Entry Task: Dec 3 rd Monday

1. Entry Task: Dec 3rd Monday Question: Define boiling point elevation and freezing point depression You have 5 minutes!

2. Agenda: • Discuss Ch. 13 sec. 5 • HW: Ch. 11-13 review ws

3. I can … • Explain how colligative properties affect the physical properties of a solution.

4. Chapter 13Properties of Solutions

5. Colligative Properties • Changes in colligative properties depend only on the number of solute particles present, not on the identity of the solute particles. • Among colligative properties are • Vapor pressure lowering • Boiling point elevation • Melting point depression • Osmotic pressure

6. Vapor Pressure • Because of solute-solvent intermolecular attraction, higher concentrations of nonvolatile solutes make it harder for solvent to escape to the vapor phase. • Therefore, the vapor pressure of a solution is lower than that of the pure solvent.

7. Raoult’s Law PA = XAPA where • XA is the mole fraction of compound A • PA is the normal vapor pressure of A at that temperature NOTE: This is one of those times when you want to make sure you have the mole fraction of the solvent.

8. 13.7problem 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? Convert 1.00 atm to torr = 760 torr 1070torr - 760 torr = 310 torr is Pa 310 torr is Pa = X 1070 torr 310 torr= X 1070 torr = 0.290

9. Boiling Point Elevation and Freezing Point Depression Nonvolatile solute-solvent interactions also cause solutions to have higher boiling points and lower freezing points than the pure solvent.

10. Boiling Point Elevation The change in boiling point is proportional to the molality of the solution: Tb = Kb  m where Kb is the molal boiling point elevation constant, a property of the solvent. Tb is added to the normal boiling point of the solvent.

11. Freezing Point Depression • The change in freezing point can be found similarly: Tf = Kf m • Here Kf is the molal freezing point depression constant of the solvent. Tf is subtracted from the normal freezing point of the solvent.

12. Note that in both equations, T does not depend on what the solute is, but only on how many particles are dissolved. Tb = Kb  m Tf = Kf m Boiling Point Elevation and Freezing Point Depression

13. 13.8 problem 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. (See Table 13.3.) Tf = Kf m 42.0 g 1 mole =0.273 mole = 0.455 m 0.600 kg 154 g 4.68 • 0.455 = 2.194 depressed -63.5 – 2.194 = -65.69 ºC

14. 13.9 problem Which of the following solutes will produce the largest increase in boiling point upon addition to 1 kg of water: 1 mol of Co(NO3)2, 2 mol of KCl, 3 mol of ethylene glycol (C2H6O2)? • Answer: 2 mol of KCl because it contains the highest concentration of particles, 2 m K+ and 2 m Cl, giving 4 m in all

15. Colligative Properties of Electrolytes Since colligative properties depend on the number of particles dissolved, solutions of electrolytes (which dissociate in solution) should show greater changes than those of nonelectrolytes.

16. Colligative Properties of Electrolytes However, a 1 M solution of NaCl does not show twice the change in freezing point that a 1 M solution of methanol does.

17. van’t Hoff Factor • One mole of NaCl in water does not really give rise to two moles of ions. • Some Na+ and Cl− reassociate for a short time, so the true concentration of particles is somewhat less than two times the concentration of NaCl.

18. van’t Hoff Factor • Reassociation is more likely at higher concentration. • Therefore, the number of particles present is concentration dependent.

19. The van’t Hoff Factor We modify the previous equations by multiplying by the van’t Hoff factor, i Tf = Kf  m i

20. 13.49 13.49 Using data from Table 13.4, calculate the freezing point and boiling points of each of the following solutions: a) 0.35 m glycerol in ethanol; (1.22) (0.35m) = 0.427 + 78.4 = 78.8 b.pt (1.99) (0.35 m) = 0.697- -114.6 = -115.3 f.pt

21. 13.49 13.49 Using data from Table 13.4, calculate the freezing point and boiling points of each of the following solutions: a) 0.35 m glycerol in ethanol; b) 1.58 mol of naphthalene, C10H8, in 14.2 mol of chloroform; 14.2 mol 119.35 g = 1694.77 g or 1.695 kg 1 mole 1.58 mole = 0.932 m 1.695 kg (3.63) (0.932m) = 3.38 + 61.2 = 64.6 b.pt (4.36) (0.932 m) = 4.36 - -63.5 = -67.9 f.pt

22. 13.49 13.49 Using data from Table 13.4, calculate the freezing point and boiling points of each of the following solutions: a) 0.35 m glycerol in ethanol; b) 1.58 mol of naphthalene, C10H8, in 14.2 mol of chloroform; c) 5.13 g KBr and 6.85 g glucose, C6H12O6, in 255 g water. 5.13 g 1 mol = 0.043 mole KBr 0.081 mole 119 g 0.255 kg 6.85 g 1 mol = 0.038 mole H2O = 0.318 m 180g (0.52) (0.318m) = 0.165 + 100.0 = 100.2 b.pt (1.86) (0.318 m) = 0.591 - -0 = -0.591 f.pt

23. 13.50 13.50 Using data from Table 13.4, calculate the freezing point and boiling point of each of the following solutions: a) 0.50 m glucose in ethanol; b) 18.0 g C10H22 in 425 g CHCl3; c) 0.45 mol ethylene glycol and 01.0 molKBr in 125 gH2O. (1.22) (0.50 m) = 0.61 + 78.4= 79.0 b.pt (1.99) (0.50 m) = 0.995 - -114.6 = -115.6 f.pt

24. 13.50 13.50 Using data from Table 13.4, calculate the freezing point and boiling point of each of the following solutions: a) 0.50 m glucose in ethanol; b) 18.0 g C10H22 in 425 g CHCl3; 18.0 g 1 mol = 0.1267 mole C10H22 0.1267 mole 142 g 0.425 kg = 0.298 m (3.63) (0.298m) = 1.08 + 61.2 = 62.3 b.pt (4.68) (0.298 m) = 1.39 - -63.5 = -64.9 f.pt