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First Mid-quarter Examination MQ-1 on Monday, Jan. 29 at 6:30 pm Covering Chapters 10, 11, and 13 PowerPoint Presentation
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First Mid-quarter Examination MQ-1 on Monday, Jan. 29 at 6:30 pm Covering Chapters 10, 11, and 13 room TBA. For students with a scheduled- class conflict with the first Mid-Quarter Exam:. (1) Send me an email today with a copy of your class schedule , and

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First Mid-quarter Examination MQ-1 on Monday, Jan. 29 at 6:30 pm Covering Chapters 10, 11, and 13


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    1. First Mid-quarter Examination MQ-1 on Monday, Jan. 29 at 6:30 pm Covering Chapters 10, 11, and 13 room TBA

    2. For students with a scheduled-class conflict with the first Mid-Quarter Exam: • (1) Send me an email today with a copy of your • class schedule, and • Let me know when you want to take the • makeup exam: • a) during the last week of classes. • b) early (5:00-6:18 pm) on Mon, Jan 29. • c) late (8:00-9:18 pm) on Mon, Jan 29.

    3. MQ-1 on Monday, Jan. 29 at 6:30 pm Covering Chapters 10, 11, and 13 Help session on Saturday, Jan 27 at 4:00 – 6:00 pm in MP1000

    4. Week 4 Sections 13.1- 13.6—Properties of Solutions 13.1 The Solution Process Energy Changes, Entropy, Rxns 13.2 Saturated Solutions and Solubility 13.3 Factors Affecting Solubility Solute-Solvent Interactions Pressure Effects Temperature Effects 13.4 Ways of Expressing Concentration Mass Percentage, ppm, and ppb Mole Fraction, Molarity, and Molality Conversion of Concentration Units 13.5 Colligative Properties 13.6 Colloids

    5. endothermic endothermic exothermic Figure 13.3

    6. Figure 13.4 e.g. MgSO4 e.g. NH4NO3

    7. Realize there is an inherent tendency for the two isolated materials to form solution, regardless of the energetics!!! This represents an “entropy” factor.

    8. Factors that FAVOR solubility: 1. Strong solute-solvent interactions 2. Weak solute-solute interactions 3. Weak solvent-solvent interactions More often we’ll settle for the solute-solvent interactions being similar to the solute-solute and solvent-solvent interactions. A general rule: Like dissolves like. i.e. polar and polar non-polar and non-polar

    9. Dissolution: solute + solvent  solution. • Crystallization: solution  solute + solvent. • Saturation: crystallization and dissolution are in equilibrium. • Solubility: amount of solute required to form a saturated solution. • Supersaturation: a solution formed when more solute is dissolved than in a saturated solution.

    10. Fig 13.12 Structure of glucose—note red O atoms in OH groups which can interact nicely with water.

    11. Consider gas solubilities in water at 20 oC with 1 atm gas pressure (~Table 13.2) Solubility/M He 0.40 x 10-3 N2 0.69 x 10-3 CO 1.04 x 10-3 O2 1.38 x 10-3 Ar 1.50 x 10-3 Kr 2.79 x 10-3 CO2 3.1 x 10-2 NH3 ~ 53

    12. Fig 13.14 Henry’s Law, Cg = k Pg Note Table 13.2 again for k

    13. To read about nitrogen narcosis, see http://www.scuba-doc.com/narked.html and about the bends, see http://www.diversalertnetwork.org/medical/articles/index.asp

    14. Solubility of GASES as a Function of Temperature

    15. And of SALTS Are these exothermic or endothermic processes?

    16. Ways of expressing concentration: a) percent, ppm, ppb usually mass/mass b) mole fraction = XA , XB sum of Xi = 1 c) molarity = M or mol/Lsolution depends on T and density of soln preparation requires dilution d) molality = m or mol/kgsolvent independent of T easily prepared

    17. (a) Prepare 0.500 L of a 0.100 M aqueous solution of KHCO3 (MW = 100.12 g) Recall M=n/L or n = (M)(L) therefore we need 5.01 g of KHCO3 dissolved and diluted to 0.500 L • Use this solution as a ‘stock’ solution to prepare • a final solution of 0.0400 M concentration. • What is the final volume of this solution? Since n = M V, M1V1 = M2V2 and V2 = M1V1/M2 = (0.100 M)(0.500 L) / (0.0400 M) = 1.25 L

    18. Consider a solution prepared by dissolving 22.4 g MgCl2 in 0.200 L of water. Assume the density of water is 1.000 g/cm3 and the density of the solution is 1.089 g/cm3. Calculate mole fraction molarity molality

    19. MQ-1 on Monday, Jan. 29 at 6:30 pm Covering Chapters 10, 11, and 13 Help session on Saturday, Jan 27 at 4:00 – 6:00 pm in MP1000

    20. For students with a scheduled-class conflict with the first Mid-Quarter Exam: • (1) Send me an email today with a copy of your • class schedule, and • Let me know when you want to take the • makeup exam: • a) during the last week of classes. • b) early (5:00-6:18 pm) on Mon, Jan 29. • c) late (8:00-9:18 pm) on Mon, Jan 29.

