w-03312009004127OaVbjTgyCl

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PRACTICE PROBLEMS

CHEM 162-2007 EXAM I

CHAPTER 12 - PROPERTIES OF SOLUTIONS

SOLUTION CONCENTRATIONS CONCEPTS (Molarity, mole fraction, etc.)

SOLUTION CONCENTRATIONS CALCULATIONS (Molarity, mole fraction, etc.)

ENERGETICS OF SOLUTIONS AND SOLUBILITY CONCEPTS

ENERGETICS OF SOLUTIONS AND SOLUBILITY CALCULATIONS

VAPOR PRESSURE OF SOLUTIONS CONCEPTS (Henry’s Law, Raoult’s Law)

VAPOR PRESSURE OF SOLUTIONS CALCULATIONS

COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE CHANGES) CONCEPTS

COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE

CHANGES) CALCULATIONS

MISCELLANEOUS

E. Tavss, PhD

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FORMULAS

Mass percent = grams of solute/100 g solution

Molarity = moles of solute/L of solution

Molality = moles of solute/kilogram of solvent

PPM = grams of solute/1,000,000 grams solution

Volume percent = volume of solute/100 mL of solution

Proof = 2 x Vol. %; e.g., 2 x 40 mL/100 mL solution = 80 proof

Mole fraction: XA = nA/(nA + nB) XA + XB = 1

Particle fraction: iXA = inA/(inA + inB) iXA + iXB = 1

Raoult’s law: Psoln = iXsolventPosolvent

iXsolvent = insolvent/(insolvent + insolute)

iXA + iXB = 1

PTotal = PA + PB = iXAPo

A + iXBPo

B

van’t Hoff factor, “i”; i = moles of particles in solution/moles of solute dissolved

Boiling-point elevation: ΔT = Tf - Ti = Kbimsolute

Freezing-point depression: ΔT = Tf - Ti = -Kfimsolute (+Kf gives absolute change in temp; -Kf gives actual T f or Ti)

Osmotic pressure: πV = inRT

π = (n/V)iRT = iMsoluteRT

ΔHsoln = ΔHsolute-solutebondbreaking + ΔHsolvent-solventbondbreaking + ΔHsolute-solventbondforming ΔHhydr = ΔHsolvent-solventbondbreaking + ΔHsolute-solventbondforming ΔHsoln = ΔHsolute-solutebondbreaking + ΔHhydr

Henry’s Law: XA=kPA; XA = mole fraction of dissolved gas in solution

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SOLUTION CONCENTRATIONS CONCEPTS

CHEM 162SG-2001 HOURLY EXAM I CHAPTER 11 - PROPERTIES OF SOLUTIONS

SOLUTION CONCENTRATIONS CONCEPTS 14. To 1000 g of water containing 0.010 mole of NaCl, 10.0 g of ethyl alcohol, C2H5OH(l)

are added. Which quantity remains the same?

A. the molarity of NaCl B. the mass percentage of NaCl

C. the mole fraction of NaCl

D. the molality of NaCl

E. the vapor pressure of water

CHEM 162-2003 HOURLY EXAM 1 + ANSWERS

CHAPTER 11 - PROPERTIES OF SOLUTIONS SOLUTION CONCENTRATION CONCEPTS

1. For which of the following conversions must you know the density of the solution? A. mass percent to molality

(g solute/100 g solution) x (mol/g) x 1000 g/kg = molality

B. mass percent to molarity

(g solute/100 g solution) x (mol/g) x (g/mL) x (1000 mL/L) = molarity C. mole fraction to mass percent

((molA x (g/mol))/(molA x (g/mol) + molB x (g/mol))) x 100 = mass percent D. mole fraction to molality

(molA)/(molA x (g/mol) + molB x (g/mol)) x 1000 g/kg = molality E. volume percent to proof mLsolute/100 mLsolution x 2 = proof

1. 1

Chem 162-2004 Exam 1 Chapter 11

SOLUTION CONCENTRATIONS (MOLARITY, MOLALITY, MOLE FRACTIONS) Molality

In a 0.1 M solution of H2O2 in water prepared at 25 ºC, which one of the following will be close to 0.1?

A. the mole fraction of H2O2

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B. the molality of H2O2

C. all of these are close to 0.1 D. the mass fraction of H2O2 E. none of these are close to 0.1

Mole fraction of H2O2:

(0.1 mol H2O2/1000 mL sol’n) x 34 g/mol = 3.4 g H2O2/1000 mL sol’n 1000 ml x 1g/ml = 1000 g solution 1000 g solution - 3.4 g H2O2 = 996.6 g H2O

mole fraction = mol H2O2/(mol H2O2 + mol H2O) = (0.1 mol/(0.1 mol + (996.6/18.02)) = 0.0018 H2O2 mole fraction

molality = mol H2O2/1000 g solvent (0.1 mol H2O2/996.6 g H2O) x 1000 g/kg = 0.100 m

mass fraction = mass H2O2/(mass H2O2 + mass H2O)

= 3.4 g H2O2/(3.4 g H2O2 + 996.6 g H2O) = 0.0034

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SOLUTION CONCENTRATIONS CALCULATIONS

23 Chem 162-2006 Hourly exam I + answers

Chapter 11 – Properties of solutions Solution concentrations calculations (mole fractions)

What is the mole fraction of SO42- in a 1.246M Na2SO4 aqueous solution. The density of

the solution is 1.157g/cm3?

A. 0.0429

B. 0.0214

C. 0.0224 D. 0.980 E. 0.0376

1.246M Na2SO4 = 1 mol Na2SO4 per liter of solution

Mole fraction = moles SO42-/(moles Na+ + moles SO4

2- + mol H2O) Na2SO4 2Na+ + SO4

2- 1.246 mol Na2SO4 (2x1.246) mol Na+ + 1.246 mol SO4

2-

1 L solution x (1157 g/L) = 1157 g solution 1.246 mol Na2SO4 x 142.04 g/mol = 176.98 g solute

gsolvent = gsolution – gsolute gsolvent = 1157gsolution – 176.98 gsolute = 980gsolvent 980g x (1 mol/18.02g) = 54.38 mol solvent

Mole fraction: 1.246 mol SO42-/(2.492 mol Na+ + 1.246 mol SO4

2- + 54.38 mol H2O) = 0.0214 = mole fraction of SO4

2-.

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19 Chem 162-2006 Hourly exam I + answers Chapter 11 – Properties of solutions Solution concentration calculations

Molality

What is the molality of a 1.562M CaCl2 solution with a density of 1.203g/cm3?

A. 1.517m

B. 1.298m C. 1.879m D. 1.562m

E. 1.475m

molarity = moles of solute per L of solution 1.562M CaCl2 = 1.562 mol CaCl2 per liter of solution molality = moles of solute per kg of solvent

MW CaCl2 = 40.08 + 35.45 + 35.45 = 110.98 g/mol 1.562 mol CaCl2 x 110.98 g/mol = 173.4 g CaCl2

1Lsoln x 1203 g/L = 1203 g solution 1203 g solution – 173.4 g solute = 1029.6 g solvent = 1.029 kg solvent 1.562 mol CaCl2/1.029 kg solvent = 1.518 mol CaCl2/kg solvent = 1.518m

CHEM 162-2003 1.5 WEEK RECITATION CHAPTER 11 - PROPERTIES OF SOLUTIONS


27. A solution is prepared by dissolving 50.0 g cesium chloride (CsCl) in 50.0 g water. The volume of the solution is 63.3 mL. Calculate the mass percent, molarity, molality, and

mole fraction of the cesium chloride.

mass percent = g solute/g solution = 50.0 g cesium chloride/(50.0 g cesium chloride + 50.0 g water) = 50 mass percent.

molarity = moles CsCl/L = (50.0 g CsCl/168.3 g/mol)/0.0633 L = 4.69 M CsCl

molality = mole/kg solvent = ((50.0 g CsCl/168.3 g/mol)/50 g water) x (1000 g/kg) = 5.94

m CsCl

mol fraction = molA/(molA + molB) mol fraction = mol CsCl/(mol CsCl + mol H2O) = (50.0 g CsCl/168.3 g/mol)/((50.0 g CsCl/168.3 g/mol) + (50.0 g H2O/18.02 g/mol)) = 0.0967

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29 Common commercial acids and bases are aqueous solutions with the

following properties:

Density Mass percent

(g/cm3) of solute

Hydrochloric acid 1.19 38

Calculate the molarity, molality, and mole fraction of the preceding reagent.

molarity of HCl = mol/liter = (38 g HCl/100 g solution) x (1 mol HCl/36.45 g

HCl) x (1.19 g/mL) x (1000 mL/L) = 12.41 mol/L = 12.41 M

molality = mol/1kg solvent = (38 g HCl/62 g solvent) x (1 mol HCl/36.45 g

HCl) x (1000 g/1 kg) = 16.82 mol/kg = 16.82 m

mol fraction of HCl = (mol HCl)/(mol HCl + mol H2O) = (38 g HCl/36.45 g

HCl/mol)/((38 g HCl/36.45 g HCl/mol) + (62 g H2O/18.02 g H2O/mol)) =

0.233 mol fraction HCl


CHAPTER 11 - PROPERTIES OF SOLUTIONS SOLUTION CONCENTRATION CALCULATIONS 7. A solution of ethanol (C2H5OH) in water is 10.0% ethanol by mass. The mole fraction of water in

the solution is

A. 0.90

B. 0.96

C. choose this choice if none of the others is correct D. 0.096

E. 0.10 10% C2H5OH

Therefore, 90% H2O = 10 g C2H5OH + 90 g H2O

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mol fraction = mol H2O/(mol H2O + mol C2H5OH)

(90/18)/((90/18) + (10/46)) = 0.958

CHEM 162-2003 HOURLY EXAM 1 + ANSWERS CHAPTER 11 - PROPERTIES OF SOLUTIONS

SOLUTION CONCENTRATION CALCULATIONS 25. A solution is prepared from 10.0 g NaCl, 5.0 g NaI, and 50.0 g H2O. The mass percent of NaCl in

the solution is A. 20.0%

B. 18.2% C. Choose this choice if none of the others is correct.

D. 15.4%

E. 30.4%

(10 g NaCl/65 g solutuon) x 100 = 15.4%

7. CHEM 162-2000 FINAL EXAM

CHAPTER 11 - PROPERTIES OF SOLUTIONS SOLUTION CONCENTRATION CALCULATIONS

What is the molarity of aluminum sulfate, Al2(SO4)3 in a 0.72% by mass aqueous Al2(SO4)3 solution with a density of 1.05 g/cm3? Molar mass of Al2(SO4)3 = 342.12 g/mol.

A. 0.022 M

B. 0.013 M C. 0.078 M

D. 1.3 M E. 0.10 M

Molarity = moles/liter moles = g/MW

Let AS = Al2(SO4)3 0.72 g AS/100 g solution x (1 mol AS/342.12 g AS) x (1.05g/cm3) x (1 cm3/1mL) x (1000 mL/1 L) = 0.022 mol/L = 0.022

A

2.

CHEM 162-2000 HOURLY EXAM I CHAPTER 11 - PROPERTIES OF SOLUTIONS

SOLUTION CONCENTRATION CALCULATIONS

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Concentrated hydrofluoric acid is a 29.2 M solution of HF in water and has a density of 1.17 g/cm3. What is the mole fraction of HF in concentrated hydrofluoric acid?

A. 0.029

B. 0.473

C. 0.578

D. 0.310 E. 0.802

mole fraction = (moles of solute)/(moles of solute + moles of solvent)

Plan: mol HF/L soln → g HF/L soln → g HF/mL soln → g HF/cm3 soln → g HF/g soln → (g HF + g H2O)/g soln; g HF → mol HF; g H2O → mol H2O; mol HF/(mol HF + mol H2O) = mol fraction

(29.2 mol/L soln) x (20.01 g HF/mol) x (1 L/1000 mL) x (1 mL/1 cm3) x (1 cm3/1.17 g soln) = 0.499 g HF/g soln = 0.499 g HF + 0.501 g H2O in 1.000 g solution

mole fraction = (moles of solute)/(moles of solute + moles of solvent) mole fraction HF = (0.499 g HF/20.01gHFmol-1)/((0.499 g HF/20.01gHFmol-1) + (0.501 gH2O/18.02 gH2Omol-1)) = 0.473 = mole fraction

3. CHEM 162-2000 HOURLY EXAM I CHAPTER 11 - PROPERTIES OF SOLUTIONS

SOLUTION CONCENTRATION CALCULATIONS A sample of ethylene glycol, C2H6O2, is dissolved in water to make a solution in which the mole fraction

of ethylene glycol is 0.236. What is the molality of ethylene glycol in this solution? A. 0.66 m

B. 1.89 m C. 13.8 m

D. 17.2 m

E. 24.7 m

mole fraction = (moles of solute)/(moles of solute + moles of solvent) molality = moles/1000 g solvent

0.236 mole fraction of EG means that 0.764 is the mole fraction of H2O. H2O is the solvent.

0.764 mol H2O x (18.02 g/mol) = 13.767 g H2O (0.236 mol EG/13.767 g H2O) x 1000 g = 17.14 mol EG/1000 g H2O = molality of EG

D

15. CHEM 162-2001 HOURLY EXAM I + ANSWERS CHAPTER 11 - PROPERTIES OF SOLUTIONS

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SOLUTION CONCENTRATION CALCULATIONS A concentration of 2.0 mg solute per 10.0 kg of solution is equivalent to

A. 0.020 percent by mass

B. 2.0 ppm C. 20 ppm D. 2.0 ppb

E. 200 ppb

0.002 g/10,000 g = 0.2 g/1,000,000 = 0.2 ppm = 200 ppb

18. CHEM 162-2001 HOURLY EXAM I + ANSWERS


SOLUTION CONCENTRATION CALCULATIONS Concentrated hydrochloric acid is 37.0% HCl by mass and has a density of 1.19 g/mL.

Calculate the molarity of HCl in concentrated hydrochloric acid. A. 2.56 M

B. 12.1 M

C. 5.34 M

D. 8.75 M E. 4.38 M

HCl solution (37.0 g HCl/100 g sol’n) x 1.19 g/mL x 1000 mL/L x (1/36.5 g/mol) = 12.06 M

3. CHEM-2002 FINAL EXAM + ANSWERS

CHAPTER 11 - PROPERTIES OF SOLUTIONS SOLUTION CONCENTRATION CALCULATIONS A solution is prepared by dissolving 17.75 g of sulfuric acid, H2SO4,

(MW = 98.078) in enough water to make 100.0 mL of solution. If the density of the solution is 1.1094 g/mL, what is the mole fraction of water in the solution?

