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Colligative Properties of Solutions

Vapor-Pressure Lowering. Boiling-Point Elevation. D T b = i K b m. 0. P 1 = X 1 P 1. Freezing-Point Depression. D T f = i K f m. p = i MRT. Osmotic Pressure ( p ). Colligative Properties of Solutions.

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Colligative Properties of Solutions

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  1. Vapor-Pressure Lowering Boiling-Point Elevation DTb = iKbm 0 P1 = X1 P 1 Freezing-Point Depression DTf = iKfm p = iMRT Osmotic Pressure (p) Colligative Properties of 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. X = mole fraction i = Vant Hoff Factor M = molarity m = molality R = gas constant T = temperature

  2. 0 DTb = Tb – T b 0 T b is the boiling point of the pure solvent 0 Tb > T b DTb = iKbm Boiling-Point Elevation T b is the boiling point of the solution DTb > 0 m is the molality of the solution Kb is the molal boiling-point elevation constant (0C/m) for a given solvent i is the van’t Hoff factor

  3. 0 DTf = T f – Tf 0 T f is the freezing point of the pure solvent 0 T f > Tf DTf = iKfm Freezing-Point Depression T f is the freezing point of the solution DTf > 0 m is the molality of the solution Kf is the molal freezing-point depression constant (0C/m) for a given solvent i is the van’t Hoff factor

  4. 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

  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 = iKfm Kf water = 1.86 oC/m = 2.41 m DTf = iKfm = (1) (1.86 oC/m) (2.41 m)= 4.48 oC = 0.00 oC – 4.48 oC = -4.48 oC

  6. Determination of Molar Mass A solution was prepared by dissolving 18.00 g glucose in 150.0 g water. The resulting solution was found to have a boiling point of 100.34°C. Calculate the molar mass of glucose. Glucose is a molecular solid that is present as individual molecules in solution. (Answer: 180g/mol) Would this be useful if we didn’t know the identify of a solute?

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