Week 14/Tu: Lecture Units ‘33 & 34’

© DJMorrissey, 2o12

Exam 3 Unit 33: Colligative Properties -- Vapor pressure of solutions -- Freezing, boiling of solutions -- Osmotic pressure Unit 34: Introduction to Equilibria -- Rate of reaction, Kinetics -- Reaction pathway -- Forward / Reverse Rxns. Unit 35: Equilibrium Constants Issues: Exam 3 results Homework Set 10 due on Saturday @ 08:00AM

0

50

100

150

200

250

300

350

10 30 50 70 90 110 130 150 170 Num

ber

of S

tude

nts

Score

CEM 141 Fall 2o12 Exam 1

Week 14/Tu: Exam Results

© DJMorrissey, 2o12

0

50

100

150

200

250

300

10 30 50 70 90 110 130 150 170

Num

ber

of S

tude

nts

Score

CEM 141 Fall 2o12 Exam 2

0 20 40 60 80

100 120 140 160

5 25 45 65 85 105 125 145 165

Num

ber

of S

tude

nts

Score

CEM 141 Fall 2o12 Exam 3

Exam 1 Exam 2 Exam 3 Count 2344 2226 2240 Mean 109.0 105.5 100.6

Median 110 107.5 100 Mode 110 120 110

SD 31.7 34.8 33.8 Range 170 170 175

Minimum 10 10 5 Maximum 180 180 180

Grade Point Estimator

Week 14/Tu: Final Exam Info

© DJMorrissey, 2o12

The final exam is a common final and has been scheduled for Thursday, December 13th 7:45 - 9:45AM The exam will have < 40 questions, cumulative, for a total of 250 pts. There will be an alternate final exam for anyone who has EITHER: 1) another final scheduled at the same time as our final OR 2) three or more finals on Thursday (i.e., two others plus our final) The alternate exam will be given the day before (Wed. at 10am) A sign-up sheet for the alternate will be available next week.

Week 14/Th: Vapor Pressure of a Solution -1-

© DJMorrissey, 2o12

Recall every liquid has a vapor pressure that depends on T. A solution will also have a vapor pressure. Constant T

However, the vapor pressure of solution will be lower according to the amount of other material.

Pres

sure

0 1 ΧBlack

Po(B)

Week 14/Th: Vapor Pressure of a Solution -1-

© DJMorrissey, 2o12

Recall every liquid has a vapor pressure that depends on T. A solution of two liquids will have partial pressures. Constant T

Pres

sure

0 1 ΧRed ΧBlack

Po(R)

Po(B)

Week 14/Th: Vapor Pressure of a Solution -1-

© DJMorrissey, 2o12

Recall every liquid has a vapor pressure that depends on T. A solution of two liquids will have partial pressures. Constant T

Pres

sure

0 1 ΧBlack

Po(R)

Recall that the mole fractions add-up: ΧBlack + XRed = 1 Xred = 1 - ΧBlack

Po(B)

Ptotal

Week 14/Th: Vapor Pressure of a Solution -2-

© DJMorrissey, 2o12

Recall every liquid has a vapor pressure that depends on T. A solution with a non-volatile will also have a vapor pressure.

P=1

TNB

Part of the phase diagram, Constant X (mole fraction)

Since the vapor pressure of a solution is lower than that of the pure liquid, it will require a higher temperature to reach a vapor pressure of one atm (or boil).

NaCl i = 2 KNO3 2 Mg(NO3)2 3 Ca3(PO4)2 5 HCl 2 CH3CO2H 1

Week 14/Th: Phase Diagram for Solution

© DJMorrissey, 2o12

Similar to the boiling point elevation, the freezing point of the solution will be depressed relative to the pure liquid. Note that all of the components in solution affect FP, BP. (“i” is called the van’t Hoff factor)

Solution

ΔTFP = KFP * mtotal * i KFP

water = -1.86 K/molal (cartoon)

© DJMorrissey, 2o12

Week 14/Tu: Practical Problem on the Horizon “Select Brand Ice Melter” is NaCl

“Pellets of Fire” are CaCl2

Table sugar is C12H22O11

Solute Solubility* Molar Mass Molarity “particles” ΔT (g/mL) (g/mol) (mol/L) i approx.** NaCl 0.356 58.44 6.09 2 -22.7 K CaCl2 0.595 110.98 5.36 3 -29.9 K Sugar 1.79 342.3 5.23 1 -9.7 K * In cold water ** using M, need density to get m

© DJMorrissey, 2o12

Week 14/Tu: Example Problem in Notes

When 1.0 mole of magnesium acetate was dissolved in water, the boiling point increased by 6.0o to 106.0oC. How much water was used to make this solution? KBP(water) = +0.50 K/molal ΔT = KBP * m * i 6.0 K = 0.50 K/molal * molality * 3 & molal = moles/kg of water 6.0 K = 0.50K/molal * 1.0 moles / kg of water * 3 kg of water = 1.0 mole * (0.50 K/molal * 3) / 6.0K = 0.25

that’s: Mg(CH3CO2)2 Molar Mass not needed, but what is “i”?

Week 14/Tu: Demo – Supersaturated NaCH3CO2

© DJMorrissey, 2o12

Week 14/Tu: Basis of Osmotic Pressure

© DJMorrissey, 2o12

Po P = xH2O Po

One beaker with an aqueous solution, water mole fraction, XH2O vapor pressure P = XH2O Po

Another beaker with pure water vapor pressure, Po

Put both under a bell jar, what happens?

Week 14/Tu: Natural Osmosis

© DJMorrissey, 2o12

Sugars (solute) Sap (solution) Water (solvent)

Plants and trees in particular rely on osmotic pressure to move water from the roots to the leaves. Trees create a sugar solution which is more concentrated than the ground water.

Week 14/Tu: Osmotic Pressure

© DJMorrissey, 2o12

Osmotic pressure is the result of a drive to form dilute solutions (generally an entropy-increasing process). There are many examples of osmotic pressure in everyday life, for example isotonic saline solution with 9.0 g/L NaCl and a density of 1.0046 kg/L is the basis for intravenous fluids.

Molarity =(9.0 g/L) / 58.44 g/mol NaCl = 0.15 moles NaCl / L Mass of solution = Mass NaCl + water 1004.6 g = 9.0 g + Mass water 995.6 g = Mass water molality = 0.15 mole NaCl / 0.9956 kg = 0.15 molal

For review show that M = m for this solution

Week 14/Tu: Reverse Osmosis

© DJMorrissey, 2o12

To “reverse osmosis” one has to apply a pressure on the solution side that is bigger than that from osmosis. Pure water will flow out of the membrane (if it can hold up).