A. 0.9953 B. 0.9762

C. 0.9908 D. 0.9135

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E. 0.9662

(17.75 g/100 ml) x (1 ml/1.1094 g) = 0.1600 g H2SO4/g solution (0.1600 g/98.078 gmol-1) = 0.001631 mol H2SO4/g solution

1.0000 g solution - 0.1600 g H2SO4 = 0.8400 g H2O/g solution (0.8400 g/18.02 gmol-1) = 0.04661 mol H2O/g solution

0.04661/(0.04661 + 0.001631) = 0.9662 = mole fraction of water This problem can also be solved in a slightly different manner by handling the mass of sulfuric acid and the volume of solution separately from the beginning:

17.75 g H2SO4/100.0 mL of solution = 17.75 g H2SO4 and 100 mL of solution 17.75 g solute/98.078 gmol^-1 = 0.1810 mol H2SO4 100.0 mL solution x 1.1094 g/mL = 110.94 g solution

110.94 g solution - 17.75 g solute = 93.19 g solvent 93.19 g solvent/18.02gsolventmol^-1 = 5.171 mol solvent = 5.171 mol H2O mol fraction of H2O = mol H2O/(mol H2O + mol H2SO4)=5.171/(5.171+0.1810)=0.9662

CHEM 162-2002 HOURLY EXAM I + ANSWERS CHAPTER 11 - PROPERTIES OF SOLUTIONS

SOLUTION CONCENTRATION CALCULATIONS 7. An alloy is prepared from 1.00 mol Ag (AW= 107.9), 1.0 mol Au (AW= 197.0), and 1.0 mol Hg (AW=200.6). The mass fraction of gold in the alloy is

Mass fraction gold = mass of gold/mass of everything Mass fraction gold = 197.0/(107.9 + 197.0 + 200.6) = 0.390 A. 0.60

B. 0.51 C. 0.44

D. 0.21

E. 0.39


SOLUTION CONCENTRATION CALCULATIONS 11. An aqueous solution of ethylene glycol (HOCH2CH2OH, MW=62.07) is 50.0% water by mass and has a density of 1.06 g/mL. Find the molarity of ethylene glycol in the solution.

M = moles/L (50 g EG/100 g soln) x (1.06 g/mL) x (1000 mL/L) = 530 g/L

moles = 530 g/62.07 = 8.539; hence, 8.539 moles/L A. 29.4 M B. 8.05 M

C. 8.54 M

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D. 6.77 M E. 10.7 M

CHEM 162-2002 HOURLY EXAM I + ANSWERS

CHAPTER 11 - PROPERTIES OF SOLUTIONS SOLUTION CONCENTRATION CALCULATIONS 20. A solution is prepared from equal masses of water, acetone (C3HO, MW=58.1), and ethanol

(C2H6O, MW=46.1). The mole fraction of water in this solution is Pick any convenient mass. I picked 58.1 g because that is exactly 1.00 mol acetone.

58.1 g H2O + 58.1 g acetone + 58.1 g EtOH 3.23 mol + 1.00 mol + 1.26 mol 3.23/(3.23 + 1.00 + 1.26) = 0.588 mol

A. 0.18

B. 0.59

C. 0.50 D. 0.67

E. 0.33 CHEM 162SG-2001 HOURLY EXAM I


1 How many grams of CuSO4·5H2O (Molar mass 249.69 g/mol) should be dissolved in a 250 mL volumetric flask to make a solution of Cu2+(aq) whose concentration is 8.00 x 10-3 M?

A. 0.235 g

B. 0.499 g

C. 0.105 g

D. 0.322 g E. 0.785 g

moles = M x volume moles Cu2+ = 8.00 x 10-3 M x 0.250 L = 0.0020 moles of Cu2+

CuSO4·5H2O → Cu2+ + SO42- + 5H2O

0.0020 mol Cu2+ x (1 mol CuSO4·5H2O/1 mol Cu2+) = 0.0020 mol CuSO4·5H2O

0.0020 mol CuSO4·5H2O x 249.69 g/mol = 0.499 g CuSO4·5H2O CHEM 162SG-2001 HOURLY EXAM I


3. What is the volume percent of 10.00 g methanol (d = 0.791 g/mL) in 75.00 g ethanol (d = 0.789 g/mL)? Assume the volumes are additive.

A. 13.5% methanol B. 7.62% methanol

C. 18.9% methanol

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D. 11.7% methanol

E. 22.5% methanol CHEM 162SG-2001 HOURLY EXAM I


5. What is the molarity of the NO3- ion if its concentration is 37 ppm?

A. 5.0 x l0-4 M

B. 6.0 x 10-4 M

C. 3.9 x l0-4 M D. 7.0 x l0-4 M

E. 2.0 x l0-4 M

37g/1000000 g Assume that this is an aqueous solution, and that the density of this dilute aqueous solution is 1.00 g/mL = 1000 g/L

(37g/1000000 g) x (1000 g/L) x (1 mol/62.01 g) = 5.97 x 10-4 M

CHEM 162SG-2001 HOURLY EXAM I


6. A 4.03 M aqueous solution of the antifreeze ethylene glycol (C2H6O2) has a density of 1.045 g/cm3. What is the mass percent of ethylene glycol in this solution? (Molar mass ofC2H6O2 is 62.7g/mol.)

A. 23.9%

B. 16.7%

C. 76.0% D. 33.6%

E. 5.9% CHEM 162SG-2001 HOURLY EXAM I

CHAPTER 11 - PROPERTIES OF SOLUTIONS SOLUTION CONCENTRATIONS CALCULATIONS

11. The concentration of a solution containing 0.131 g of an unknown solid in 25.4 g of water is 0.056 molal. What is the molar mass of the unknown substance?

A. 56 g/mol

B. 72 g/mol C. 87 g/mol D. 98 g/mol

E. 92 g/mol

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25.


SOLUTION CONCENTRATIONS (PROOF) The term “proof” is defined as twice the percent by volume of ethanol in aqueous solution. (So 86 proof is 43% ethanol by volume). What is the molarity of ethanol

(C2H6O) in a 92 proof solution?

Density of ethanol = 0.800 g/mL

A. 0.92 M B. 17 M

C. 0.46 M D. 0.80 M

E. 8.0 M

MW C2H6O = 46

92 proof is 46% ethanol by volume. Hence, it is 46 mL ethanol/100 mL solution. Plan: mL ethanol/mL solution → g ethanol/mL solution → mol ethanol/mL solution

→ mol ethanol/1000 mL solution (46 mL ethanol/100 mL solution) x (0.800 g/mL) x (1 mol/46 g) x 1000 = 8.0 mol ethanol/1000 mL soln = 8.0 Methanol

7.

Chem 162-2004 Exam 1

Chapter 11 SOLUTION CONCENTRATIONS (MASS PERCENT)

What is the mass percent of CuSO4 in a solution whose density is 1.30 g/mL and whose molarity is 1.22 M in CuSO4 ?

A. 12.4%

B. 22.1% C. choose this choice if none of the others is correct. D. 31.6%

E. 15.0%

Plan: We’re starting with mole/L sol’n and want to go to g/100 g. mole/Lsol’n → mole/mLsol’n → mol/gsol’n → g/gsol’n → g/100 g sol’n

MW CuSO4 = 63.55 + 32.06 + 64.00 = 159.6 gmol-1 (1.22 mol/Lsol’n) x (1L/1000 mL) x (1mL/1.30g) x (159.6g/mol) x 100 = 14.98 mass

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percent

21. Chem 162-2004 Final Exam + Answers

Chapter 11 - Properties of Solutions

Solution Concentrations Concentrated hydrochloric acid is 37.5% HCl by mass, and has a density of 1.19 g/cm3. Calculate the molarity of the concentrated HCl.

A. 18.4 M B. 10.5 M C. 9.20 M

D. 12.2 M E. 4.60 M Molarity = moles per liter of solution

Plan: g/100 g soln → g/g soln → g/mL soln → g/L soln → mol/L soln (37.5 g HCl/100 g soln) x (1.19 g/mL) x (1000 mL/L) x (1 mol/36.46 g) = 12.24 M


CHAPTER 11 - PROPERTIES OF SOLUTIONS SOLUTION CONCENTRATIONS CALCULATIONS(MOLARITY.)

A solution of MgCl2 is prepared by adding 0.154 mol of the salt to enough water to make 500. mL of solution. A 50.0 mL sample of this solution is then diluted to 250.0 mL. The concentration of Cl- in this new solution is

A. 0.0154 M

B. 0.123 M

C. 0.308 M

D. 0.0616 M E. 0.616 M

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0.154 mol/0.500 L = 0.308 mol/l = 0.308 M MgCl2 solution Dilution of 50.0 mL to 250.0 mL is a five fold dilution.

Hence final concentration of MgCl2 is 0.308/5 = 0.0616 M MgCl2 → Mg2+ + 2Cl-

Therefore, 0.0616 M MgCl2 dissociates into 0.1232 M Cl- solution


SOLUTION CONCENTRATIONS CALCULATIONS (MASS PERCENT) Concentrated nitric acid (HNO3) is 16.0 M in nitric acid . The mass percent of nitric acid

in this solution is 70.0%. The density of the solution is

A. 1.61 g/cm3

B. 1.84 g/cm3 C. 1.07 g/cm3 D. 1.19 g/cm3

E. 1.44 g/cm3

Begin: 16 mol HNO3/1000 mL solution End: 70 g HNO3/100 g solution

16 mol HNO3/1000 mL solution x (63.02 g/mol) = 1.0083 g HNO3/mL solution Set up a proportion:

1.0083 g HNO3/mL solution = 70 g HNO3/X mL solution X mL = 69.42 mL

70 g HNO3/69.42 mL solution = 70 g HNO3/100 g solution 69.42 mL solution = 100 g solution D = 100 g/69.42 mL = 1.44 g/mL

Note to ET: There is probably a more direct way to do this.

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6. Chem 162-2005 Final Exam + Answers Chapter 11 - Properties of Solutions

Solution Concentrations calculations

What is the molality of 3.75 M H2SO4 solution that has a density of 1.230 g

mL ?

A. 1.50 m

B. 3.15 m C. 2.55 m

D. 3.75 m

E. 4.35 m

molality = moles/kg solvent molarity = mol/L solution

3.75 mol/L x (1L/1000 mL) x (1 mL/1.230 g) x (1000 g/kg) = 3.0488 mol H2SO4/kg soln

But we want mol H2SO4/kg solvent. 3.0488 mol H2SO4 x (98g/mol) = 298.78 g H2SO4 = 0.29878 kg H2SO4

1 kg soln - 0.29878 kg H2SO4 = 0.7012 kg solvent 3.0488 mol H2SO4/0.7012 kg solvent = 4.35 mol H2SO4/kg solvent = 4.35 m

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ENERGETICS OF SOLUTIONS CONCEPTS


Chapter 11 – Properties of solutions Energetics of solutions and solubility concepts

Ionic salts tend not to dissolve in non-polar solvents because

A. The enthalpy of solvation is very large.

B. The large difference between the enthalpy to expand the solvent and the enthalpy to expand the solute.

C. The enthalpy of solvation is not large enough to overcome the enthalpy to expand

the solute.

D. The solvent repels the solute. E. The enthalpy to expand the solute is too small.

A. The enthalpy of solvation is always a negative number (i.e., small), so by definition a negative number cannot be large. Perhaps that’s the wrong way to look at the word

“large”. Let’s try it from an absolute sense. The enthalpy of solvation is a combination of the enthalpy of solvent-solvent bond breaking and solute-solvent bond formation. Since solvent-solvent bond breaking is the breaking of London forces for a non-polar

solvent-nonpolar solvent bond, which is very small and endothermic, and the forming of London forces for bond formation between ions and a non-polar solvent, which is very

small and exothermic, then the enthalpy of solvation is very small. B. Although there is a large difference between the enthalpy to expand the solvent and the enthalpy to expand the solute, this difference is not relevant. What is relevant is how

much energy you get back when the solute forms a bond with the water. C. Since the reaction is the sum of the enthalpy to expand the solute (endothermic solute-

solute bond breaking) and the enthalpy of solvation (the sum of the enthalpy to expand the solvent and the enthalpy of solvent-solute bond formation) then the enthalpy of solvation has to be large enough to overcome the enthalpy of solute expansion so that the sum of

the two is negative or approximately zero. D. The solvent doesn’t repel the solute. The solvent actually attracts the solute. It just

doesn’t attract it strongly enough to make up for the endothermic enthalpies of the solute-solute bond breaking and the solvent-solvent bond breaking. E. Not true. The enthalpy to expand an ionic salt is large.

19

21 Chem 162-2006 Hourly exam I + answers Chapter 11 – Properties of solutions Energetics of solutions concepts

Which statement about solubilities in water is best?

A. Substances are more soluble at higher temperature. B. Polar substances are more soluble at higher temperatures; non-polar substances are less soluble.

C. Ionic compounds are more soluble at higher temperatures; other compounds have little or no temperature dependence of their solubilities.

D. Non-polar compounds will not dissolve at any temperature. E. Most salts are more soluble at higher temperatures, while most gases are less soluble at higher temperatures.

A. This is not a bad statement, because most substances are more soluble at higher

temperatures. However, gases are less soluble at higher temperatures. B. This is nonsense. Most substances are more soluble at higher temperatures whether they are polar or not.

C. Whether the substances are ionic or not, as long as the substances are not gases they are generally more soluble at higher temperatures.

D. Even non-polar substances will dissolve in water, but very slightly. Their solubility will increase at a higher temperature. The statement “Non-polar compounds will not dissolve at any temperature” may be considered as being acceptable if there are no better

statements. E. This is a very good general rule. Most salts are more soluble at higher temperatures,

while most gases are less soluble at higher temperatures.


ENERGETICS OF SOLUTIONS CONCEPTS

39 Which solvent, water or carbon tetrachloride, would you choose to dissolve

the following?

Water is a polar solvent and CCl4 is a non-polar solvent. Polar solutes will

dissolve in water; non-polar solutes will dissolve in CCl4.

c. CH3CCH3; CH3COCH3 is polar, so water is the solvent of choice.

