Freezing Point. Common Applications of Freezing Point Depression. Ethylene glycol – deadly to small animals. Propylene glycol. Freezing point depression. 0. D T f = T f – T f. 0. T f is the freezing point of the pure solvent.

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2-3-2011

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  1. 2-3-2011

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

  3. Freezing point depression

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

  5. 0 DTf = T f – Tf moles of solute m= mass of solvent (kg) = 3.202 kg solvent 1 mol 62.01 g 478 g x 0 Tf = T f – DTf 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. DTf = Kfm Kf(water) = 1.86 0C/m = 2.41 m DTf = Kfm = 1.86 0C/m x 2.41 m = 4.48 0C = 0.00 0C – 4.48 0C = -4.48 0C

  6. Osmotic Pressure (p) 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 (p) is the pressure required to stop osmosis.

  7. more concentrated dilute

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

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

  10. Vapor-Pressure Lowering Boiling-Point Elevation DTb = Kbm 0 P1 = X1 P 1 Freezing-Point Depression DTf = Kfm p = MRT Osmotic Pressure (p) 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.

  11. Molecular mass from colligative properties

  12. 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? DTb = Kbm 3.5 oC = (0.52 oC/m)* m m = 6.73

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

  14. A 5.50 g of a newly synthesized compound was dissolved in 250 g of benzene )kf = 5.12 oC/m) and the freezing point depression was found to be 1.2oC. Find the molecular mass of the compound. DTf  = kfm 1.2oC 5.12 = oC/m * m Molality = 0.199 m

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

  16. Example A solution of 0.85 g of a compound in 100 g of benzene has a freezing point of 5.16 oC. What are the molality and the molar mass of the solute? The normal freezing point of benzene is 5.5 oC, kf = 5.12 oC/m. DTf  = kfm m = (5.5 – 5.16)/5.12 = 0.066 mol/kg benzene

  17. 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

  18. 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 0C. Find the molar mass of the polymer. Can calculate the molarity, where: p = MRT

  19. 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

  20. 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.

  21. actual number of particles in soln after dissociation van’t Hoff factor (i) = number of formula units initially dissolved in soln 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 solution 0.2 m ions in solution i should be 1 nonelectrolytes 2 NaCl CaCl2 3

  22. Boiling-Point Elevation DTb = iKbm Freezing-Point Depression DTf = i Kfm p = iMRT Osmotic Pressure (p) Colligative Properties of Electrolyte Solutions

  23. Freezing Point Depression At what temperature will a 5.4 molal solution of NaCl freeze? Assume i = 2 Solution ∆Tf = Kf • m • i ∆Tf = (1.86 oC/m) • 5.4 m • 2 ∆Tf = 20.1oC FP = 0 – 20.1 = -20.1oC

  24. What is the freezing point depression of a 0.15 m aqueous solution of Al2(SO4)3? (kf = 1.86 oC/m)? Solution Al2(SO4)3 = 2 Al3+ + 3 SO42- Therefore, the overall effective molality of the solution is: m = 0.15 * 5 = 0.75, neglecting interionic attractions DTf  = {1.86 oC/m}*0.75 m = 1.4 oC

  25. Vapor pressure lowering A solution of CaCl2 (FW = 111 g/mol) was prepared by dissolving 25.0 g of CaCl2 in exactly 500 g of H2O. What is the expected vapor pressure at 80 oC (Powater at 80 oC = 355 torr)? What would the vapor pressure of the solution be if CaCl2 were not an electrolyte? Psoln = xsolvent Posolvent Therefore, calculate the number of moles of water and CaCl2 nwater = 500g/{18.0 g/mol} = 27.8

  26. nCaCl2 = 25.0 g/{111 g/mol} = 0.225 nspecies = 3 * 0.225 = 0.675 xwater = 27.8/{27.8 + 0.675} = 0.975 Psoln = 0.975 * 355 torr = 346 torr If CaCl2 were not an electrolyte, CaCl2 will not dissociate and the number of moles of all species of solute will be just 0.225 Xwater = 27.8/{27.8 + 0.225} = 0.993 Psoln = 0.993 * 355 torr = 352 torr

  27. Boiling point elevation Find the boiling point elevation of a 0.100 m MgSO4  aqueous solution (kb = 0.51 oC/m, i = 1.3). DTb = ikb * m DTb = 1.3 * 0.51 oC/m * 0.100m = 0.066 oC If MgSO4 was not an electrolyte DTb = 0.51 oC/m * 0.100 m = 0.051 oC.

  28. 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.

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