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13.5 Colligative properties

13.5 Colligative properties. Dissolving solute in pure liquid will change all physical properties of liquid, Density, Vapor Pressure, Boiling Point, Freezing Point, Osmotic Pressure Colligative Properties are properties of a liquid that change when a solute is added.

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13.5 Colligative properties

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  1. 13.5 Colligative properties • Dissolving solute in pure liquid will change all physical properties of liquid, Density, Vapor Pressure, Boiling Point, Freezing Point, Osmotic Pressure • Colligative Properties are properties of a liquid that change when a solute is added. • The magnitude of the change depends on the number of solute particles in the solution, not on the identity of the solute particles.

  2. Lowering the vapor pressure A liquid in a closed container will establish equilibrium with its vapor. When that equilibrium is reached, the pressure exerted by the vapor is called the vapor pressure The presence of a non-volatile solute means that fewer solvent particles are at the solution’s surface, so less solvent evaporates

  3. Raoult’s LawDescribes vapor pressure lowering mathematically. • The lowering of the vapour pressure when a non-volatile soluteis dissolved in a volatile solvent (A) can be described by Raoult’s Law: • PA =XAPA PA = vapour pressure of solvent A above the solution XA = mole fraction of the solvent A in the solution PA = vapour pressure of pure solvent A only the solvent (A) contributes to the vapour pressure of the solution

  4. Example: What is the vapor pressure of water above a sucrose (MW=342.3 g/mol) solution prepared by dissolving 158.0 g of sucrose in 641.6 g of water at 25 ºC? The vapor pressure of pure water at 25 ºC is 23.76 mmHg. = 0.462 mol Moles C12H22O11 = ( ( ( ( ( ( mol H2O 1 mol H2O 1 mol C12H22O11 35.6 (641.6 g H2O) (158 g C12H22O11) 18 g H2O 342.3 g C12H22O11 mol H2O + mol C12H22O11 35.6+ 0.462 Moles H2O= = 35.6 mol X = = 0.987 H2O= P =X P P =(0.987)(23.76 mmHg) = 23.5 mmHg H2O H2O soln soln

  5. Example: Glycerin (C3H8O3) is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25C of solution made by adding 50.0 mL of glycerin to 500.0 mL of water. The vapor pressure of pure water at 25C is 23.5 torrr and its density is 1.0 g/mL. Calculate the vapor pressure lowering Moles C3H8O3 = = 0.684 mol ( ( ( ( ( ( ( ( ( ( ( ( 1 mol H2O 1.0 g H2O 1 mol C3H8O3 1.26 g C3H8O3 mol H2O 27.8 (50 mL C3H8O3) (500 mLH2O) 1 mLH2O 1 mL C3H8O3 27.8+ 0.684 mol H2O + mol C3H8O3 18 g H2O 92.1 g C3H8O3 Moles H2O= = 27.8 mol X = = 0.976 H2O= P =(0.976)(23.8 torr) = 23.2 torr P =X P H2O H2O H2O H2O

  6. Example: The vapor pressure of pure water at 110C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.00 atm at 110C. Assuming that Raoult’s law is obeyed, what is the mole fraction of ethylene glycol in the solution? P = 1 atm = 760 torr P = 1070 torr H2O H2O P 760 torr X H2O = = 0.710 H2O= P 1070 torr H2O + = 1 = 1 0.71 = 0.290 C2H6O2 X P =X P X X H2O H2O C2H6O2 H2O H2O

  7. Mixtures of Volatile LiquidsBoth liquids evaporate & contribute to the vapor pressure

  8. Raoult’s Law: Mixing Two Volatile Liquids • Since both liquids are volatile and contribute to the vapour, the total vapor pressure can be represented using Dalton’s Law: • PT = PA + PB • The vapor pressure from each component follows Raoult’s Law: • PT = XAPA + XBPB   • Also, XA + XB = 1 (since there are 2 components)

  9. Benzene and Toluene A two solvent (volatile) system • The vapor pressure from each component follows Raoult's Law. • Benzene - Toluene mixture: • Recall that with only two components, XBz + XTol = 1 • Benzene: when XBz = 1, PBz = PBz = 384 torr & when XBz = 0 , PBz = 0 • Toluene: when XTol = 1, PTol = PTol = 133 torr & when XTol = 0, PBz = 0  

  10. Example: A mixture of benzene (C6H6) and toluene (C7H8) containing 1.0 mol of benzene and 2.0 mol of toluene. What is the total vapor pressure of the solution? [vapor pressures of pure benzene and toluene are 75 torr and 22 torr, respectively]   PT = XAPA + XBPB X X = 0.33 = 0.67 C6H6= C7H8= 1+2 1+2 1 2 PT = [(0.33)(75 torr)] + [(0.67)((22 torr)] = 24.75 torr + 14.75 torr = 39.5 torr