║

20

O


ENERGETICS OF SOLUTIONS CONCEPTS 10. Based on enthalpy alone and ignoring the effects of disorganization, the best explanation for the observation that a nonpolar solute is more soluble in a nonpolar solvent than in a polar solvent is

based on

A. a combination of solute-solute interactions in the pure solute and solvent-solvent interactions in the pure solvent B. a combination of solvent-solvent interactions in the pure solvent and solute-solvent interactions

C. solute-solute interactions in the pure solute

D. solvent-solvent interactions in the pure solvent

E. solute-solvent interactions

Let U = solute V = solvent

NP = non-polar P = polar BB = bond-breaking

BM = bond-making Three steps: solute bond breaking; solvent bond-breaking; solute-solvent bond making.

Non-polar solute dissolving in a non-polar solvent

Steps ΔH UNP → UNPBB small + VNP → VNPBB small +

UNP + VNP → UNPVNPBM small -

Non-polar solute dissolving in a polar solvent Steps ΔH UNP → UNPBB small +

VP → VPBB large + UNP + VP → UNPVPBM small -

The major difference is the solvent bond-breaking.


CHAPTER 11 - PROPERTIES OF SOLUTIONS ENERGETICS OF SOLUTIONS CONCEPTS

21

11. Consider the following phenomena I. Contraction of scurvy by sailors

II. The cleaning action of detergents III. The “bends”

IV. The solubility of AgCl in water The ones that can be explained by the “like dissolves like” rule are

A. I and III

B. II and IV

C. I, II, and IV

D. all of them E. I, II, and III

Scurvy is caused by an insufficient quantity of vitamin C in the diet. Vitamin C is polar as is the H2O in the blood, making vitamin C highly water soluble and therefore excreted rapidly, and therefore must be

replaced regularly. In detergents, the non-polar dirt likes the non-polar long-chain of the detergent, while the polar end of the

detergent likes the polar H2O. Nitrogen is nonpolar whereas the aqueous blood system is polar. Hence, there is no like dissolves like issue. The issue with bends is related to a change in pressure.

AgCl is considered to be insoluble in water. However, a small amount of AgCl is soluble in water. To the extent that it is soluble, it is soluble because silver and chloride ions have a positive and negative

charge, while H2O has a partial positive and negative charge. This part of the question is very poor and misleading. The student would do well to ignore part IV.

4. CHEM 162-2000 HOURLY EXAM I


Which of the following would you expect to be more soluble

in water than in nonpolar solvents?

I. C2H2 II. NH4Cl III. NH3 IV. CCl4

A. I and II

B. II and III

C. II and IV D. I and III

E. III and IV

B

22

Polar molecules dissolve polar molecules; non-polar molecules dissolve non-polar molecules; i.e., like dissolves

like. C2H2 and CCl4 are non-polar molecules and therefore are soluble in non-polar solvents. NH4Cl and NH3 are polar

molecules and therefore would be expected to be soluble in

the polar molecule, water.

12.

CHEM 162-2000 HOURLY EXAM I CHAPTER 11 - PROPERTIES OF SOLUTIONS

ENERGETICS OF SOLUTIONS CONCEPTS Arrange NaF, H2O, and C2H4 in order of increasing normal boiling point (i.e., substance with the lowest boiling point listed first).

A. C2H4 < NaF < H2O

B. H2O < NaF < C2H4

C. C2H4 < H2O < NaF

D. H2O < C2H4 < NaF E. NaF < C2H4 < H2O

C 13.

CHEM 162-2000 HOURLY EXAM I CHAPTER 11 - PROPERTIES OF SOLUTIONS ENERGETICS OF SOLUTIONS CONCEPTS

Which of the following would increase the molar solubility of a gas in a liquid?

A. Decrease the temperature.

B. Decrease the pressure of the gas above the liquid.

C. Increase the surface area of the liquid. D. Decrease the volume of the liquid. E. Increase the volume of the liquid.

A

25. CHEM 162-2000 HOURLY EXAM I CHAPTER 11 - PROPERTIES OF SOLUTIONS ENERGETICS OF SOLUTIONS CONCEPTS

Which of the following statements concerning solubility is false?

C

23

A. The relative strengths of solute-solute, solvent-solvent, and solute-solvent attractions

play an important role in determining the solubility of a solute in a solvent. B. The solubility of many salts in water increases with increasing temperature.

C. A solute can be highly soluble in a solvent only if the solution process is exothermic.

D. Ionic and highly polar solutes are hydrophilic.

E. Compounds that can form hydrogen bonds are more likely to be soluble in water than similar compounds that cannot form hydrogen bonds.



ENERGETICS OF SOLUTIONS CONCEPTS Which of the following statements is false regarding solubility?

A. A saturated solution can be in dynamic equilibrium with undissolved solute. All chemical equilibria is in dynamic equilibrium. The molecules don’t sit still. They

move out of solution and back into it.

B. The solubility of a solute is independent of the nature of the solvent.

The solubility of a solute is highly dependent on the nature of the solvent. Like dissolves like.

C. Solubilities of substances vary with temperature. Generally increasing temperature results in increased solubility. D. Ionic compounds are not normally very soluble in nonpolar solvents.

Like dissolves like. E. Nonpolar compounds are generally more soluble in nonpolar solvents than in water.

Like dissolves like.

23. CHEM 162-2001 HOURLY EXAM I + ANSWERS CHAPTER 11 - PROPERTIES OF SOLUTIONS ENERGETICS OF SOLUTIONS CONCEPTS

Which of the following liquids would you expect to be largely insoluble in water? W. C2H5OH X. CCl4 Y. CS2 Z. H3C - C - CH3 (acetone)

| |

O

A. W and X only

B. X and Y only

C. Y and Z only

24

D. W and Z only E. X and Z only

X and Y are non-polar


ENERGETICS OF SOLUTIONS CONCEPTS Many ionic solids such as NaCl are highly soluble in water because of

A. weak dispersion forces between water molecules. B. strong hydrogen bonds between water molecules.

C. a strong ionic lattice in the solid.

D. ion-dipole forces between water and the ions.

E. strong O-H covalent bonds within the water molecule.


CHAPTER 11 - OF SOLUTIONS ENERGETICS OF SOLUTIONS CONCEPTS 25. The solubility of CaI2 (FW=294) in water is 209 g of CaI2 in 100 mL of aqueous solution. A

solution is prepared whose concentration is 8.1 mmol in 1.0 mL of solution. This solution is called A. saturated

B. dilute C nonconducting D unsaturated

E. supersaturated

209/294 = 0.711 mol CaI2 soluble in 100 mL 8.1 mmol in 1.0 mL = 0.8 mol in 100 mL, a supersaturated solution.



2. Many ionic solids are quite soluble in water because of Water dipoles are attracted to the cations and anions of ionic solids.

A. strong O-H covalent bonds within the water molecule

B. strong hydrogen bonds between water molecules. C. a strong ionic lattice in the solid.

25

D. weak dispersion forces between water molecules.

E. ion-dipole forces between water and the ions



16. Which statement is correct?

A. A saturated solution contains a high concentration of solute.

B. A saturated solution involves a dynamic equilibrium between dissolving and

crystallization. C. The solubility of solids always increases with temperature. D. The solubility of a gas usually increases with temperature.

E. A supersaturated solution involves a dynamic equilibrium between so lute and solution.


ENERGETICS OF SOLUTIONS CONCEPTS 17. When an ionic salt dissolves in water, the solute-solvent interaction is

A. hydrogen bonding B. London forces C. ion- ion forces

D. ion-dipole

E. dipole-dipole

CHEM 162SG-2001 HOURLY EXAM I CHAPTER 11 - PROPERTIES OF SOLUTIONS ENERGETICS OF SOLUTIONS CONCEPTS

22. Which one of the following solutes would you expect to be most soluble in water?

A. CH3OH

B. C2H6 C. CHCl3

D. CCl4

E. CH4

26

C2H6, CCl4 and CH4 are all non-polar molecules, and therefore would

not be expected to be soluble in water. CHCl3 and CH3OH are both polar molecules and are likely to form dipole-dipole bonds with water,

and therefore be soluble. However, CH3OH is also capable of strong hydrogen bonding with water, which should result in greater solubility

of CH3OH in water. Chem 162-2003 Final exam + answers


Energetics of solutions and solubility concepts (enthalpy of hydration, enthalpy of solution, etc.)

21. Which of the following gases are very soluble in water?

W. CH4 X. O2 Y. HCl Z. NH3

A. W and Z B. W and X

C. Y and Z

D. X and Y

E. X and Z Both CH4 and O2 are non-polar, i.e., they have no dipole moment. Since H2O is polar, and like

dissolves like, then CH4 and O2 will be only very slightly soluble in H2O. However, both HCl and NH3 are both polar molecules. Hence, they will form relatively strong dipole-dipole bonds with

the polar water. In addition, the HCl dissociates into ions, forming strong ion dipole bonds with H2O, while NH3 forms strong hydrogen bonds with water.

20.


ENERGETICS OF SOLUTIONS AND SOLUBILITIES

When a solid dissolves in water, heat energy is released if A. choose this choice if none of the others is correct.

B. the absolute value of the hydration energy is greater than the absolute value of

the lattice energy

C. the hydration energy is negative D. the solution is ideal

E. the lattice energy is positive The sum of the enthalpy of solute-solute bond breaking (which is the same as the lattice

energy but opposite in sign, and since the lattice energy is always negative this is always

27

positive) and the enthalpy of hydration (which is always negative) must be negative to release heat energy. Therefore, the absolute value of the hydration energy must be

greater than the absolute value of the lattice energy, making “B” the correct answer. “C” is ruled out because the positive enthalpy of solute-solute bond breaking might be

larger than the hydration energy, which will result in an endothermic reaction. “D” is ruled out, because the solution being ideal is irrelevant. I’m not certain about “E” because if the lattice energy is positive then the enthalpy of solute-solute bond breaking

will be negative. Since the enthalpy of hydration is always negative then this combination will result in a negative total enthalpy, or heat release. I guess that this is

not a good choice because the lattice energy is never positive.

11.

Chem 162-2004 Exam 1 CHAPTER 11

ENERGETICS OF SOLUTIONS AND SOLUBILITIES Rank the following compounds in order of increasing solubility in water

I. CH3CH2CH2CH3 II. CH3CH2-O-CH2CH3

III. CH3CH2CH2CH2OH IV. CH3OH

A. III<II<IV<I

B. II<III<I<IV

C. I<II<III<IV

D. choose this choice if none of the others is correct. E. II<I<IV<III

CH3CH2CH2CH3 is a hydrocarbon. Since there is virtually no difference in electronegativity between carbons and hydrogen in hydrocarbons, then all hydrocarbons

are non-polar. However, even if there was a difference in electronegativity, the VSEPR shows vectorial symmetry. The symmetry would have killed the polarity. Non-polar

molecules are not soluble in water (like dissolves like rule). CH3CH2-O-CH2CH3 contains oxygen, carbon and hydrogen. Since there is a large difference in electronegativity between oxygen and carbon or hydrogen, and since the

VSEPR geometry shows that this molecule is not symmetrical, then this molecule is polar. Furthermore, it is capable of hydrogen bonding with water. Hence, this molecule

should be soluble in water. CH3CH2CH2CH2OH contains oxygen, carbon and hydrogen. Since there is a large difference in electronegativity between oxygen and carbon or hydrogen, and since the

VSEPR geometry shows that this molecule is not symmetrical, then this molecule is polar. Furthermore, it is capable of strong hydrogen bonding with water because not

28

only can the hydrogen from water hydrogen bond with the oxygen on this molecule, but the hydrogen on this molecule can hydrogen bond with the oxygen in water. This is

different than the situation with CH3CH2-O-CH2CH3 in which the hydrogen in the water can hydrogen bond with it, but it does not have the type of hydrogen required to

bond with the oxygen in water. Hence, CH3CH2CH2CH2OH should be more soluble in water than CH3CH2-O-CH2CH3. CH3OH has similar behavior to CH3CH2CH2CH2OH. It is capable of strong hydrogen

bonding with water because not only can the hydrogen from water hydrogen bond with the oxygen on this molecule, but the hydrogen on this molecule can hydrogen bond

with the oxygen in water. However, whereas CH3CH2CH2CH2OH contains a backbone of four carbon atoms, making that part of the molecule non-polar, CH3OH contains only a small non-polar segment. This makes CH3OH a more polar molecule than

CH3CH2CH2CH2OH, and therefore more soluble in water.

This ranking is from most insoluble to most soluble. I < II < III < IV

29

Chem 162-2004 Final Exam + Answers Chapter 11 - Properties of Solutions

Energetics of Solutions Many ionic solids such as NaCl are highly soluble in water because of: A. a strong ionic lattice in the solid

B. strong O-H covalent bonds within the water molecule C. weak dispersion forces between water molecules D. strong hydrogen bonds between water molecules E. ion-dipole forces between water and the ions

(A) A strong ionic lattice, in itself, decreases the solubility in water because the ion-ion bonds are difficult to break. (B) The bonding within the water molecule is irrelevant. (C) Although the dispersion forces between water molecules are weak, that isn’t relevant.

(D) Strong hydrogen bonds between water molecules, by themselves, decrease solubility of NaCl because the hydrogen bonds are difficult to break. (E) The energy released due to the strong ion-dipole forces between water and the ions is the main driving force for the high solubility of

NaCl in water.


CHAPTER 11 - PROPERTIES OF SOLUTIONS ENERGETICS OF SOLUTIONS CONCEPTS (ENTHALPY OF HYDRATION,

ENTHALPY OF SOLUTION, ETC.) AND SOLUBILITIES The reason that heptane and octane (both nonpolar liquids) are miscible is

A. Absence of significantly favorable solute-solute interactions B. Favorable solute-solvent interactions

C. Disorganization increases in the solution D. Choose this choice if none of the other choices are correct

E. Choose this choice if two of the other choices are correct

A. Absence of significantly favorable solute-solute interactions is saying the same thing

as the presence of significantly unfavorable solute-solute interactions. The presence of significantly unfavorable solute-solute interactions never helps miscibility (i.e.,

dissolution); if anything, if the solute-solute interaction is so unfavorable, i.e., if the ΔH is so highly positive, it will tend to inhibit miscibility. B. Favorable solute-solvent interactions is a factor. The word favorable means that the

30

negative enthalpy of the solute-solvent interaction should be large enough to either (a) compensate for the sum of the positive enthalpies of the solute-solute bond breaking and

the solvent-solvent bond breaking, or (b) come close to compensating for the sum of the positive enthalpies of the solute-solute bond breaking and the solvent-solvent bond

breaking. C. Disorganization increasing upon dissolution is a factor. (Another term for this is an increase in entropy; I describe this effect as a decrease in confinement.)

