1. 13.5 Colligative Properties Charles Jie, Brendan Buckbee, Rajeen Amin, and Chris Swisher

2. Some Basics Electrolytes dissolve in solution, Nonelecrolytes do not. Thus, electrolytes dissolved in water would ideally have twice as many particles dissolved in solution as nonelectrolytes. Raoult’s Law: PA = vapor pressure with solute PA° = vapor pressure without solute XA = mole fraction of solvent in solution A PA=(XA)(PA°) Ideal Solution obeys Raoult's Law

3. Boiling Point and Freezing Point info Kb=Molal boiling-point-elevation constant DTb= change in boiling point m=molality DTb=(Kb)(m) Kf =molal freezing-point-depression constant m=molality DT f = change (decrease) in freezing point DT f= (Kf)(m)

4. Osmosis Osmosis: the movement of a solvent from low solute concentration to high solute concentration Pi=(n/V)RT Pi=MRT Pi=osmostic pressure M= Molarity R= Ideal-gas Constant T= Temperature in Kelvin

5. More info Differences between observed and Expected changes in Temperature points Ratio of actual value of colligative property to calculated value as a nonelectrolyte i= (T f measured)/(T f nonelectrolyte) affected by dilution and the magnitudes of the charges on the ions i =van’t Hoff factor ( measure of the extent of ion dissociation)

6. Problem 45 A. An ideal solution obeys Raoult's law (P A =X A P o A ) B. at 60 o C P o H2O = 149 torr assume 1 mole each of water and ethylene glycol X H2O = 1 mol H 2 O/(1 mol H 2 O + 1 mol ethylene) = 0.5 P H2O =(.5)(149 torr) = 74.5 torr However, 67 torr is the actual P H2O. Therefore, the solution is not ideal according to Raoult's law.

7. Problem 47 A. at 65.3 C P o H2O = 187.5 torr 100.0g H 2 O x 1 mol H 2 O/ 18.02g H 2 O = 5.55 mol H 2 O 15.0g C 12 H 22 O 11 x 1 mol/ 342.34 g =.044 mol C 12 H 22 O11 P H2O = X H2O P o H20 X H2O = 5.55 mol H 2 O/ (5.55 mol H 2 O +.044 mol) X H2O = 0.992 P H2O =(.992)(187.5 torr)= 186 torr B. at 40 o C P o H2O = 55.3 torr 500g H 2 O x 1 mol H 2 O/18.02g= 27.7 mol P H2O = X H2O P o H20 (55.3 torr- 4.60 torr)=X H2O (55.3 torr) X H2O = 0.917 = 27.7mol H 2 O/(27.7 mol + xmol C 3 H 8 O 2 ) x= 2.51 mol C 3 H 8 O 2 x 76.11g/1 mol C 3 H 8 O 2 x= 192 g C 3 H 8 O 2

8. Problem 49 A) assume the prescence of 10g H 2 O and C 2 H 5 OH 10.00g H 2 O x 1mol H 2 O/18.02 g H 2 O = 0.5549mol H 2 O 10.00g C 2 H 5 OH x 1mol C 2 H 5 OH/46.08g C 2 H 5 OH= 0.2170mol C 2 H 5 OH 0.5549 mol+ 0.2170 mol =0.7719 mol solution 0.2170 mol 0.7719 mol =0.2811 B) P total =X a P o a + X b P o b P total =(.2811) (400 torr) +(0.7189) (175 torr) =238 torr C) X Eth = (P Eth )/(P total ) = (.2811*400torr) =0.472 238 torr

9. Problem 51 a) NaCl has a higher boiling point than glucose because it is a strong electrolyte and disassociates in water, which would produce twice as much dissolved particles as one mole of glucose (nonelectrolyte). Thus, NaCl would have more moles of dissolved particles and thus would have a higher boiling point. b) NaCl DeltaT = K b m DeltaT = (2)(0.512C/m)(0.10m) = 0.10 C T(b) = 100 + 0.10 C T(b)=100.10 C C 6 H 12 O 6 DeltaT = (0.512C/m)(0.10m) = 0.051 C T(b) =100+ 0.051 C T(b)=100.1C

10. Problem 53 ΔT F = K F · m · i  ΔT F, the freezing point depression  K F, the molal freezing point depression constant  m is the molality (mol solute per kg of solvent)molality  i number of solute particles per mol ΔT F =.040 m glycerin (C 3 H 8 O 3 ) x 1.86 C/m x 1 = -.0744 C ΔT F =.020 m KBr x1.86 C/m x 2 = -.0744 C ΔT F =.030 m phenol (C 6 H 5 OH) x 1.86 C/m x 1 = -.0558 C.030 m phenol >.040 m glycerin =.020 m KBr

11. Problem 57 Solve for Molarity of Aspirin Molarity= moles solute liters solution Solve for Osmotic Pressure π =MRT M= Molarity R= Ideal-gas Constant T= Temperature in Kelvin

12. 0.64g of Adrenaline elevates the boiling point of 36.0g CCl 4 by 0.49 o C. What is the molar mass of adrenaline? 0.49 o C x 1m x 1kg x 36.0gCCl 4 = 0.0035mol 5.02 o C 1000g 0.64g x 1 = 180g/mol 0.0035mol Problem 59

13. Problem 61 0.150g of Lysozyme in 210mL of solution has an osmotic pressure of 0.953torr at 25 o C. What is the molar mass of Lysozyme? 25oC + 273 = 298K n = Vπ RT n =.210L x 0.953torr = 1.1x10 -5 mol 62.36(L-torr/K-mol) x 298K 0.150g x 1 = 1.4x10 4 mol 1.1x10 -5 mol