Colligative Properties of Nonvolatile Solutes 01. Colligative Properties of Nonvolatile Solutes 02. Colligative Properties of Nonvolatile Solutes 04

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1 Colligative Properties of Nonvolatile Solutes 01 Colligative Properties of Nonvolatile Solutes 02 Colligative Properties: Depend on the amount not on the identity There are four main colligative properties: 1. Vapor pressure lowering 2. Freezing point depression 3. Boiling point elevation 4. Osmotic pressure When solute molecules displace solvent molecules at the surface, the vapor pressure drops since fewer gas molecules are needed to equalize the escape rate and capture rates at the liquid surface. Colligative Properties of Nonvolatile Solutes 03 Colligative Properties of Nonvolatile Solutes 04 X solv = moles of solvent Total moles Raoult s Law: P soln = P solv X solv *Think of total moles in the above equation as total moles of particles in solution Van t Hoff factor i = moles of particles in solution moles of solute dissolved For example: If we dissolve moles of NaCl in moles of water, what is the mole fraction of water? i = 2 for NaCl (if completely dissociated) 0.313moles of NaCl x 2 = total moles of particles of NaCl X H2O = 27.75moles H 2 O = moles of H 2 O moles of NaCl 1

2 Vapor Pressure of a Solution and a Nonvolatile Solute 05 Colligative Properties of a Volatile Solute 06 Why is the vapor pressure of a solution with a nonvolatile solute lower than for the pure solvent? Entropy What happens if both components are volatile (have measurable vapor pressures)? The vapor pressure has a value intermediate between the vapor pressures of the two liquids. P T = P A + P B = X A P A + X B P B = X A P A + (1 X A )P B Colligative Properties of a Volatile Solute 07 Boiling-Point Elevation 08 The following diagram shows a close-up view of part of the vapor pressure curves for two pure liquids and a mixture of the two. Which curves represent pure liquids, and which the mixture? Boiling-Point Elevation ( T b ): The boiling point of the solution (T b ) minus the boiling point of the pure solvent (T b): T b = T b T b T b is proportional to concentration: T b = K b mi K b = molal boiling-point elevation constant. 2

3 Freezing-Point Depression 09 Freezing-Point Depression 10 Freezing-Point Depression ( T f ): The freezing point of the pure solvent (T f) minus the freezing point of the solution (T f ). T f = T f T f T f is proportional to concentration: T f = K f mi K f = molal freezing-point depression constant. van t Hoff Factor, i: This factor equals the number of ions produced from each molecule of a compound upon dissolving. i = 1 for CH 3 OH i = 3 for CaCl 2 i = 2 for NaCl i = 5 for Ca 3 (PO 4 ) 2 For compounds that dissociate on dissolving, use: T b = K b mi T f = K f mi Freezing-Point Depression 11 Entropy of a Solution 12 3

4 Why is the Boiling Point of a Solution Elevated? 13 Why is the Freezing Point of a Solution Depressed? 14 ENTROPY! ENTROPY! Freezing-Point Depression 15 Freezing-Point Depression 16 The phase diagram shows a close-up of the liquid vapor phase transition boundaries for pure chloroform. a) Estimate the boiling point of pure chloroform. a) Estimate the molal concentration of the nonvolatile solute. (See Table 11.4 for K b ). 4

5 Osmosis and Osmotic Pressure 17 Osmosis and Osmotic Pressure 18 Osmosis: The selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one. Osmotic pressure (! or ): The pressure required to stop osmosis (obtain equilibrium)! = i MRT M = molar concentration of solute particles R = (L atm)/(mol K) Osmosis and Osmotic Pressure 19 Example 20 The average osmotic pressure of seawater is about 30.0 atm at 25 C. Calculate the molar concentration of an aqueous solution of urea [(NH 2 ) 2 CO].! = imrt M =!/irt = 30.0atm/(1( Latm/molK)(298K)) =1.23 mol/l 5

6 Example 21 Uses of Colligative Properties 22 Reverse Osmosis What is the osmotic pressure (in atm) of a M sucrose solution at 16 C? Desalination: T = 16 C = 289K! =(0.884mol/L)( Latm/molK)(289K) = atm Uses of Colligative Properties Molecular Mass Calculation 23 Example 24 Any of the four colligative properties can be used to calculate molecular mass In lab we used freezing point depression, but the degree of depression is very small Most accurate value of molecular mass is to use osmotic-pressure measurements A 202 ml benzene solution containing 2.47 g of an organic polymer has an osmotic pressure of 8.63 mm Hg at 21 C. Calculate the molar mass of the polymer. M =!/irt =(8.63mmHg x 1atm/760mm Hg) = 4.71 x 10-4 mol/l (1)( Latm/molK)(294K) 4.71 x 10-4 mol/l = x / 0.202L benzene x = x 10-5 moles of polymer MM organic polymer = 2.47g/ x 10-5 moles = g/mol 6

7 Example 25 Uses of Colligative Properties 26 What is the molar mass of sucrose if a solution of g of sucrose in ml of water has an osmotic pressure of 149 mm Hg at 298 K? M =!/irt = (149mm Hg x 1 atm/760mm Hg) = mol/L (1)( Latm/molK)(298K) Fractional Distillation is the separation of volatile liquid mixtures into fractions of different composition mol/L = x / 0.300L x = moles of sucrose MM sucrose = 0.822g/0.0024moles = 341.8g/mol Uses of Colligative Properties 27 Example 28 Fractional distillation can be represented on a phase diagram by plotting temperature against composition. Two miscible liquids, A and B, have vapor pressures of 250 mm Hg and 450 mm Hg, respectively. They were mixed in equal molar amounts. What is the total vapor pressure of the mixture and what are their mole fractions in the vapor phase? P mix = P A X A + P B X B = (250mm Hg)(0.5) + (450mm Hg)(0.5) = 350 mm Hg P A = P A X A = (250mm Hg)(0.5) = 125mm Hg X A = 125mm Hg/350mm Hg = P B = P B X B = (450mm Hg)(0.5) = 225mm Hg X B = 225mm Hg/350mm Hg =

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