20.


CHAPTER 11 - PROPERTIES OF SOLUTIONS ENERGETICS OF SOLUTIONS CONCEPTS (ENTHALPY OF HYDRATION,

ENTHALPY OF SOLUTION, ETC.) AND SOLUBILITIES The lattice energy of NaBr is –728 kJ/mol and its enthalpy of solution in water is –18 kJ/mol. Its enthalpy of hydration is

A. + 746 kJ/mol

B. –746 kJ/mol

C. –36 kJ/mol

D. + 710 kJ/mol E. –710 kJ/mol ΔHsoln = ΔHsolute-solutebondbreaking + ΔHsolvent-solventbondbreaking + ΔHsolute-solventbondforming ΔHhydr = ΔHsolvent-solventbondbreaking + ΔHsolute-solventbondforming ΔHsoln = ΔHsolute-solutebondbreaking + ΔHhydr If lattice energy = -728 kJ/mol, then ΔHsolute-solutebondbreaking = +728 kJ/mol. ΔHsoln = ΔHsolute-solutebondbreaking + ΔHhydr

-18 kJ/mol = +728 kJ/mol + ΔHhydr ΔHhydr = -746 kJ/mol

31


Energetics of Solutions Concepts

When NaCl dissolves in water, H of solution is approximately zero. This observation

implies which of the following: A. the entropy of the dissolved ions equals than the entropy of the ionic solid.

B. the lattice energy of NaCl is relatively weak. C. the vapor pressure of the solution is equal to outside pressure.

D. the reaction is nonspontaneous.

E. the hydration energy of the ions is numerically equal to the lattice energy of NaCl

A. H of the reaction is not related to the entropy of the reaction.

B. H consists of the energy necessary to break the Na+Cl- bonds (lattice energy) the energy to break the H2O-H2O bonds and the energy to form the H2O-ion bonds. So, in the

absence of the other contributing values, we can’t isolate the lattice energy and say that that is relatively weak. Hence, “B” is false. Another way to answer this question is to realize that lattice energy is always/generally a huge value, so the lattice energy is not

likely to be relatively weak. C. Dissolution of a salt in water has nothing to do with its vapor pressure. The relationship

of vapor pressure and outside pressure is for the boiling point.

D. Since H is approximately zero, and S will have some positive value (going from one

mole of a salt to two moles of a solution is always positive, and going from solid to aqueous is always positive), then the reaction is spontaneous.

E. Since H is the sum of the hydration energy (the sum of the energy needed to break the

solvent bonds and the energy released due to bonding of the solvent and the salt ions) and

the lattice energy, and H is approximately zero, then the hydration energy must be

approximately equal to the lattice energy.

32

ENERGETICS OF SOLUTIONS CALCULATIONS

33

17 Chem 162-2006 Hourly exam I + answers Chapter 11 – Properties of solutions Energetics of solutions and solubility calculations

Which compound is most likely to dissolve in cyclohexane (C6H12)? A. NaCl

B. POCl2 C. SO2 D. SF6

E. CH3OH Like dissolves like. Cyclohexane is a hydrocarbon and is therefore non-polar. Hence,

only non-polar compounds will dissolve in cyclohexane. A. NaCl is a salt. Due to the large difference in electronegativity between Na and Cl the Na is positively charged and the Cl is negatively charged. Hence it is polar and should

not dissolve in cyclohexane. B. POCl2 has atoms varying in electronegativity and is not symmetrical. Therefore it is

polar and should not dissolve in cyclohexane. . . :O:

│• P

: Cl: :Cl: . . . .

C. SO2 SO2 has atoms varying in electronegativity, and is not symmetrical. Therefore, it is polar, and should not be soluble in cyclohexane.

. . : O = S ― O : Bent; 120o; polar . . . . . .

D. SF6 Although SF6 has polar bonds, it is symmetrical, and is therefore not polar, and should dissolve in cyclohexane.

. . . . : F : : F : . . . .

: F ― S ― F : . . . .

: F : : F : . . . . E. The C-O-H bonds in CH3OH are polar, and the C-O-H moiety is not symmetrical.

Therefore, it is a polar molecule and should not be soluble in cyclohexane. H

| . . H ― C ― O ― H | . .

H

34


ENERGETICS OF SOLUTIONS CALCULATIONS

35 The lattice energy of KCl is -715 kJ/mol, and the enthalpy of hydration is -684 kJ/mol. Calculate the enthalpy of solution per mole of solid KCl. Describe the process to which

this enthalpy change applies.

ΔHsoln = ΔHsolutebondbreaking + ΔHsolventbondbreaking + ΔHsolute-solventbondformation ΔHhydration = ΔHsolventbondbreaking + ΔHsolute-solventbondformation

Therefore, ΔHsoln = ΔHsolutebondbreaking + ΔHhydration ΔHsolutebondbreaking = the negative of the lattice energy = +715 kJ/mol

ΔHsoln = ΔHsolutebondbreaking + ΔHhydration ΔHsoln = +715 -684 = 31 kJ/mol

The enthalpy of solution change applies to a process in which the KCl solid salt is

separated into K+ + Cl

-, the H2O solvent is dissociated into individual H2O molecules, and

then the individual K+ & Cl

- ions bond with the individual H2O molecules.

ΔHsolution

K-Cl → K+ + Cl

- ΔHsolutebondbreaking = +715 kJ/mol

H2O --- H2O → H2O + H2O ΔHsolventbondbreaking ΔHhydration = -684 kJ/mol K

+ + Cl

- + H2O + H2O → H2O---K

+ + Cl

----H2O ΔHsolute-solventbondformation

K-Cl + H2O --- H2O → H2O---K+ + Cl

----H2O ΔHsoln=ΔHubb+ΔHvbb+ΔHuvbf=ΔHubb+ΔHhydr

35

VAPOR PRESSURE OF SOLUTIONS CONCEPTS


Chapter 11 – Properties of solutions Vapor pressure of solutions concepts (Henry’s law)

The concentration of helium in water is found to be 1.83×10-6g/L. If the partial pressure of He is 1.2×10-3atm, what is the henry’s law constant kH for helium?

A. 3.8×10-4 M/atm

B. 1.5×10-3M/atm

C. 1.8×109M/atm D. 4.6×10-3M/atm E. 2.5×10-6M/atm

Henry’s Law: MA = kPA

1.83 x 10-6 g/L x (1 mol/4.00g) = k x 1.2 x 10-3 atm kH = 3.81 x 10-4 M/atm


CHAPTER 11- PROPERTIES OF SOLUTIONS

VAPOR PRESSURE OF SOLUTIONS CONCEPTS Which one of the following statements is false?

A. The vapor pressure of a liquid depends on the temperature. B. One can boil a liquid above, below or at its normal boiling point.

C. At the same temperature, a volatile substance has a higher vapor pressure than a nonvolatile substance.

D. The vapor pressure of a pure liquid depends on the volume of liquid present.

E. The vapor pressure of a liquid remains the same if the surface area of evaporation becomes larger.

A. True. As the temperature is raised more energy is provided to break

intermolecular forces in the liquid, thereby allowing molecules to escape the liquid

phase and go into the vapor phase. B. True. The “normal” boiling point is defined as the boiling point at 1 atm. pressure.

Above 1 atm pressure the liquid will boil at a higher temperature. Below 1 atm pressure the liquid will boil at a lower temperature.

C. True. The difference between a volatile substance and a nonvolatile substance is

the strength of the intermolecular forces. A volatile substance has weaker

D

36

intermolecular forces, so more molecules can escape into the vapor phase, resulting in a higher vapor pressure.

D. False. The vapor pressure of a pure substance is independent of the volume. Hence, it makes no difference whether 100 mL or 100 L of H2O is brought to a

boil. At 100oC both boil, and have vapor pressures of 760 mm. Increasing the volume would tend to increase the surface area, but increasing the surface area doesn’t result in an increase in vapor pressure (See “E” below.)

E. True. One may argue that a substance having greater surface area would have a greater number of molecules in the vapor phase. Whereas this is true, pressure is

force per unit area, so increasing the surface area doesn’t affect the force, and therefore doesn’t affect the pressure.



VAPOR PRESSURE OF SOLUTIONS CONCEPTS

The normal boiling point of acetone is 56.2C and that of benzene is 80.1C. At 33C,

which of the following would you predict concerning the vapor pressures of liquid acetone and liquid benzene?

A. Acetone would have the greater vapor pressure.

B. Benzene would have the greater vapor pressure. C. The vapor pressures would be equal. D. The vapor pressure of benzene would be 760 Torr.

E. The relative vapor pressures are not predictable from the data given.

A


CHAPTER 11 - PROPERTIES OF SOLUTIONS VAPOR PRESSURE OF SOLUTIONS CONCEPTS

The vapor pressure of a pure liquid depends on which of the following? W. the strength of intermolecular attractions in the liquid

X. the volume of liquid present Y. the surface area of the liquid

Z. temperature

A. W, X, Y, and Z B. X, Y, and Z only

C. W, Y, and Z only

D. W and Z only

D

37

E. Z only

16.


Chapter 11 VAPOR PRESSURE OF SOLUTIONS - RAOULT’S LAW

A solution of a salt in water sits in an open beaker. Assuming constant temperature, what will happen to the vapor pressure of the solution with time?

A. decreases over time

B. need to know the temperature and pressure to answer this question

C. need to know which salt is in solution to answer this question D. increases over time

E. stays the same over time Over time some water will evaporate, decreasing the concentration of the water in the

solution, consequently reducing the vapor pressure. It is important to note that as a result of the beaker being open, the evaporated water does not build up in the gas phase.

It leaves the system, so the solution continues to become less and less concentrated with respect to water and the vapor pressure goes down according to Raoult’s law.

38


Vapor Pressure of Solutions Concepts

When tap water is heated in an uncovered pot, bubbles may be seen long before the water boils. This observation can be explained by:

A. the temperature dependence of the mole fraction in Raoult’s Law B. the dependence of the boiling point on external pressure.

C. changes in atmospheric pressure. D. the decomposition of a small amount of water

E. the decreased solubility of gases at higher temperatures.

A. There is no temperature dependence of mole fraction. B. Although the boiling point is dependent on external pressure, this is irrelevant in this case because the water is heated in an uncovered pot, so the external pressure doesn’t

change. C. In an uncovered pot the atmospheric pressure doesn’t change.

D. The implication is that water decomposes into hydrogen and oxygen. Although hydrogen and oxygen spontaneously form water, water doesn’t spontaneously form hydrogen and oxygen. Therefore, the bubbles aren’t hydrogen and oxygen.

E. The solubility of gases in water is inversely related to the temperature. Hence, as the temperature rises, the bubbles formed is air, which is not soluble in water at elevated

temperatures.


39

16 Chem 162-2006 Hourly exam I + answers Chapter 11 – Properties of solutions Vapor pressure of solutions calculations (Raoult’s Law)

What is the vapor pressure of a solution formed by mixing 192.0 g of C6H12O6 in 250.00

mL of water at 25 °C? The density of water is 1.00g/cm3 and its vapor pressure at 25 °C is 23.8 Torr.

A. 13.5 Torr

B. 24.2 Torr C. 21.4 Torr

D. 23.1 Torr

E. 22.1 Torr

Raoult’s law: Psoln = iXsolventP

osolvent

192.0 g C6H12O6 x (1 mol/180.18g) = 1.0656 mol C6H12O6 250 mL H2O x 1g/mL x (1 mol/18.02 g) = 13.87 mol H2O

Psoln = ((1 x 13.87)/((1 x 13.87) + (1 x 1.0656))) x 23.8 = 22.1 torr



45 The solubility of nitrogen in water is 8.21 x 10-4

mol/L at 0oC when the N2

pressure above water is 0.790 atm. Calculate the Henry’s law constant for N2

in units of Latm/mol for Henry’s law in the form P = kC, where C is the gas

concentration in mol/L. Calculate the solubility of N2 in water when the partial

pressure of nitrogen above water is 1.10 atm at 0oC.

Henry’s law: PA = kCA

0.790 atm = k x 8.21 x 10-4

mol/L

k = 962 atmL/mol

1.10 atm = 962 atmL/mol x X

X = 0.00114 mol/L

CHEM 162-2003 1.5 WEEK RECITATION

40



47 A solution is prepared by mixing 50.0 g glucose (C6H12O6) with 600.0 g

water. What is the vapor pressure of this solution at 25oC? (At 25

oC the vapor

pressure of pure water is 23.8 torr. Glucose is a nonelectrolyte.)

Psoln = XsolventPosolvent

X = molesH2O/(molesH2O + molesglucose)

X = (600 g x (1 mol/18 g))/ ((600 g x (1 mol/18 g)) + (50.0 g x (1 mol/180 g)))

= 0.9917

Psoln = 0.9917 x 23.8 torr

Psoln = 23.6 torr



51 Pentane (C5H12) and hexane (C6H14) form an ideal solution. At 25oC the

vapor pressures of pentane and hexane are 511 and 150 torr, respectively. A

solution is prepared by mixing 25 mL pentane (density, 0.63 g/mL) with 45 mL

hexane (density, 0.66 g/mL).

a. What is the vapor pressure of the resulting solution?

PTotal = PA + PB = XAPo

A + XBPo

B

moles of pentane = 25 mL x (0.63 g/mL) x (1 mol/72 g) = 0.2188 mol

moles of hexane = 45 mL x (0.66 g/mL) x (1 mol/86 g) = 0.3453 mol

Xpentane = 0.2188/(0.2188 + 0.3453) = 0.3879

Xhexane = 0.3453/(0.2188 + 0.3453) = 0.6121

PP = XPPo

P

PP = 0.3879 x 511 torr = 198 torr

41

PH = XHPo

H

PH = 0.6121 x 150 torr = 92 torr

PTotal = PP + PH = 198 + 92 torr = 290 torr



49 At a certain temperature, the vapor pressure of pure benzene (C6H6) is

0.930 atm. A solution was prepared by dissolving 10.0 g of a nondissociating,

nonvolatile solute in 78.11 g of benzene at that temperature. The vapor

pressure of the solution was found to be 0.900 atm. Assuming the solution

behaves ideally, determine the molar mass of the solute.

