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Chemistry 102(01) Spring 2013. Instructor: Dr. Upali Siriwardane e-mail : upali@coes.latech.edu Office : CTH 311 Phone 257-4941 Office Hours : M ,W 8:00-9:00 & 11:00-12:00 am; Tu,Th,F 9:30 - 11:30 am. Test Dates :. September 27 , 2013 (Test 1): Chapter 12 & 13
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Chemistry 102(01) Spring 2013 Instructor: Dr. UpaliSiriwardane e-mail: upali@coes.latech.edu Office: CTH 311 Phone257-4941 Office Hours: M,W 8:00-9:00 & 11:00-12:00 am; Tu,Th,F9:30 - 11:30 am. Test Dates: September 27, 2013 (Test 1): Chapter 12 & 13 April 24, 2013 (Test 2): Chapter 14 & 15 May13, 2013 (Test 3) Chapter 16 & 17 May 15, 2012 (Make-up test) comprehensive: Chapters 12-17 9:30-10:45:15 AM, CTH 328
REQUIRED: Textbook:Principles of Chemistry: A Molecular Approach, 2nd Edition-Nivaldo J. Tro - Pearson Prentice Hall and also purchase the Mastering Chemistry Group Homework, Slides and Exam review guides and sample exam questions are available online: http://moodle.latech.edu/ and follow the course information links. OPTIONAL: Study Guide: Chemistry: A Molecular Approach, 2nd Edition-Nivaldo J. Tro 2nd Edition Student Solutions Manual: Chemistry: A Molecular Approach, 2nd Edition-Nivaldo J. Tro2nd Text Book & Resources
Chapter 12.Solutions 12.1 Thirsty Solutions: 12.2 Types of Solutions and Solubility 12.3 Energetics of Solution Formation 12.4 Solubility Equilibrium and Factors Affecting solution Formation 12.5 Expressing Solution Concentration 12.6 Colligative Properties: Vapor Pressure, Freezing Point, Boiling Point, Osmatic Pressure 12.7 Colligative properties of Strong Electrolytes
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)
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.
2) What’s Rauolts Law? How it applies to • a volatile & nondissociating, • b) nonvolatile & nondissociating, • c) nonvolatile & dissociating solutes in a solution of volatile solvent.
3) What is the vapor pressure (atm) of a solution of a nonvolatile, nondissociating solute(mole fraction 0.25) in water at 50oC? The vapor pressure of water at 50oC is 0.122 atm.
4)What is the total pressure at 25oC of a solution of 2.90 moles of C6H14 and 5.94 moles of C6H12 at 25oC if the vapor pressures of the pure solvents are 151 and 98 mm Hg respectively at 25oC?(Atomic weights: C = 12.01, H = 1.008, Cl = 35.45).
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
Ideal, Negative, Positive Behavior ofVapor Pressure of Two Volatile Liquids
5)What is ideal, positive and negative behavior applying Raoult's Law. • Ideal: • Positive: • Negative:
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
Boiling Point Elevation & Freezing point Depression Constants
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
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
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
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- isolute) Psolvent DTf = i kf x m DTb = i kb x m P = i MRT
Osmotic Pressure P = MRTi where P => osmotic pressure M => concentration R => gas constant T => absolute Kelvin temperature i => number of particles per formula unit
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
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
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
6)Which of the following solutes dissolved in 1000 g of water estimate the number particles in the solution? Use Vant Hoff factor. • 0.030 moles urea, CO(NH2)2 (a covalent compound) • 0.030 moles acetic acid, CH3COOH(weak acid) • 0.030 moles ammonium nitrate, NH4NO3(soluble) • 0.030 moles calcium sulfate, CaSO4 (insoluble) • 0.030 moles aluminum chloride, AlCl3 (soluble)
7)Determine the molecular weight of acetic acid if a solution that contains 30.0 grams of acetic acid per kilogram of water freezes at -0.93oC. Do these results agree with the assumption that acetic acid has the formula CH3CO2H? Kf(water) = 1.86.
8)Explain why a 0.100 m solution of HCl dissolved in benzene has a freezing point depression of 0.512oC, while an 0.100 m solution of HCl in water has a freezing point depression of 0.352oC. Kf(benzene) = 5.5, Kf(water) = 1.86
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
a) True solutionsb) Colloids (Tyndall effect)c) Suspensions. Types of Solutions
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
Hard Water • natural water containing relatively high concentrations of Ca+2, Mg+2,Fe+3, or Mn+2 cations and CO3-2 and HCO3-1 anions