chemistry 102(01) spring 2013

12-1CHEM 102, Spring 2013 LA TECH

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:

Chemistry 102(01) Spring 2013

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. Tro 2nd

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 Propertiesa) Vapor Pressure Loweringb) Freezing Point Depressionc) Boiling Point Elevationd) Osmotic Pressure

Two types of solutes affect colligative properties differentlya) Volatile solutes (covalent)b) nonvolatile solutes (ionic)

1) What are colligative properties?

Vapor Pressure ofPure Water vs. Sea Water

Vapor Pressure LoweringRaoult’s Law

P1 = X1P1o

Psol = solvent Psolvent

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.

Vapor Pressure Lowering

2) What’s Rauolts Law? How it applies to a) 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).

Acetone/water(CH3)2CO/H2O

Ethanol(C2H5OH)/hexane(C6H14)

Benzene (C6H6)/toluene CH3C6H5

Deviations from Raoult’s LawIntermolecular forces between components in a

dissolved solution cause deviations from the adjustment to vapor pressure.

Pvap A

Pvap B

Ideal, Negative, Positive Behavior of Vapor Pressure of

Two Volatile Liquids

5) What is ideal, positive and negative behavior applying Raoult's Law.

Ideal:

Positive:

Negative:

Effect on Boling and Freezing point

Boiling Point Elevation

Boiling Point ElevationDTb = Tfinal - Tinitial

(DTb = bpsolution - bppure solvent)DTb = kb x mwhere kb => boiling point elevation constant

m => molality of all solutes in solutionFreezing 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

Solvent Freezing

Colligative Properties ofElectrolytesNumber 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 ElectrolytesIonic 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 unitPsol = (1- i solute) Psolvent

DTf = i kf x mDTb = i kb x mP = i MRT

Osmosis

Measuring Osmotic Pressure

Osmosis and the Cell

Osmotic Pressure

P = MRTiwhere 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 Na3PO4 b) 0.2 M C6H12O6 (glucose)c) 0.3 M Al2(SO4)3 d) 0.3 M CaCl2 e) 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

Normal vs. Reverse Osmosis

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. SuspensionSmaller 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 nmProperties * 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

Tyndall Effect

Surfactants

Soaps and Detergents

CH 3CH 2CH2CH 2CH 2CH 2CH2CH 2CH 2CH 2CH2CH 2CH 2CH 2CH2CH 2CH 2

Hydrophobic end Hydrophilic end

sodium stearate

C O-Na+

CH 3CH 2CH2CH 2CH 2CH 2CH2CH 2CH 2CH 2CH2CH 2OS O-3Na +

sodium lauryl sulfate

Cleaning Action

Earth’s Water Supply

Treatment of Drinking Water

Hard Waternatural water containing relatively high

concentrations of Ca+2, Mg+2, Fe+3, or Mn+2 cations and CO3

-2 and HCO3-1 anions

Common HazardousHousehold Chemicals

chemistry 102(01) spring 2013

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