Psoln = XsolventPosolvent

0.900 atm = Xbenzene x 0.930 atm

Xbenzene = 0.9677

0.9677 = molbenzene/(molbenzene + molsolute)

0.9677 = (78.11 g x (1 mol/78.11 g))/((78.11 g x (1 mol/78.11 g)) + (10.0 g x

(1 mol/X g)))

X = molecular weight of solute = 300 g


CHAPTER 11 - PROPERTIES OF SOLUTIONS VAPOR PRESSURE OF SOLUTIONS CALCULATIONS

20. A solution containing 0.40 mol of benzene (C6H6) and 0.80 mol of toluene (C7H8) has a total vapor pressure of 200 mm Hg. If the vapor pressure of pure toluene is 120 mm Hg at this temperature, then the vapor pressure of pure benzene at this temperature is

A. choose this choice if none of the others are correct

B. 80 mm Hg C. 160 mm Hg D. 260 mm Hg

E. 360 mm Hg

42

Ptotal = XBzP

oBz + XToPo

To

200 = ((0.4/1.2) x Po) + ((0.8/1.2) x 120) Po

Bz = 360



24. The partial pressure of O2 in the atmosphere at sea level is 0.12 atm. A saturated solution of O2 at 273 K under these conditions has a mole fraction of O2 of 5.4 x 10-6. The Henry’s Law constant at 273 K for O2 is

A. 9.0 x 10-5 atm-1

B. 4.5 x 10-5 atm-1

C. 2.2 x 104 atm-1

D. 1.1 x 104 atm-1 E. choose this choice if none of the others is correct

Henry’s Law: Henry’s Law can be written several ways. One way is:

XA = kPA (5.4 x 10-6) = k x 0.12

k = 4.5 x 10-5 CHEM 162-2003 HOURLY EXAM 1 + ANSWERS


5. Alcohol A and alcohol B both have the same molar mass. A solution of the two that is 25% alcohol A by mass has a total vapor pressure of 0.0900 atm. A solution of the two that is 75% alcohol A by mass has a total vapor pressure of 0.123 atm. The vapor pressures of the two pure

alcohols at this temperature are

A. 0.140 atm for alcohol A and 0.0735 atm for alcohol B

B. cannot be calculated without knowing the density

C. 0.0735 atm for alcohol A and 0.140 atm for alcohol B D. cannot be calculated without knowing the actual molecular formula

E. cannot be calculated without knowing the temperature Simultaneous equations:

PTotal = PA + PB PTotal = XAPo

A + XBPoB

0.0900 = 0.25X + 0.75Y 0.123 = 0.75X + 0.25Y

43

Multiply first equation by 3:

0.2700 = 0.75X + 2.25Y 0.123 = 0.75X + 0.25Y

Subtract: 0.147 = 2.00Y

Y = 0.073

Now multiply the bottom equation by 3: 0.0900 = 0.25X + 0.75Y 0.369 = 2.25X + 0.75Y

Subtract:

0.279 = 2X 0.140 = X

19. CHEM 162-2000 HOURLY EXAM I CHAPTER 11 - PROPERTIES OF SOLUTIONS VAPOR PRESSURE OF SOLUTIONS CALCULATIONS

A solution is prepared by dissolving the nonvolatile nonelectrolyte solute glucose (C6H12O6, molar mass 180 g/mol) in water. If the mass percent of glucose in this solution

is 9.70%, calculate the vapor pressure of the solution at 22C. The vapor pressure of pure

water at 22C is 19.83 Torr.

A. 19.83 Torr

B. 19.81 Torr C. 19.73 Torr

D. 19.62 Torr

E. 19.53 Torr

Psoln = iXsolventPo

solvent Psoln = (9.70/180)/((9.70 g/180) + (90.30/18.02))/19.83

Psoln = ((90.30/18.02)/((90.30/18.02) + (9.70 g/180))) x 19.83 P = 19.62 torr

D

7. CHEM 162-2001 HOURLY EXAM I + ANSWERS CHAPTER 11 - PROPERTIES OF SOLUTIONS VAPOR PRESSURE OF SOLUTIONS CALCULATIONS Suppose that at 298 K, a maximum of 46 mg of O2 can dissolve in 100 g of water under a

pressure of 10.0 atm of O2. What is the maximum mass of O2 in air that can dissolve in

100 g of water at 298 K? Assume that the pressure of air is 1.00 atm and the mole fraction

44

of O2 in air is 0.21.

A. 1.8 mg

B. 4.4 mg

C. 0.97 mg

D. 3.7 mg E. 2.2 mg

Even though it is the convention to have concentration as mole fraction, concentration is not required to be in mole fraction, as long as what the concentration is is defined. In this

case the concentration is mg of solute/100 g solvent. Csolute = kP We are given the concentration of solute (oxygen) and its headspace pressure, so we can

find the constant, k. Then we’ll use k and the new oxygen pressure to find the new concentration of the solute, oxygen.

(46 mg/100 g solvent) = k x 10.0 k = 0.046 C = 0.046 x P

Pressure of gases is directly related to moles of gases. Combination gas law: P1V1/n1T1 = P2V2/n2T2

At constant V and T, P1/n1 = P2/n2 P1/P2 = n1/n2 Although I don’t know how to demonstrate this algebraically, P1 is to the total pressure as

moles1 is to the total moles. P1/PTotal = n1/nTotal.

P1/1.00 atm = 0.21/1 P1 = 0.21 atm C = 0.046 x 0.21 = 0.00966 mg O2/g solvent

0.00966 x 100 = 0.966 mg O2/100 g solvent


CHAPTER 11 - PROPERTIES OF SOLUTIONS VAPOR PRESSURE OF SOLUTIONS CALCULATIONS How many grams of naphthalene (C10H8, molar mass 128.2 g/mol) must be dissolved in

2.00 moles of benzene, C6H6, in order to lower the vapor pressure of benzene by 10.0%?

A. 19.2 g

B. 28.5 g

C. 32.4 g

45

D. 38.2 g E. 48.6 g

Raoult’s Law: Psolv = Xsolv x Po

solv

0.9 = Xsolv x 1.0 Xsolv = 0.9 0.9 = 2.00/2.00 + X

X = 0.22 moles napthalene 0.22 mol napthalene = g/128.2

g = 28.2 g napthalene



Benzene and toluene form an ideal solution. At 25C, the vapor pressure of pure benzene

is 95.1 Torr and the vapor pressure of pure toluene is 28.4 Torr. If a solution contains 1.0 mol of benzene and 2.0 mol of toluene, what is the mole fraction of benzene in the

vapor of this solution at 25C?

A. 0.25 B. 0.31

C. 0.51

D. 0.63

E. 0.82 Psolv = Xsolv x Po

solv

Benzene Toluene Pure 95.1 torr Pure 28.4 torr

1.0 mol 2.0 mol Psolv = (1.0/3.0) x 95.1 2/3 x 28.4 = 31.70 vapor pressure = 18.93 vapor pressure

31.70/(31.70 + 18.93) = 0.626

6. CHEM-2002 FINAL EXAM + ANSWERS CHAPTER 11 - PROPERTIES OF SOLUTIONS


The vapor pressure of pure benzene (C6H6) and of pure toluene (C7H8)

at 25C are 85.1 mmHg and 28.4 mmHg, respectively. A solution is prepared with a mole fraction of toluene of 0.750. Assuming that the

46

solution is an ideal solution, calculate the total vapor pressure above

the solution?

A. 19.3 mmHg B. 94.6 mmHg

C. 75.1 mmHg

D. 83.2 mmHg

E. 42.6 mmHg

Raoult’s Law: Psolv = Xsolv x Posolv

Pbenzene = (1-0.750) x (85.1) = 21.275

Ptoluene = (0.750 x 28.4) = 21.300 Ptotal = Pbenzene + Ptoluene

= 42.575

PTotal = XBzPoBz + XToP

oTo

PTotal = ((1-0.750) x 85.1) + (0.750 x 28.4)

PTotal = 42.575

7.

CHEM-2002 FINAL EXAM + ANSWERS CHAPTER 11 - PROPERTIES OF SOLUTIONS VAPOR PRESSURE OF SOLUTIONS CALCULATIONS

How many moles of a solid nonelectrolyte must be added to 44.5

moles of water at 70C to produce a solution with a vapor pressure of

215.0 mmHg? The vapor pressure of pure water at 70C is 233.7 mmHg.

A. 3.87 moles

B. 2.53 moles

C. 5.10 moles D. 4.93 moles

47

E. 1.27 moles

Raoult’s Law: Psoln = Xsolv x Posolv

215.0 = X x 233.7

X = 0.920 = mole fraction of water

1-0.920 = 0.080 = mole fraction of solid nonelectrolyte Call solid nonelectrolyte “SNE”

0.080 = (YSNE)/(YSNE + 44.5H2O)

YSNE = 3.87


CHAPTER 11 - PROPERTIES OF SOLUTIONS VAPOR PRESSURE OF SOLUTIONS CALCULATIONS 6. What is the mole fraction of benzene in the vapor above a solution prepared from benzene and

toluene, in which the mole fraction of benzene in the solution is 0.333? The temperature is 298 K.; and at this temperature the vapor pressure of pure benzene is 0.125 atm and of pure toluene is 0.0374 atm.

Psolv = Xsolv x Posolv

PB = 0.333 x 0.125 = 0.04163 PT = 0.667 x 0.0374 = 0.02495

XB = 0.04163/(0.04163 + 0.02495) = 0.625 A. 0.333

B. 0.667 C. 0.0416 D. 0.375

E. 0.625

CHEM 162-2002 HOURLY EXAM I + ANSWERS CHAPTER 11 - PROPERTIES OF SOLUTIONS VAPOR PRESSURE OF SOLUTIONS CALCULATIONS

48

16. At a pressure of O2 of 1.00 atm, the mole fraction of dissolved O2 in water is 1.6 x 10-5 at 333 K. What mass of O2 (MW=32.0) can dissolve in 250 g of water under 1.00 atm of air pressure at 333 K.? The

mole percent of O2 in air is 21%. Henry’s law: Cgas = kP

There are several ways to solve this problem. Some important factors: (1) “C” of the gas in water can be expressed in numerous ways, including mole

fraction. (2) Note that the initial condition is O2 at 1.00 atm of pressure above the water. However, the requested information does not pertain to O2 being at 1.00 atm of pressure above the water, but air being

at 1.00 atm of pressure above the water. Since air is 21% O2, then the question is asking about the solubility of O2 in water when the O2 is at 0.21 atm pressure above the water. Plan: C and P are given; find k. Then multiply k by the new P to find the new C. From the new C find

the mass of O2 dissolved in 250 g water.

O2 1 atm mole fraction O2 = 1.6 x 10-5

?g O2 in 250 g H2O under 0.21 atm pressure

1.6 x 10-5 = k x 1 atm k = 1.6 x 10-5 At 0.21 atm: C = 1.6 x 10-5 x 0.21

C = 3.36 x 10-6 = mole fraction of O2 in H2O If mole fraction of O2 is 3.36 x 10-6, then mole fraction of H2O is 1 - (2.36 x 10-6) = 0.999997 = 1

Therefore there is 3.36 x 10-6 mol O2 dissolved in 1 mol H2O. 3.36 x 10-6 mol O2 x 32 g/mol = 1.0752 x 10-4 g O2 1 mol H2O x 18.02 g/mol = 18.02 g H2O

1.0752 x 10-4 g O2/18.02 g H2O x 250 g H2O = 1.492 x 10-3 g O2 0.00149 g O2 dissolved in 250 g H2O at 0.21 atm

Another way of solving this problem is as follows: Henry’s law states that the concentration of a gas in a solution is directly proportional to the pressure of the gas above the

solution.

P = kC P = 1 atm O2 C, expressed as a mole fraction, = 1.6 x 10-5

1 = k x 1.6 x 10-5 k = 6.25 x 104

New conditions: 1 atm of air = 0.21 atm O2.

0.21 = (6.25 x 104) x mole fraction of O2 0.21 = (6.25 x 104) x ((g/32)/((g/32) + (250/18)))

g = 0.00149

49

A. 1.5x10-3 g

B. 9.8x10-5 g C. 55.5 g

D. 4.0x10-6 g E. 5.1x10-4 g

CHEM 162SG-2001 HOURLY EXAM I CHAPTER 11 - PROPERTIES OF SOLUTIONS VAPOR PRESSURE OF SOLUTIONS CALCULATIONS

8. The vapor pressure of pure benzene at 25°C is 95.1 mm Hg and that of toluene 28.4 mm Hg. If 0.40 mole of benzene and 0.60 mol of toluene are mixed in a container, what

is the mole fraction of benzene in the vapor phase?

A. 0.21 B. 0.60

C. 0.78

D. 0.69

E. 0.82


CHAPTER 11 - PROPERTIES OF SOLUTIONS VAPOR PRESSURE OF SOLUTIONS CALCULATIONS 9. If 27.0 g of acetylene, C2H2, dissolve in 1.00 L of acetone at 1.0 atmosphere pressure,

what is the mass of acetylene dissolved in 5.00 L of water if the pressure of acetylene is increased to 10.0 atm?

A. 1.35 x 103 g

B. 2.70 x 103 g

C. 2.70 x 102 g D. 4.63 x 103 g

Chem 162-2003 Final exam + answers


Vapor pressure of solutions calculations (Henry’s L=law; Raoult’s law) 6. What mass of benzene, (C6H6, molar mass 78.1), must be added to 2.00 moles of carbon

tetrachloride, CCl4, in order to reduce the vapor pressure of carbon tetrachloride by 20%.

A. 52.0 g B. 45.2 g

50

C. 39.0 g

D. 27.5 g E. 17.3 g

Raoult’s law: Psoln = XsolventP

osolvent

If CCl4 was pure, that is, if the mole fraction was 1, then the vapor pressure of the solution would be equal to the vapor pressure of the pure CCl4 solvent. If the vapor pressure is reduced by 20%, then:

Psoln = 0.80 x Posolvent

XB + XC = 1 XB + 0.80 = 1

XB = 0.20 If 2 moles corresponds to a mole fraction of 0.80, then 0.5 moles corresponds to a mole fraction of 0.20. mass = 0.5 mol benzene x 78.1 g/mol = 39.05 g benzene.

