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Chemistry 102(01) Spring 2012

Chemistry 102(01) Spring 2012. CTH 328 9:30-10:45 am Instructor : Dr. Upali Siriwardane e-mail : upali@latech.edu Office : CTH 311 Phone 257-4941 Office Hours : M,W 8:00-9:00 & 11:00-12:00 am; Tu,Th,F 8:00 - 10:00 am.. Exams: 9 :30-10:45 am, CTH 328.

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Chemistry 102(01) Spring 2012

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  1. Chemistry 102(01) Spring 2012 CTH 328 9:30-10:45 am Instructor: Dr. UpaliSiriwardane e-mail: upali@latech.edu Office: CTH 311 Phone 257-4941 Office Hours: M,W 8:00-9:00 & 11:00-12:00 am; Tu,Th,F8:00 - 10:00 am.. Exams: 9:30-10:45 am, CTH 328. March 26 , 2012 (Test 1): Chapter 13 April 18 , 2012 (Test 2): Chapter 14 &15 May 14 , 2012 (Test 3):Chapter 16 &18 Optional Comprehensive Final Exam: May 17, 2012 : Chapters 13, 14, 15, 16, 17, and 18

  2. Chapter 15. The Chemistry of Solutes and Solutions 15.1 Solubility and Intermolecular Forces 15.2 Enthalpy, Entropy, and Dissolving Solids 15.3 Solubility and Equilibrium 15.4 Temperature and Solubility 15.5 Pressure and Dissolving Gases in Liquids: Henry's Law 15.6 Solution Concentration: Keeping Track of Units 15.7 Vapor Pressures, Boiling Points, and Freezing Points of Solutions 15.8 Osmotic Pressure of Solutions 15.9 Colloids 15.10 Surfactants 15.11 Water: Natural, Clean, and Otherwise

  3. Solution Concentration Units a) Molarity (M) b) Molality (m) c) Mole fraction (Ca) d) Mass percent (% weight) e) Volume percent (% volume) f) "Proof" g) ppm and ppb

  4. 1) Define following solution concentration units: • a) Molarity (M) b) Molality (m) • c) Mole fraction (Ca) d) Mass percent (% weight) • e) Volume percent (% volume) • f) "Proof" g) ppm and ppb

  5. Molarity The number of moles of solute per liter of solution. molarity M moles of solute M = liter of solution units  molar = moles/liter = M

  6. Molarity Calculation An aqueous solution 58.5 g of NaCl and 2206g water has a density of 1.108 g/cm3. Calculate the Molarity of the solution. 58.5 g  1 mole Solution volume 58.5 g + 2206 g in L                                         1.00 mole NaCl Molarity of NaCl solution = ------------------------- = 0.489 M                                                    2.044 L solution

  7. Molality • number of moles of solute particles (ions or molecules) per kilogram of solvent #moles solute m = #kilograms of solvent Calculate the molality of C2H5OH in water solution which is prepared by mixing 75.0 mL of C2H5OH and 125 g of H2O at 20oC. The density of C2H5OH is 0.789 g/mL.

  8. Molarity Calculation 125 g of H2O = 0.125 kg H2O                       1.284 mole C2H5OH Molality(m) = ------------------------ = 10.27 m                          0.125 kg H2O

  9. Mole Fraction #moles of component i Xi = total number of moles Calculate the mole fraction of benzene in a benzene(C6H6)-chloroform(CHCl3) solution which contains 60 g of benzene and 30 g of chloroform.M.W. = 78.12 (C6H6) M.W. = 119.37 (CHCl3)

  10. Mole Fraction Calculation                                  moles of a                na Mola Fraction(ca) = -------------------  =     --------------                             moles of na + moles nbna + nb             a = C6H6             b = CHCl3                                      nC6H6 Mola Fraction(ca) = ------------------                                    nC6H6 + nCHCl3 m.w (C6H6) = 78.12 g/mole m.w (CHCl3) = 119.37 g/mole 60/78.12 = 0.768 mole C6H6 30/119.37 = 0.251 mole CHCl3 ca(C6H6) = 0.768/ 0.786+ 0.251 =0.754 Ca(CHCl3) = 0.0.251/ 0.786+ 0.251 = 0.246 1.000

  11. Weight Percent #g of solute wt % =  102 #g of solution Volume Percent #L of solute Vol % = 102 #L of solution Proof proof = Vol % x 2

  12. Problem What is the mole fraction of ethanol, C2H5OH, in ethanol solution that is 40.%(w/w) ethanol, C2H5OH, by mass?a. 0.40 b. 0.46 c. 0.21 d. 0.54

  13. Parts per Million #g of solute #mg of solute ppm =  106 = #g of solution #kg of solution #mL solute ppm = #L of solution Parts per Billion #g of solute #micro-g of solute ppb =  109 = #g of solution #kg of solution

  14. ppm and ppb conversions 1 ppm = (1g/ 1x 106g) 1x 106 = (1/1000 g) x 1x 106 1x 106 / 1000g = mg/ 1x 103 g = mg/ L 1 ppb = (1g/ 1x 109g) 1x 109 = (1/1000000 g) 1x 109/1000000g = mg/ 1x 103 g = mg/ L

  15. Problem A solution of hydrogen peroxide is 30.0% H2O2 by mass and has a density of 1.11 g/cm3. The MOLARITY of the solution is: 7.94 M b) 8.82 M c) 9.79 M d) 0.980 e) none of theseM.W. = 34.02 (H2O2)