    21. A 9.386 M solution of H2SO4 has a density of 1.509 g/cm3. Calculate molality % by mass mole fraction of H2SO4

    22. Week 4 Sections 13.1- 13.6 13.3 Factors Affecting Solubility 13.4 Ways of Expressing Concentration Mass Percentage, ppm, and ppb Mole Fraction, Molarity, and Molality Conversion of Concentration Units 13.5 Colligative Properties Lowering the Vapor Pressure Boiling-Point Elevation Freezing-Point Depression Osmosis Determination of Molar Mass 13.6 Colloids

    23. Colligative Properties Solution properties that depend only on the total # of ‘particles’ present. Vapor Pressure Boiling Point Freezing Point Osmotic Pressure

    24. Note that VP of a solution is lower than that of pure solvent.

    25. A fascinating and somewhat surprising observation:

    26. Raoult’s Law PA = XA PAo PA = vapor pressure over solution XA = mole fraction of component A (solvent) PAo = vapor pressure of pure component A (solvent) also PA = (1 – XB) PAo where XB = mol fraction of B (solute) (Recall also Dalton’s Law: PA = XA Ptotal )

    27. At first, we consider only nonvolatile solutes.

    28. At 20 oC, the vapor pressure of benzene (cmpd A, MW=78) is 0.1252 atm,i.e. PAo = 0.1252 atm. If 6.40 g of naphthalene (cmpd B, C10H8, 128.17 g/mol) is dissolved in 78.0 g of benzene calculate the vapor pressure of benzene over the solution.

    29. As shown on page 549, we also can consider solutions with two volatile components. Consider a liquid soln containing 1.0 mol benzene and 2.0 mol of toluene at 20 oC. This yields Xbenzene = 0.33 and Xtoluene = 0.67 This can be coupled with the fact that Pobenzene = 75 torr Potoluene = 22 torr Apply Raoult’s law to each separately, to obtain Pbenzene = XbenzenePobenzene = 25 torr Ptoluene = Xtoluene Potoluene = 15 torr and PT = Pb + Pt = 40 torr

    30. But with Pb = 25 torr and Pt = 15 torr We also can calculate the concentrations of the two in the gas phase! Xbgas = 25/40 = 0.63 and Xtgas = 15/40 = 0.37

    31. Boiling Point Elevation and Freezing Point Depression (directly related to Raoult’s Law)

    32. Boiling Point Elevation • ΔTb = Tbfinal– Tbinitial= Kb m => + quantity • where m is the molal concentration Freezing Point Depression ΔTf = Tffinal – Tfinitial= - Kf m => - quantity where m is the molal concentration. [note the definition and the negative sign!!!] for H2O, Kb = 0.052 oC/m and Kf = 1.86 oC/m

    33. Consider a water solution which has 0.500 mol of sucrose in 1.000 kg of water. Therefore it has a concentration of 0.500 molal or 0.500 mol/kg. recall Kb = 0.52 oC/m and Kf = 1.86 oC/m What is the boiling point and freezing point of this solution?

    34. (a) When 5.50 g of biphenyl (C12H10, 154.2 g/mol) is dissolved in 100 g benzene (C6H6, 78.0 g/mol), the BP increases by 0.903 oC. Calculate Kb for benzene.

    35. (a) When 5.50 g of biphenyl (C12H10, 154.2 g/mol) is dissolved in 100 g benzene (C6H6, 78.0 g/mol), the BP increases by 0.903 oC. Calculate Kb for benzene.  Kb = 2.53 K kg/mol (b) When 6.30 g of an unknown hydrocarbon is dissolved in 150.0 g of benzene, the BP of the solution increases by 0.597 oC. What is the MW of the unknown substance?

    36. But recall we said Colligative Properties depend on the total concentration of ‘species’. A sample of sea water contains the following in 1.000 L of solution. Estimate the freezing point of this solution. Na+ = 4.58 mol Cl- = 0.533 mol Mg2+ = 0.052 mol SO42- = 0.028 mol Ca2+ = 0.010 mol HCO3- = 0.002 mol K+ = 0.010 Br- = 0.001 mol neutral species = 0.001 mol Sum of species = 1.095 mol

    37. List the following aqueous solutions in increasing order of their expected freezing points. 0.050 m CaCl2  0.05 m Ca+2 and 0.10 m Cl-1 0.15 m NaCl  0.30 m total 0.10 m HCl  0.20 m total 0.050 m HOAc  between 0.05 and 0.10 m total 0.10 m C12H22O11  0.10 m total

    38. List the following aqueous solutions in increasing order of their expected freezing points. 0.050 m CaCl2 x 3 = 0.150 0.15 m NaCl x 2 = 0.30 0.10 m HCl x 2 = 0.20 0.050 m HOAc x 1 = 0.050 0.10 m C12H22O11 x 1 = 0.10 These calculations assume total dissociation of the salts and zero dissociation of the last two. The van’t Hoft “i factor”

    39. This effect of the dissociation of electrolytes is usually taken into account through the van’t Hoff i factor, which can be stated formally as ΔTb = i Kb m Note that i may be defined as ΔTf(actual) Kf meffective meffective i = -------------- = ------------- = ------------ ΔTf(ideal) Kf mideal mideal In real systems, these i factors are NOT integers, but rather fractions whose values depend on concentration.

    40. Focus, for example on NaCl. Notice the limiting value as well as the values at higher concentrations.