2.

Chem 162-2004 Exam 1 Chapter 11 VAPOR PRESSURE OF SOLUTIONS - HENRY’S LAW

What is the amount of CO2 released when a bottle containing 0.50 L of club soda bottled at a pressure of 1.1 atm of CO2 is opened at 298 K. The Henry’s Law constant for CO2

at 298 K = 6.1 x 10-4 atm-1, the partial pressure of CO2 in the atmosphere is 3.0 x 10-4

atm, and the density of the solution is 1.0 g/mL.

A. 0.088 mol CO2

B. 0.0041 mol CO2 C. 0.0069 mol CO2 D. 0.0095 mol CO2

E. 0.019 mol CO2

Club soda 0.50 L

1.1 atm CO2 = P1

T = 298 K k = 6.1 x 10-4atm-1

3.0 x 10-4 atm = P2

X = kPa The concentration must be in mole fraction, because only mole fraction would account

for the units: No units = atm-1 x atm

(molCO2)/(molCO2 + mol H2O) = kPa After bottling:

51

Y/(Y + (500g/18.02gmol-1)) = 6.1 x 10-4 x 1.1 atm Y = 0.0186 mol CO2 dissolved after bottling

After opening: Y/(Y + (500/18.02)) = 6.1 x 10-4 x 3.0 x 10-4

Y = 5.08 x 10-6 mol CO2 dissolved Mol lost = 0.0186 - (5.08 x 10-6) = 0.0186 mol

6.


VAPOR PRESSURE OF SOLUTIONS - HENRY’S LAW A solution of benzene and toluene at 20ºC has a mole fraction of benzene = 0.53. What

is the mole fraction of benzene in the vapor above the solution. Vapor pressures at 20ºC benzene = 76 torr; toluene = 21 torr

A. 0.28 B. 0.47

C. 0.50 D. 0.53

E. 0.80

PBzsoln = XBzP

oBz

PBzsoln = 0.53 x 76 PBzsoln = 40.28 torr

PTosoln = XToPo

To PTosoln = XToPo

To = (1 - 0.53) x 21

PTosoln = 9.87 torr

Moles of benzene and toluene in the vapor is directly proportional to the pressure of benzene and toluene in the vapor.

XBz = 40.28/(40.28 + 9.87) = 0.80

52


Vapor Pressure of Solutions

The vapor pressure of pure water at 80 o C is 355 Torr. What is the vapor pressure of a

solution prepared by adding 0.300 mol of sucrose to 200 g of water at 80 o C?

A. 335 Torr

B. 316 Torr C. 325 Torr

D. 346 Torr

E. 353 Torr

Raoult’s Law: Psoln = XsolventP

osolvent

mol H2O = 200g/18.02gmol-1 = 11.099 mol

Psoln = ((11.099 mol H2O)/((11.099 mol H2O + 0.300 molsucros))) x 355 Torr = 345.7 Torr

11.


CHAPTER 11 - PROPERTIES OF SOLUTIONS VAPOR PRESSURE OF SOLUTIONS CALC’N- RAOULT’S LAW

At 313 K the vapor pressure of pure methanol is 303 mmHg and the vapor pressure of pure propanol is 44.6 mmHg. A solution of these two substances is prepared and it has a vapor pressure of 174 mmHg at 313 K. The mole fraction of methanol in the solution is

A. 0.151 B. 0.574

C. 0.501

D. 0.871 E. 0.750

Psoln = iXsolventMPo

solventM + iXsolventPPosolventP

XM + XP = 1 Let X = mole fraction of methanol, and 1-X = mole fraction of propanol. 174 = (X x 303) + ((1 - X) x 44.6)

X = Xmethanol = 0.501


VAPOR PRESSURE OF SOLUTIONS CALC’N- RAOULT’S LAW The mass of naphthalene (C10H8) that must be dissolved in 175 g of heptane (C7H16 ) to

53

lower its vapor pressure by 25.0% is

A. 701 g B. 525 g

C. 175 g

D. 74.6 g

E. 101 g Naphthalene is non-volatile.

Psoln = iXsolventPo

solvent XA = nA/(nA + nB)

nheptane = 175g x (1mol/100.23g) = 1.746 mol heptane According to Raoult’s law, when the mole fraction of heptane equals 1, and the pure solvent has a certain vapor pressure, then the pressure above the solution will be equal to

the vapor pressure of the pure solvent. Therefore, if the vapor pressure above the solution is only 75% of what it was originally (i.e., it was lowered by 25%), then the

mole fraction of heptane must equal 0.75. 0.75 = (1.746 mol heptane)/(1.746 mol heptane + mol napthalene) 0.75 = (1.746 mol heptane)/(1.746 mol heptane + (gnapthalene/MWnapthalene))

0.75 = (1.746 mol heptane)/(1.746 mol heptane + (gnapthalene/128.18)) X = 74.60 g napthalene


VAPOR PRESSURE OF SOLUTIONS CALC’N - HENRY’S LAW What is the amount of dissolved CO2 in a sealed bottle containing 0.50 L of club soda bottled at a pressure of 1.1 atm of CO2 at 298 K. The Henry’s Law constant for CO2 at

298 K = 6.1 x 10-4 atm-1, and the density of the solution is 1.0 g/mL.

A. 0.0069 mol CO2 B. 0.088 mol CO2

C. 0.00067 mol CO2 D. 0.0095 mol CO2

E. 0.019 mol CO2

Henry’s Law: XA=kPA mol CO2/(mol CO2 + mol H2O) = 6.1 x 10-4 x 1.1 Y/(Y + (0.50L x 1000 g/L x (1 mol/18.02 g))) = (6.1 x 10-4 x 1.1)

Y = 0.0186 mol CO2

54

Based on the very small constant, the above calculation assumed that the moles of CO2 dissolved in the club soda was insignificant, i.e., that all of the 0.50 L of club soda is

water. Proof that this assumption is valid is that 0.0186 mol CO2 is only 0.07% of the 0.50 L of H2O. However, it wasn’t necessary to make this assumption. In the

calculation the moles of CO2 could have been subtracted from the moles of H2O : ((Y)/(Y + (0.50 x 1000 x (1/18.02) – Y))) = (6.1 x 10-4 x 1.1) Y = 0.0186 mol CO2


Chapter 11 - Properties of Solutions Vapor Pressure of Solutions Calculations What amount of urea must be dissolved in 5.00 moles of water in order to reduce the vapor pressure of water by 10.0%?

A. 0.748 moles B. 0.280 moles C. 0.315 moles D. 0.556 moles

E. 0.772 moles Raoult’s law: Psoln = iXsolventPosolvent Since urea is an organic molecule with a van’t Hoff factor of 1, then we can use a mole fraction formula instead of a particle fraction

formula. Raoult’s law: Psoln = XsolventPosolvent Since we are reducing the vapor pressure (i.e., Psoln) by 10%, and Posolvent doesn’t change, then the mole fraction of the so lvent must

be reduced by 10%. Initial X = molH2O/(molH2O + molurea) = 5/(5+0) = 1 Mole fraction reduced by 10%: (0.9 x 1) = 5/(5 + X)

X = 0.556 mol of urea


55

49 CHEM 162-2006 FINAL EXAM + ANSWERS CHAPTER 11 – PROPERTIES OF SOLUTIONS VAPOR PRESSURE OF SOLUTIONS CONCEPTS (RAOULT’S LAW)

Which one of the following statements is true? X. A negative deviation from Raoult’s law means that the observed pressure will be

lower than that predicted by Raoult’s law. Y. Negative deviations from Raoult’s law are expected when ΔHsol’n is positive

(endothermic).

Z. Strong interactions between solute and solvent result in negative deviations from Raoult’s law.

A. X only B. X and Y only C. Y and Z only

D. Z only E. X and Z only

A good example of a negative deviation from Raoult’s law is a solution of acetone and chloroform. Acetone molecules have a dipole-dipole bond formation with each other.

This is a moderate interaction. Chloroform molecules have a dipole-dipole bond formation with each other. This is a moderate interaction. But due to the strong inductive

effect of three chlorine atoms in chloroform, combined with the presence of an oxygen atom in acetone, the chloroform-acetone bond formation is that of hydrogen bonding, which is a particularly strong dipole-dipole bond formation. (This is a deviation from the

rule that only H-F, H-O, or H-N molecules are capable of hydrogen bonding, but three chlorines have a strong inductive effect.) This results in a lower vapor pressure than

expected by the relatively weak acetone-acetone or chloroform-chloroform bond formation. The increased strength of the bond results in so much heat being given off (exothermic reaction), that mixing acetone and chloroform can self- ignite.

X. True. A negative deviation from Raoult’s law means that the observed pressure will

be lower than predicted by Raoult’s law. Y. False. Negative deviations from Raoult’s law are expected when the bond between the solute and the solvent are stronger than the solute-solute or solvent-solvent bond,

resulting in an exothermic reaction, not an endothermic reaction. Z. True. Strong interactions between solute and solvent result in lower vapor pressure

than expected, which is a negative deviation from Raoult’s law.

56

7 Chem 162-2006 Hourly exam I + answers Chapter 11 – Properties of solutions Colligative properties (Osmotic pressure)

Reverse osmosis is

A. Allowing solute through a membrane instead of solvent. B. The diffusion of a non-aqueous solvent across a semipermiable membrane. C. The calculation of molar mass from osmotic pressure.

D. Using external pressure to force solvent out of a solution across a membrane, increasing the concentration of the solution.

E. Allowing both solvent and solute across a semipermiable membrane. A. Reverse osmosis is allowing a solvent through a membrane instead of a solute, not the

other way around. B. Reverse osmosis is not the diffusion of a non-aqueous solvent across a semipermeable

membrane, but the diffusion of any solvent across a semipermeable membrane. C. Reverse osmosis is a procedure, not a calculation. Also, it has nothing to do with the determination of molar mass.

D. Reverse osmosis is using external pressure to force solvent out of a solution across a membrane, increasing the concentration of the solution within the membrane.

It is the opposite procedure of osmosis. Osmosis is a procedure whereby the natural osmotic pressure of the solution forces solvent out of a solution across a membrane, into the salt solution, decreasing the concentration of the salt solution within the

membrane. E. Reverse osmosis is only allowing the solvent to cross a semipermeable membrane.

57

2 Chem 162-2006 Hourly exam I + answers Chapter 11 – Properties of solutions Colligative Properties Concepts

Boiling Point

Colligative properties (freezing point depression, boiling point elevation, etc.) depend on A. The properties of the solvent and the solute. B. Only the properties of the solute.

C. Only the properties of the solvent.

D. The properties of the solvent and the number of solute particles.

E. The properties of the solvent and the number of solvent particles.

Use boiling point elevation as an example. ∆T = Tf – Ti = -Kfimsolute Kf is a property of the solvent. That is, it is different for each solvent. “i” is a measure of

the number of particles of solute. “i x m” is the number of particles of solute and is independent of the properties of the solute. For example, if i x m has the same value for

NaCl and urea (e.g., 0.05m NaCl and 0.10m urea), the effect on boiling point constant is the same, regardless of the fact that the properties of NaCl and urea are vastly different.

Answer: The properties of the solvent and the number of solute particles.

58

13 Chem 162-2006 Hourly exam I + answers Chapter 11 – Properties of solutions Colligative properties concepts (Osmotic pressure)

Why is osmotic pressure more commonly used to measure molar mass than freezing point depression?

A. The molar mass has no effect on freezing point.

B. The van’t Hoff factor i doesn’t affect osmotic pressure, but does affect freezing point.

C. A small amount of solute will produce a measurable osmotic pressure, but only a

negligible change in freezing point.

D. The osmotic pressure experiment is less expensive. E. It takes too long to freeze solutions.

A. The molar mass has an effect on freezing point since freezing point includes a molality term, and molality includes a mole term, and moles = g/molar mass, i.e.,

moles includes a molar mass term. B. Both osmotic pressure and freezing point are colligative properties. Therefore, they both use the van’t Hoff factor i.

C. A small amount of solute will produce a measurable osmotic pressure but only a negligible change in freezing point.

Let’s use 0.01m, a small value of m, so that M and m are approximately the same. Freezing point depression: ∆T = Tf – Ti = -Kfimsolute ∆T = Tf – 273 = -1.86 x 1 x 0.01 = 272.98K (i.e., 0.02o lower than the normal boiling

point) Osmotic pressure: п = iMsoluteRT

п = 1 x 0.01 x 0.08205 x 273 = 0.224 atm = 170 mm (i.e., easily measurable) D. Neither osmotic pressure nor freezing point experiments are costly, so cost is not a relevant factor.

E. Freezing solutions does not take long.

59

20 Chem 162-2006 Hourly exam I + answers Chapter 11 – Properties of solutions Colligative properties concepts

Which of the following is a non electrolyte in aqueous solution?

A. HCl

B. CH3OCH3

C. NH3 D. NaCl E. NaOH

A. HCl is a strong acid. Strong acids are strong electrolytes.

B. CH3OCH3 is an organic molecule. Organic molecules, other than organic acids, don’t dissociate in aqueous solution. Therefore, CH3OCH3 is a non-electrolyte. C. NH3 + H2O NH4

+ + OH-

NH3 is a weak base and is therefore a weak electrolyte because only approximately 1% of NH3 gets converted into its salt.

D. NaCl is a salt, and totally dissociates into its ions. It is therefore a strong electrolyte. E. NaOH is a strong base, totally dissociating into its ions, and is therefore a strong electrolyte.

25 Chem 162-2006 Hourly exam I + answers Chapter 11 – Properties of solutions Colligative properties concepts

Freezing points

Which aqueous solution will have the lowest freezing point? A. 0.0100m LiF B. 0.0100m Li2CO3

C. 0.0080m Fe(NO3)3

D. 0.0200m C6H12O6

E. 0.0080m KOH

Tf – Ti = -Kfimsolute The substance with the lowest freezing point is the one with the largest im. LiF Li+ + F- 2 x 0.0100m = 0.0200

Li2CO3 2Li+ + CO32- 3 x 0.0100m = 0.0300

Fe(NO3)3 Fe3+ + 3NO3- 4 x 0.0080m = 0.0320

C6H12O6 C6H12O6 1 x 0.0200m = 0.0200 KOH K+ + OH- 2 x 0.0080m = 0.0160

An aqueous solution of Fe(NO3)3 has the lowest freezing point.