  16. Comparison of Concentration Terms

  17. Effect of Solutes on SolutionColligative Properties • Colligative Properties: Depend on the number of particles not on the identity of the particles • Solution Colligative Properties a) Vapor Pressure Lowering b) Freezing Point Depression c) Boiling Point Elevation d) Osmotic Pressure • Two types of solutes affect colligative properties differently a) Volatile solutes (covalent) b) nonvolatile solutes (ionic)

  18. Vapor Pressure ofPure Water vs. Sea Water

  19. Vapor Pressure Lowering Raoult’s Law P1 = X1P1o • Psol= solventPsolvent • Psol= (1-solute) Psolvent The vapor pressure above a glucose-water solution at 25oC is 23.8 torr. What is the mole fraction of glucose (non-dissociating solute) in the solution. The vapor pressure of water at 25oC is 30.5 torr.

  20. Vapor Pressure Lowering

  21. Effect on Boling and Freezing point

  22. Boiling Point Elevation

  23. Boiling Point Elevation DTb = Tfinal - Tinitial (DTb = bpsolution - bppure solvent) DTb = kb x m where kb => boiling point elevation constant m => molality of all solutes in solution Freezing Point Depression (DTf = fppure solvent - fpsolution) DTf = kf x m where kf=> freezing point depression constant m => molality of all solutes in solution For electrolytes multiply i => number of particles per formula unit

  24. Boiling Point Elevation & Freezing point Depression Constants

  25. What is the freezing point of a 0.500 m aqueous solution of glucose? (Kf for H2O is 1.86 oC/m)(DTf = fppure solvent - fpsolution)DTf = kf x m Freezing Point Depression Problem

  26. Calculation of Molecular Weight • A 2.25g sample of a compound is dissolved in 125 g of benzene. The freezing point of the solution is 1.02oC. What is the molecular weight of the compound? Kf for benzene = 5.12 oC/m, freezing point = 5.5oC. DTf = kf x mm = moles/ kg of solventMW = g/moles

  27. Solvent Freezing

  28. Colligative Properties ofElectrolytes • Number of solute particles in the solution depends on dissociation into ions expressed as Van’t Hoff facotor(i) • Van’t Hoff facotor (i) moles of particles in solution moles of solutes dissolved

  29. Colligative Properties of Electrolytes Ionic vs. covalent substances vpwater > vp1M sucrose > vp1M NaCl > vp 1M CaCl2 1 mole sucrose = 1 mole molecules (i = 1) 1 mole NaCl = 2 mole of ions (i = 2) 1 mole CaCl2 = 3 moles ions (i = 3) i => number of particles per formula unit Psol = (1- isolute) Psolvent DTf = i kf x m DTb = i kb x m P = i MRT

  30. Osmosis

  31. Measuring Osmotic Pressure

  32. Osmosis and the Cell

  33. Osmotic Pressure P = MRTi where P => osmotic pressure M => concentration R => gas constant T => absolute Kelvin temperature i => number of particles per formula unit

  34. Calculate the osmotic pressure in atm at 20oC of an aqueous solution containing 5.0 g of sucrose (C12H22O11), in 100.0 mL solution.M.W.(C12H22O11)= 342.34P = MRT R = 0.0821 L-atm/mol K = 62.4 L-torr/mol K Calculation

  35. Calculate the osmotic pressure in torr of a 0.500 M solution of NaCl in water at 25oC. Assume a 100%dissociation of NaCl. Calculation

  36. Define the Van't Hoff factor (i). Which of the following solutions will show the highest osmotic pressure: a) 0.2 M Na3PO4b) 0.2 M C6H12O6 (glucose)c) 0.3 M Al2(SO4)3d) 0.3 M CaCl2e) 0.3 M NaCl Which one has higher Osmotic Pressure

  37. Normal vs. Reverse Osmosis

  38. Deviations from Raoult’s Law Intermolecular forces between components in a dissolved solution cause deviations from the adjustment to vapor pressure. Pvap A Vapor Pressure Pvap B A

  39. Ideal, Negative, Positive Behavior ofVapor Pressure of Two Volatile Liquids

  40. Predict the type of behavior (ideal, negative, positive) based on vapor pressure of the following pairs ofvolatile liquids and explain it in terms of intermolecular attractions: a) Acetone/water(CH3)2CO/H2Ob) Ethanol(C2H5OH)/hexane(C6H14) c) Benzene (C6H6)/toluene CH3C6H5. Ideal, Negative, Positive Behavior of Vapor Pressure

  41. Acetone/water(CH3)2CO/H2O

  42. Ethanol(C2H5OH)/hexane(C6H14)

  43. Benzene (C6H6)/toluene CH3C6H5

  44. a) True solutionsb) Colloids (Tyndall effect)c) Suspensions. Types of Solutions

  45. Solution vs. Dispersion vs. Suspension Smaller particles => Larger particles Colloidal True solution dispersion Suspension Particles Ions & molecules Colloids Large-sized particles Particle size 0.2-2.0 nm 2-2000 nm >2000 nm Properties * Don’t settle out * Don’t settle out * Settle out on on standing on standing on standing * Not filterable * Not filterable * Filterable Example Sea water Fog River silt

  46. Tyndall Effect

  47. Surfactants

  48. Soaps and Detergents

  49. Cleaning Action

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