  11. Boiling-point elevation and Freezing-point depression • In a solution of a nonvolatile solute, boiling and freezing points differ from those of the pure solvent Boiling point is elevated when solute inhibits solvent from escaping. • The boiling point of the solution is higher than that of the pure liquid Freezing point is depressed when solute inhibits solvent from crystallizing. • The freezing point of the solution is lower than that of the pure liquid

  12. The diagram below shows how a phase diagram is affected by dissolving a solute in a solvent. • The black curve represents the pure liquid and the blue curve represents the solution. • Notice the changes in the boiling & freezing points. Phase diagrams for a pure solvent and for a solution of nonvolatile solute

  13. The increase of boiling point, Tb is directly proportional to the concentration of the solution expressed by its molality, m. Boiling-point elevation  Tb = (Tb –Tb ) = kbm Where, Tb = BP. Elevation Tb = BP of solvent in solution Tb° = BP of pure solvent m= molality , kb = BP Constant • The boiling-point elevation is proportional to the concentration of solute particles, regardless of whether the particles are molecules or ions • A 1 m aqueous solution of NaCl is 1 m Na+ and 1 mCl-, making 2 m in total solute particles • The boiling-point of elevation of a 1 m aqueous solution of NaCl is (2m)(0.51 C/m) = 1C.

  14. The decrease of freezing point, Tf is directly proportional to the concentration of the solution expressed by its molality, m. Freezing-point depression  Tf = (Tf–Tf) = kfm Where, Tf = FP depression Tf = FP of solvent in solution Tf°= FP of pure solvent m= molality kf = FP depression constant

  15. The van 't Hoff factor, i : The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved, and the concentration of a substance as calculated from its mass. • For most non-electrolytes dissolved in water, the van' t Hoff factor is essentially 1. • For most ionic compounds dissolved in water, the van't Hoff factor is equal to the number of discrete ions in a formula unit of the substance. Tf (measured) i = Tf (calculated for nonelectrolyte) For NaCl, van’t Hoff factor is 2, because NaCl consists of one Na+ and on Cl- per formula unit

  16. Example: Automotive antifreeze consists of ethylene glycol, (CH2(OH)CH2(OH), a nonvolatile noneletrolyte. Calculate the boiling point and freezing point of a 25.0 mass % solution of ethylene glycol in water.[kb=0.51 (C/m) and kf=1.86 (C/m). Let us consider we have 1000 g of solution: Mass of ethylene glycol = 250 g Mass of water = 750 g     Tf= (Tf –Tf ) = kfm Tb= (Tb –Tb ) = kbm Tb=Tb +Tb = 2.7 C +100 C Tf=Tf – Tf= 0 C – 10 C Molality= = ( ( ( ( ( ( ( ( 1 mol C2H6O2 1000 g H2O 250 g C2H6O2 moles C2H6O2 = 5.37 m 750 g H2O 62.1 g C2H6O2 1 kg H2O Kg H2O = (0.51 C/m) (5.37 m) = (1.86 C/m) (5.37 m) = 2.7 C = 10.0 C = 102.7 C = – 10.0 C

  17. Example: Calculate the freezing point of a solution containing 0.600 kg of CHCl3 and 42.0 g of eucalyptol (C10H18O), a fragrant substance found in the leaves of eucalyptus trees [kf=4.68 (C/m) and Tf=  63.5 C for chloroform].   Tf = (Tf –Tf) = kfm Tf=Tf – Tf= – 63.5 C – 2.1 C Molality= = = 0.45 m ( ( ( ( ( ( moles C10H18O 1 mol C10H18O 42 g C10H18O Kg CHCl3 154 g C10H18O 0.60 kg CHCl3 = (4.68 C/m) (0.45 m) = 2.1 C = – 65.6 C

  18. Example: List the following aqueous solutions in order of their expected freezing point : 0.050 m CaCl2, 0.15 m NaCl, 0.10 mHCl, 0.050 m CH3COOH, 0.10 m C12H22O11 CaCl2, NaCl and HCl are stronger electrolytes CH3COOH is week electrolyte C12H22O11 is nonelectrolyte 0.050 m CaCl2 0.050 m in Ca2+and 0.10 m in Cl- 0.15 m in particles 0.15 m NaCl0.15 m in Na+and 0.15 m in Cl- 0.30 m in particles 0.10 m HCl0.10 m in H+and 0.10 m in Cl- 0.20 m in particles 0.050 m CH3COOH between 0.050 m and 0.10 m in particles 0.050 m C12H22O11 0.10 m in particles 0.15 m NaCl(lowest freezing-point ), 0.10 m HCl , 0.050 m CaCl2, 0.050 m C12H22O11, 0.050 m CH3COOH (highest freezing-point )

