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13.5 Colligative Properties

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

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13.5 Colligative Properties

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  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 DTf= change (decrease) in freezing point DTf=(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= (Tf measured)/(Tfnonelectrolyte) 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 (PA=XAPoA)  B. at 60oC PoH2O= 149 torr  assume 1 mole each of water and ethylene glycol  XH2O= 1 mol H2O/(1 mol H2O + 1 mol ethylene) = 0.5 PH2O=(.5)(149 torr) = 74.5 torr However, 67 torr is the actual PH2O. Therefore, the solution is not ideal according to Raoult's law.

  7. Problem 47 A. at 65.3 C PoH2O = 187.5 torr  100.0g H2O x  1 mol H2O/ 18.02g H2O = 5.55 mol H2O 15.0g C12H22O11 x 1 mol/ 342.34 g = .044 mol C12H22O11 PH2O= XH2O PoH20      XH2O= 5.55 mol H2O/ (5.55 mol H2O + .044 mol)                                   XH2O= 0.992 PH2O=(.992)(187.5 torr)= 186 torr B. at 40oC PoH2O= 55.3 torr   500g H2O x 1 mol H2O/18.02g= 27.7 mol PH2O= XH2O PoH20 (55.3 torr- 4.60 torr)=XH2O(55.3 torr)             XH2O= 0.917 = 27.7mol H2O/(27.7 mol + xmol C3H8O2) x= 2.51 mol C3H8O2 x 76.11g/1 mol C3H8O2     x= 192 g C3H8O2

  8. Problem 49 A) assume the prescence of 10g H2O and C2H5OH 10.00g H2O x 1mol H2O/18.02 g H2O = 0.5549mol H2O 10.00g C2H5OH x 1mol C2H5OH/46.08g C2H5OH= 0.2170mol C2H5OH  0.5549 mol+ 0.2170 mol =0.7719 mol solution 0.2170 mol     0.7719 mol      =0.2811 B) Ptotal=Xa Poa    + Xb PobPtotal=(.2811) (400 torr)   +(0.7189) (175 torr) =238 torr C) XEth = (PEth)/(Ptotal) = (.2811*400torr)        =0.472                                          238 torr           

  9. 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. Problem 51 b)  NaCl DeltaT = Kbm DeltaT = (2)(0.512C/m)(0.10m) = 0.10 C T(b) = 100 + 0.10 C T(b)=100.10 C C6H12O6 DeltaT = (0.512C/m)(0.10m) = 0.051 C  T(b) =100+ 0.051 C T(b)=100.1C

  10. ΔTF = KF · m · i ΔTF, the freezing point depression KF, the molal freezing point depression constant  m is the molality (mol solute per kg of solvent) i number of solute particles per mol ΔTF =.040 m glycerin (C3H8O3) x 1.86 C/m x 1 = -.0744 CΔTF =.020 m KBr x1.86 C/m x 2 = -.0744 CΔTF =.030 m phenol (C6H5OH) x 1.86 C/m x 1 = -.0558 C .030 m phenol > .040 m glycerin = .020 m KBr Problem 53

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

  12. 0.64g of Adrenaline elevates the boiling point of 36.0g CCl4 by 0.49oC. What is the molar mass of adrenaline? 0.49oC x  1m         x     1kg    x 36.0gCCl4 = 0.0035mol                 5.02oC        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 25oC. What is the molar mass of Lysozyme? 25oC + 273 = 298K     n = Vπ                                           RT n = .210L x 0.953torr                 = 1.1x10-5mol       62.36(L-torr/K-mol) x 298K 0.150g x          1         = 1.4x104mol                1.1x10-5mol

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