60

CHAPTER 11 - PROPERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE

CHANGES) CONCEPTS 16. The units of Kf for water are

A. K mol (solute)/L (water) B. mol (solute)/K kg (water)

C. K mol (solute)/kg (water) D. mol (solute)/L (water) K

E. K kg (water)/mol (solute)

ΔT = Tf - Ti = -Kfm Kf = -(Tf - Ti)/m

Kf = K/(mol solute/kg solvent) Kf = K kg/mol solute



COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE CHANGES) CONCEPTS A 0.150 M aqueous solution of KBr is diluted by adding water. Which of the

following properties of the solution will increase when the water is added? X. the boiling point of the solution

Y. the freezing point of the solution Z. the value of the van't Hoff factor i

A. Y and Z only*

B. X and Y only C. X and Z only. D. Y only

E. X, Y, and Z

Addition of water will dilute the solution. X. The boiling point of the solution will increase if the solution becomes more concentrated, but will decrease (i.e., become closer to pure water) if the solution

becomes more diluted. Y. The freezing point of the solution will decrease if the solution becomes more

concentrated, but will increase (i.e., become closer to pure water), if the solution becomes more diluted. Z. In concentrated solutions KBr doesn’t completely dissociate. Hence, its van’t Hoff

factor is not 2.0. Perhaps it is 1.9. However, as the solution becomes more dilute the KBr will dissociate further, approaching 2.0. That is, the van’t Hoff factor will

increase with dilution.

A

61

*Correctness of answer needs to be verified.




Arrange LiF, C2H6, and CH3OH in order of increasing normal boiling point (i.e., substance with the lowest boiling point listed first).

A. C2H6 < LiF < CH3OH B. CH3OH < LiF < C2H6

C. CH3OH < C2H4 < LiF D. C2H6 < CH3OH < LiF E. LiF < C2H6 < CH3OH

D

62


CHAPTER 11 - PROPERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC

PRESSURE CHANGES) CONCEPTS Which of the following aqueous solutions should have the highest normal boiling point?

A. 1.0 m sodium chloride B. 1.0 m sucrose

C. 1.0 m hydrochloric acid

D. 1.0 m aluminum nitrate

E. 1.0 m calcium chloride The boiling point is related to vapor pressure; the lower the vapor pressure the higher the

normal boiling point. The vapor pressure is governed by Raoult’s law. Psoln = vapor pressure of the solution.

Psoln = iXsolventPo

solvent

iXsolvent = insolvent/(insolvent + insolute)

iXsolvent is the particle fraction of the solvent.

“i” is the van’t Hoff factor Since the solvent is water in all cases, then the Po

solvent will be the same in all cases, and

therefore will not be a factor in solving this problem. Therefore, the relative vapor pressures of the solutions are equal to the relative mole fractions of the solvents. Psoln = insolvent/(insolvent + insolute)

“i” for H2O = 1 Since the moles of solvent are the same in all cases, let’s arbitrarily assign it a value of “1”.

A. NaCl → Na+ + Cl- i = 2 Psoln = (1 x 1)/((1x1) + (2x1)) = 0.33 B. Sucrose → Sucrose i = 1

Psoln = (1 x 1)/((1x1) + (1x1)) = 0.50 C. HCl → H+ + Cl- i = 2

Psoln = (1 x 1)/((1x1) + (2x1)) = 0.33 D. Al(NO3)3 → Al3+ + 3NO3

- i = 4 Psoln = (1 x 1)/((1x1) + (4x1)) = 0.20

E. CaCl2 → Ca2+ + 2Cl- i = 3 Psoln = (1 x 1)/((1x1) + (3x1)) = 0.25

Al(NO3)3 has the lowest vapor pressure and therefore has the highest boiling point.

D

63



If water is added to an aqueous NaCl solution, making the solution more dilute, then A. the boiling point of the solution will increase.

B. the freezing point of the solution will decrease. C. the vapor pressure of the solution will decrease.

D. the osmotic pressure of the solution will increase.

E. the value of the van't Hoff factor i will increase.

E


CHAPTER 11 - PROPERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE CHANGES) CONCEPTS

Which of the following solutions will have the largest value of the van't Hoff factor i?

FeCl3 ----> Fe3+

+ 3 Cl-; i.e., 4 particles

BaCl2 ----> Ba2+

+ 2 Cl-; i.e., 3 particles

C6H12O6 ----> NR; i.e., 1 particle KCl ----> K

+ + Cl

-; i.e., 2 particles

HCl ----> H+ + Cl

-; i.e., 2 particles

A. 0.01 m FeCl3

B. 0.01 m BaCl2

C. 0.01 m sucrose. D. 0.01 m KCl

E. 0.01 m HCl

The van’t Hoff factor is the number of particles that a

molecule will dissociate into in solution.

64




Fractional distillation is a method used to separate two liquids based on differences in their

A. boiling point.

B. freezing point. C. osmotic pressure.

D. solubility in water. E. electrical conductivity.


CHAPTER 11 - PROPERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE CHANGES) CONCEPTS

Which one of the following aqueous ionic solutions would have a maximum (ideal) i value of 4, where “i” refers to the van’t Hoff factor?

A. CaCl2 B. C6H12O6

C. Fe2(SO4)3 D. CH3COOH

E. Na3PO4

9. CHEM 162-2001 HOURLY EXAM I + ANSWERS CHAPTER 11 - PROPERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC

PRESSURE CHANGES) CONCEPTS Which colligative property would be most useful for determining the molar mass of a

protein (large molar mass)?

65

A. osmotic pressure

B. freezing point depression C. boiling point elevation D. vapor pressure lowering

E. All colligative properties would be equally useful.

15.

CHEM-2002 FINAL EXAM + ANSWERS CHAPTER 11 - PROPERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC

PRESSURE CHANGES) CONCEPTS Given that each of the following aqueous solutions is at the same temperature, which

has the largest osmotic pressure? (Assume ideal behavior)

A. 0.0030 M lithium phosphate, Li3PO4. 4 particles x 0.0030 = 0.0120

B. 0.0060 M magnesium chloride, MgCl2. 3 particles x 0.0060 = 0.0180

C. 0.0020 M aluminum sulfate, Al2(SO4)3. 5 particles x 0.0020 = 0.0100 D. 0.0070 M ammonium nitrate, NH4NO3. 2 particles x 0.0070 = 0.0140

E. 0.014 M CH3OH, (methanol). 1 particle x 0.014 = 0.0140 πV = inRT

π = (inRT/V) R = 0.08205 for all solutions, so R is the same for all solutions.

T is the same for all solutions. Since the concentrations are given in molarity, which is moles/L, consider the volume as being the same for all solutions.

Hence, we can use π = in to determine relative values of osmotic pressure. “i” is the van’t Hoff factor, which is used in all colligative property formulas.

A. i x n = 4 particles x 0.0030 moles = 0.0120 B. i x n = 3 particles x 0.0060 moles = 0.0180 C. i x n = 5 particles x 0.0020 moles = 0.0100

D. i x n = 2 particles x 0.0070 moles = 0.0140 E. i x n = 1 particle x 0.014 moles = 0.0140

B has the largest value of ixn. Therefore, B has the largest osmotic pressure.

66



21. The units of Kb are (T2 - T1) = Kb x m (T2 - T1) = Kb x (mol/kg)

K = Kb x (mol/kg) Kb = (K x kg)/mol

A. K kg/mol

B. m/K

C. mol/K D. K/M E. it has no units



19. Which statement is not true regarding colligative properties?

A. The magnitude depends on the concentration. B. The magnitude depends on whether the solute is an electrolyte or not.

C. The magnitude depends on the identity of the solute.

D. Raoult's Law describes the vapor pressure above a solution.

E. Since the vapor pressure of the solvent is lowered by a nonvolatile solute, the boiling point of the solution is higher.


COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE CHANGES) CONCEPTS 21. Which solution has the highest boiling point?

A. 0.1 m urea B. 0.06 m HCl C. 0.05 m CaCl2

D. 0.04 m (NH4)3PO4

E. 0.07 m NaCl


67

CHAPTER 11 - PROPERTIES OF SOLUTIONS VAN’T HOFF FACTOR CONCEPT

The value of i decreases with increasing concentration of a salt because of

A. ion pair formation in solution

B. the increase in the density of the solution C. repulsions between ions of like charge D. increasing dissociation of the salt

E. precipitation occuring

i is the number of particles that a salt separates into. If a salt, such as NaCl, totally separates, than i will be 2.00. If the NaCl doesn’t separate into ions at all, then i will be 1.00. If NaCl is only partially separated into its ions, that is, if there are many ion pairs,

then i will be somewhere between 1.00 and 2.00.

When the solution is very dilute, the H2O bonding with the ions competes with the binding of the ions to each other, so that the ions are highly solvent separated, and the value of i is high (2.00 in the case of NaCl). However, when the salt concentration

increases, i.e., when there is less water to compete with the binding of the ions to each other, the binding of the ions to each other is high, resulting in the formation of many

ion pairs and a corresponding low value of i (e.g., 1.50 in the case of NaCl).

68

COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING

POINT, OSMOTIC PRESSURE CHANGES) CALCULATIONS

48 CHEM 162-2006 FINAL EXAM + ANSWERS CHAPTER 11 – PROPERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES (FREEZING POINT) CALCULATIONS

What is the expected freezing point of 0.0022m MgCl2(aq)? Kf for water = 1.86 °C kg/mol

A. -0.0040 °C B. -0.0080 °C C. -0.010 °C

D. -0.012 °C

E. -0.020 °C

MgCl2 → Mg2+ + 2Cl-

i = 3 ∆T = Tf – Ti = -Kfimsolute (X – 0) = -1.86 x 3 x 0.0022 = -0.0123oC

69

20 CHEM 162-2006 FINAL EXAM + ANSWERS CHAPTER 11 – PROPERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES (OSMOTIC PRESSURE) CALCULATIONS

What is the osmotic pressure of 0.500M HOCN at 25.0 °C? Ka= 3.5×10-4 Hint: i is not 1

A. 1272 atm B. 32.78 atm

C. 12.55 atm

D. 0.323 atm E. 12.23 atm

HOCN + H2O H3O+ + OCN-

HOCN + H2O H3O+ + OCN-

Initial 0.500 0 0

Change -X +X +X

Equilibrium 0.500-X +X +X

X2/(0.500-X) = (3.5 x 10-4) X2/(0.500) = (3.5 x 10-4) X = 0.0132

Use simpler numbers: 1.00 M HOCN dissociated into 0.0264 M H3O+ and 0.0264 M OCN-

If the HOCN didn’t dissociate at all, i = 1. If it dissociated completely into 1M H3O+ and 1M OCN-, then i = 2. If it dissociated 50% into 0.500M H3O+ and 0.500M OCN-, then i = 1.5

Formula: Add: 1 + dissociation = 1 + 0.0264 = i i = 1.0264

πV = inRT π = (inRT)/V π = (iMRT)

π = (1.0264 x 0.500 x 0.08205 x 298) = 12.55 atm

70

9 Chem 162-2006 Hourly exam I + answers Chapter 11 – Properties of solutions Colligative properties calculations

Boiling point

What is the boiling point of a solution made by adding 10.51g of P4 to 150.0g of CHCl3? Kb of CHCl3 = 3.63 °C kg/mol. Tb of CHCl3 = 61.2 °C.

A. 2.1 °C

B. 63.3 °C

C. 45.3 °C

D. 50.7 °C E. 59.2 °C

Boiling point elevation: Tf – Ti = Kbimsolute 10.51gP4 x (1 mol/123.88 g) = 0.08484 mol P4

0.08484 mol P4/0.1500 kg CHCl3 = 0.5656m P4 Tf – 61.2 = 3.63 x 1 x 0.5656

Tf = 63.25oC


Chapter 11 – Properties of solutions Colligative properties calculations Osmotic pressure

What is the osmotic pressure of a solution made by dissolving 18.00g of Na3PO4 in

enough water to make 250.00mL of solution at 20.0 °C?

A. 42.2 atm

B. 10.6 atm C. 96.2 atm D. 73.0 atm

E. 0.44 atm

MW Na3PO4 = (22.99x3) + 30.97 + (16.00 x 4) = 163.94 g/mol Na3PO4 3Na+ + PO4

3- i = 4

пV = inRT пV = i(g/MW)RT

п = (i(g/MW)RT)/V п = (4 x (18.00/163.94) x 0.08205 x 293)/0.250 п = 42.23 atm

71

18 Chem 162-2006 Hourly exam I + answers Chapter 11 – Properties of solutions Colligative properties calculations

Osmotic pressure

A new compound produces an osmotic pressure of 0.0964 atm when 3.15 g of the compound is dissolved in enough water to make 150.0 mL at 25 °C. What is the molecular weight of the compound?

A. 447g/mol B. 45,300 g/mol

C. 49.5 g/mol D. 1690 g/mol

E. 5330 g/mol

Osmotic pressure: пV = inRT

пV = i(g/MW)RT п = (i(g/MW)RT)/V

Assume i = 1. 0.0964 = (1 x (3.15/MW) x 0.08205 x 298)/0.1500 MW = 5326 g/mol


COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE CHANGES) CALCULATIONS

59 A solution is prepared by dissolving 4.9 g sucrose (C12H22O11) in 175 g

water. Calculate the boiling point of this solution. Sucrose is a nonelectrolyte.

Boiling-point elevation: ΔT = Tf - Ti = Kbmsolute

m = moles solute/1000 g solvent

m = ((4.9 g x (1 mole/342 g))/175 g) x (1000 g/kg) = 0.0819

Tf - 100.00 = 0.51 x 0.0819

Tf = 100.04oC

CHEM 162-2003 1.5 WEEK RECITATION

72

CHAPTER 11 - PROPERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE


73 modified

Calculate the boiling point of an aqueous 1 molal sucrose solution.

ΔT = Tf - Ti = Kbmsolute

Sucrose → Sucrose

Tf - 100.00 = 0.51 x 1

Tf = 100.51oC

Calculate the boiling point of an aqueous 1 molal sodium chloride solution.


NaCl → Na+ + Cl

-

Tf - 100.00 = 0.51 x (2 x 1)

Tf = 101.02oC

Calculate the boiling point of an aqueous 1 molal Na3PO4 solution.