  19. Osmosis • Osmosis is the spontaneous movement of water across a semi-permeable membrane from an area of low solute concentration to an area of high solute concentration • Osmotic Pressure - The pressure that must be applied to stop osmosis

  20. The osmotic pressure obeys a law similar in form to the ideal gas law  V= nRT  =(n/V)RT = MRT , osmotic pressure of soln V, volume soln n, number of moles of solute R, ideal-gas constant M, molarity of soln • Two solutions having identical osmotic pressure are isotonic • Solution of lower osmotic pressure is hypotonic with respect to more concentrated soln • Solution of more concentrated solution is hypertonic with respect to the dilute soln

  21. Example: The average osmotic pressure of blood is 7.7 atm at 25 C. What molarity of glucose (C6H12O6) will be isotonic with blood?  = M RT T = 273 + 25 = 298 K R = 0.0821 L.atm/mol.K  M =  = 7.7 atm RT M = ? 7.7 atm M = (0.0821 L.atm/mol.K)(298 K) = 0.31 atm

  22. Example: What is the osmotic pressure at 20 C of a 0.0020 M sucrose (C12H22O11) solution? T = 273 + 20 = 293 K R = 0.0821 L.atm/mol.K  = M RT M = 0.0020 M = 0.002 (mol/L)  = ?  = (0.0020 (mol/L))(0.0821 L.atm/mol.K)(293 K) = 0.048 atm

  23. The colligative properties of solutions provides a useful means experimentally determining molar mass. Example: A solution of an unknown nonvolatile nonelectrolyte was prepared by dissolving 0.25 g of the substance in 40.0 g of CCl4. The boiling point of the resultant solution was 0.357 C higher than that of the pure solvent. Calculate the molar mass of the solute. Tb = kbm ( ( Tb ( ( 0.357 C 0.0711 mol solute 0.25 g = = 0.0711 m Molality = Kb 5.02 C/m 0.00284 mol Kg CCl4 Number of mole of solute in the solution = (0.040 kg CCl4) = 0.00284 mol solute Molar mass = = 88 g/mol

  24. Example: Camphor (C10H16O) melts at 179.8 C, and it has a particularly large freezing-point-depression constant, kf =40.0 C/m. When 0.186 g of an organic substance of unknown molar mass is dissolved In 22.01 g of liquid camphor, the freezing point of the mixture is found to be 176.7 C. What is the molar mass of the solute? Tf = 179.8 – 176.7 = 3.1 C Tf = kfm = = 0.0775 m Molality = ( ( Tf ( ( 3.1 C 0.0775 mol solute 0.186 g Kf 40.0C/m 0. 0.0017 mol Kg C10H16O Number of mole of solute in the solution = (0.02201 kg C10H16O) = 0.0017 mol solute Molar mass = = 110 g/mol

  25. Example: The osmotic pressure of an aqueous solution of a certain protein was measured to determine the protein’s molar mass. The solution contained 3.50 mg of protein dissolved in sufficient water to form 5.00 mL of solution. The osmotic pressure of the solution at 25Cwas found to be 1.54 torr. Treating the protein as a nonelectrolyte, calculate its molar mass. T = 273 + 25 = 298 C R = 0.0821 L.atm/mol.K  = 1.54 torr =1.54/760 = 0.002026 atm  = M RT ( ( 0.002026 atm 0.0035 g (0.0821 L.atm/mol.K)(298 K) Molarity = 4.14  10-7 mol = 8.28  10-5 mol /L (8.28  10-5 mol /L) (5.00  10-3 L) Mole of solute in the solution = = 4.14  10-7 mol Molar mass = = 8.45  103 g/mol

  26. Example: A sample of 2.05 g of polystyrene of uniform polymer chain length was dissolved in enough toluene to form 0.100 L of solution. The osmotic pressure of this solution was found to be 1.21 kPa at 25C. Calculate the molar mass of the polystyrene. T = 273 + 25 = 298 C R = 8.314 kg.m2/S2.mol.K  = 1.21 kPa=1210/101325 = 0.01194 atm  = M RT ( ( 0.01194 atm 2.05 g (0.0821 L.atm/mol.K)(298 K) Molarity = 4.88  10-5 mol = 4.88  10-4 mol /L (4.88  10-4 mol /L) (0.10 L) Mole of solute in the solution = = 4.88  10-5 mol Molar mass = = 4.20  104 g/mol

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