Na3PO4 → 3Na+ + PO4

3-

Tf - 100.00 = 0.51 x (4 x 1)

Tf = 102.04oC

CHEM 162-2003 1.5 WEEK RECITATION CHAPTER 11 - PROPERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE


63 (modif.) Calculate the freezing point of an antifreeze solution that is 50.0%

by mass of ethylene glycol (HOCH2CH2OH) in water. Ethylene glycol is a

nonelectrolyte.

Freezing-point depression: ΔT = Tf - Ti = -Kfmsolute

m = moles solute/1000 g solvent

m = ((50 g x (1 mole/62 g))/50 g) x (1000 g/kg) = 16.13

Tf - 0.00 = -(1.86 x 16.13)

73

Tf = -30.00oC



69 If the human eye has an osmotic pressure of 8.00 atm at 25oC, what

concentration of solute particles in water will provide an isotonic eyedrop

solution (a solution with equal osmotic pressure)?

Osmotic pressure: πV = nRT

π = (n/V)RT = MRT

8 atm = M x 0.08205 latmdeg-1

mol-1

x 298.15 K

M = 0.327


COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE CHANGES) CALCULATIONS 2. A solution is prepared from a complex inorganic salt of molar mass 317 by dissolving 4.76 g of

the salt in 100 g of water. The boiling point of the solution is found to be 100.31oC and the boiling point elevation constant of water is 0.51 (in the usual units). This complex salt in solution

A. dissociates into 2 ions

B. dissociates into 4 ions

C. dissociates into 3 ions

D. remains an undissociated molecule E. forms dimers

(Tf - Ti) = Kbim (100.31 - 100.00) = 0.51 x i x ((4.76 g/317 g mol-1)/100 g) x 1000 g/kg

i = 4.05


74


6. A 53.5 mL-volume of a solution of a nonelectrolyte of molecular weight 425 in benzene has an osmotic pressure of 50.6 atm at 300 K. The mass of the nonelectrolyte in the solution is

A. 46.7 g

B. 4.25 g

C. 5.89 g D. 0.0158 g

E. 3.84 g πV = inRT

50.6 x 0.0535 = 1 x (g/(425 g mol-1)) x 0.08205 x 300 g = 46.7



15. The difference between the boiling point and the freezing point of water is 100oC. Antifreeze, actually ethylene glycol (MW 62.07 g), is added to water to raise its boiling point and lower its freezing point. (The Kb and Kf of water are 0.51 and 1.86 respectively) The difference between

the boiling point and freezing point of a solution that is 50% ethylene glycol and 50% water by mass is

A. 151.1oC

B. 138.2oC

C. 129.9oC

D. choose this choice if none of the others is correct E. 108.2oC

BP: Tf - Ti = Kbim Tf - 100 = 0.51 x ((50/62.07)/50) x 1000

Tf = 108.2oC FP: Tf - Ti = -Kbim

Tf - 0oC = -1.86 x ((50/62.07)/50) x 1000

Tf = -29.95oC

108.2oC - (-29.95oC) = 138.2oC

75



COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE CHANGES) CALCULATIONS Which of the following aqueous solutions should have the lowest normal freezing

point?

A. 1.0 m potassium fluoride

B. 1.0 m sodium phosphate

C. 1.0 m ammonia D. 1.0 m sucrose E. 1.0 m barium chloride

KF → K+ + F- i = 2 i x m = 2.0

Na3PO4 → 3Na+ + PO43- i = 4 i x m = 4.0

NH3 → NH3 i = 1 i x m = 1.0 sucrose → sucrose i = 1 i x m = 1.0

BaCl2 → Ba2+ + 2Cl- i = 3 i x m = 3.0 Since the freezing point depression is directly related to the size of the particle

molality, then Na3PO4 would be expected to have the lowest freezing point.

B

11. CHEM 162-2000 FINAL EXAM CHAPTER 11 - PROPERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE CHANGES) CALCULATIONS

When 0.64 g of the heart stimulant epinephrine was dissolved in 36.00 g of benzene,

the normal freezing point of the solution was found to be 5.03C. What is the

approximate molar mass of epinephrine? For benzene, the normal freezing point is

5.53C and the molal freezing point constant is 5.12C kg mol-1.

A. 390 g/mol

B. 970 g/mol C. 510 g/mol D. 180 g/mol

E. 96 g/mol

D

76



PRESSURE CHANGES) CALCULATIONS

The osmotic pressure of a solution of 2.33 g of a protein in 250 cm3 of solution at a temperature of 298 K is 0.0147 atm. What is the molar mass of this protein?

A. 6.57 x 104 g/mol

B. 8.73 x 103 g/mol

C. 1.55 x 104 g/mol

D. 3.92 x 104 g/mol

E. 3.88 x 103 g/mol

C


COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE CHANGES) CALCULATIONS Assuming complete dissociation and ideal behavior, what mass of KNO3 would have to be

added to 100.00 g H2O in order to lower the freezing point by 0.500C? (The freezing point

constant kf is 1.86C kg/mol for water and the molar mass of KNO3 is 101.1 g/mol.)

A. 3.19 g

B. 2.72 g C. 1.98 g

D. 1.60 g

E. 1.36 g

E


COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE CHANGES) CALCULATIONS A 0.10 g sample of the protein myoglobin was dissolved in enough water to make 100.0

mL of solution. The osmotic pressure of the solution was found to be 1.08 Torr at 20C. Calculate the approximate molar mass of myoglobin.

πV = nRT

πV = gRT/MW ((1.08/760) x 0.1) = (0.10 x 0.08205 x 298.15)/MW MW = 17214

A. 35,000 g/mol

77

B. 9,000 g/mol C. 23,000 g/mol

D. 64,000 g/mol

E. 17,000 g/mol



PRESSURE CHANGES) CALCULATIONS A 1.09 g sample of substance X is dissolved in 26.8 g of benzene. The freezing point of

the solution is 3.76C. The freezing point of pure benzene is 5.53C. Kf = 5.12 C kg/mol

for benzene. What is the molar mass of X?

A. 118 g

mol

B. 67.2 g

mol

C. 153 g

mol

D. 88.5 g

mol

E. 87 g

mol

ΔTf = -Kf x m -(5.53 - 3.76) = -5.12 x m 0.3457 = m

0.3457 mol/1000 g benzene = X mol/26.8 g benzene X = 0.00926 mol of substance

0.00926 = 1.09/MW MW = 117.7

16. CHEM-2002 FINAL EXAM + ANSWERS CHAPTER 11 - PROPERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC

PRESSURE CHANGES) CALCULATIONS

The normal freezing point of benzene is 5.53C and the normal boiling point of benzene

is 80.10C. What is the boiling point of a solution in which benzene is the solvent,

given that the freezing point of the solution is 1.00C? For benzene, Kf = 5.12C m-1

and Kb = 2.53C m-1.

78

Typical problem in which there is a common unknown in two equations; you solve for the unknown in one equation, substitute it into the second equation, and then solve the

second equation.

Tsol’n - Tsolvent = -Kfm (1.00 - 5.53) = -5.12 x m m = 0.8848

Tsol’n - Tsolvent = -Kbm

T - 80.10 = 2.53 x 0.8848 T = 82.34oC

A. 80.93C

B. 86.78C

C. 78.25C

D. 82.34C

E. 83.45C



The freezing point of ethanol, C2H6O is –114.6C. The molal freezing point depression

constant for ethanol is 2.00Cm-1. What is the freezing point of a solution prepared with

50.0 g of glycerol, C3H8O3, dissolved in 200.0 g of ethanol? C3H8O3 (MW = 92.093)

∆Tf = Tfsolution - Tfsolvent = -Kf x molality

molality of glycerol Molality = moles of solute in 1000 g solvent 50.0 g glycerol in 200.0 g ethanol = 250.0 g glycerol in 1000 g ethanol

250.0 g/92.093 gmol-1 = 2.715 m (X - -114.6) = -(2.00 x 2.715)

X = -120.03oC

A. -5.440C

B. -105.2C

C. -109.2C

D. -120.0C

79

E. -128.2C


COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE CHANGES) CALCULATIONS 5. Given that the boiling point of pure water is 373.15 K, that Kb for water is 0.512 in the usual units,

and assuming complete dissociation of KI (FW= 166.0); the mass of KI that must be dissolved in 100 g of water to raise its boiling point to 374.20 K is

∆t = i x KB x m (374.20 - 373.15) = 2 x 0.512 x m m = 1.025 moles/1000 g = 0.1025 moles/100 g

g = 0.1025 x 166.0 = 17.0 g KI in 100 g

A. 17.0 g

B. 0.205 g C. 36.9 g

D. 34.1 g E. 68.1 g

CHEM 162-2002 HOURLY EXAM I + ANSWERS CHAPTER 11 - PROPERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE

CHANGES) CALCULATIONS 10. What is the osmotic pressure of a solution prepared from 9.389 g of MgS04 (FW=120.4) and enough

water to make 780 mL of solution at 1.0°C? For this magnesium sulfate solution, i is l.21. πV = inRT n = 9.389/120.4 = 0.07798

π x 0.787 = 1.21 x 0.07798 x 0.08205 x 274.15 π = 2.69 atm

A. 2.42 atm

B. 2.72 atm

C. 0.00821 atm D. 22.4 atm E. 2.25 atm

CHEM 162-2002 HOURLY EXAM I + ANSWERS CHAPTER 11 - PROPERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE


80

15. A 2.26-g sample of an unknown compound is dissolved in 25.00 mL benzene (d = 0.874 g/mL). The solution freezes at 277.40 K. Pure benzene freezes at 278.68 K. and its Kf is 5.12 (in the usual units).

What is the molar mass of the unknown?

∆t = -kf x m (277.40 - 278.68) = -5.12 x m m = 0.250

m = moles/1000 g solvent m = (g/MW)/1000 g solvent

25.00 mL solvent x (0.874 g/mL) = 21.85 g solvent 0.250 = ((2.26/MW)/21.85) x 1000 = moles/1000 g solvent MW = 413.7

A. 36.2 g/mol

B. 265 g/mol C. 252 g/mol

D. 414 g/mol

E. 473 g/mol

CHEM 162SG-2001 HOURLY EXAM I CHAPTER 11 - PROPERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE

CHANGES) CALCULATIONS 23. An aqueous solution is prepared by dissolving 2.05 g of hemoglobin in 100.0 mL of

solution. The solution has an osmotic pressure of 5.69 mm Hg at 298 K. What is the molar mass of hemoglobin?

PV = inRT ((5.69/760) x 0.100) = (1 x (2.05 g/MW) x 0.08205 x 298)

MW = 6695 g

A. 6.7 x 104 g/mol

B. 3.3 x l03 g/mol C. 3.3 x 104 g/mol

D. 3.8 x 103 g/mol E. 5.6 x 102 g/mol



81

24. A 1.0 m aqueous solution of glucose boils at 100.51°C. What is the theoretical boiling point elevation of 1.0 m aqueous solution of aluminum nitrate Al(NO3)3(aq)?

A. 0.51°C

B. 1.02°C C. 1.53°C D. 2.55°C

E. 2.04°C


COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE CHANGES) CALCULATIONS 25. Find the mass of the nonvolatile solute naphthalene, C10H8, which must be added to

200.0 g benzene to produce a boiling point elevation of 1.0 K. (kb for benzene is 2.53 K. kg mol-1; molar mass of naphthalene is 128 g/mol).

A. 10.1 g

B. 22.4 g

C. 50.6 g D. 18.3 g E. 30.8 g


Chapter 11 - Properties of Solutions Colligative properties calculations (Boiling point elev., freezing point elev., osmotic pressure changes, van’t Hoff factor)

11. When 0.200 mole of substance X is dissolved in 500 grams of water, the freezing point of the resultant solution is 270.92 K. Pure water freezes at 273.15 K and has K f = 1.86 (in the usual units). Substance X could be which one of the following choices?

A. CaCl2

B. Mg3(PO4)2

C. C6H12O6 D. NaCl

E. Na3PO4 ΔT = Tf - Ti = -Kfimsolute

(270.92 - 273.15) = -1.86 x i x (0.200 mol/0.500 kg) i = 2.997

A. CaCl2 CaCl2 Ca2+ + 2Cl- Van’t Hoff factor = 3

B. Mg3(PO4)2 Mg3(PO4)2 3Mg2+ + 2(PO4)3-

2 Van’t Hoff factor = 5

C. C6H12O6 C6H12O6 NR Van’t Hoff factor = 1

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D. NaCl NaCl Na+ + Cl- Van’t Hoff factor = 2

E. Na3PO4 Na3PO4 3Na+ + (PO4)3- Van’t Hoff factor = 4



Colligative properties calculations (Boiling point elev., freezing point elev., osmotic pressure changes,

van’t Hoff factor) 16. A 12.5 gram mass of a molecular compound is dissolved in 250 mL of total solution at

274 K. The osmotic pressure of this solution is 0.100 atm. What is the molar mass of this solute?

A. 224 g/mol

B. 11,200 g/mol

C. 85,000 g/mol

D. 850 g/mol E. 15700 g/mol

πV = inRT πV = i(g/MW)RT

0.100 x 0.250 = 1 x (12.5/MW) x 0.08205 x 274 MW = 1.12 x 104 g/mol

15.


Chapter 11 OSMOTIC PRESSURE What is the molarity of solute in an aqueous solution whose osmotic pressure at 21ºC is

0.132 atm? (Assume the solute neither dissociates nor associates)

A. 0.010 M B. 0.0081 M

C. 0.0033 M

D. 0.0055 M

E. choose this choice if none of the others is correct.

πV = inRT π = i(n/V)RT

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π = iMsoluteRT 0.132 atm = 1 x M x 0.08205 x 294 deg

M = 5.47 x 10-3

21.


FREEZING POINT DEPRESSION When a 20.0-g sample of an unknown nonelectrolyte is dissolved in 500. g of benzene, the freezing point of the resulting solution is 3.77ºC. The freezing point of pure benzene

is 5.48ºC and Kf for benzene is 5.12ºC/m. What is the molar mass of the unknown?

A. 120. g/mol

B. 140. g/mol C. 80.0 g/mol D. 100. g/mol

E. 160. g/mol ΔT = Tf - Ti = -Kfimsolute

Solute = 20 g Benzene = 500 g Tf = 3.77

Ti = 5.48 Kf = 5.12

(3.77 - 5.48) = -5.12 x 1 x ((20g/MW)/500 g) x 1000 g/kg MW = 119.8 g

10.


Chapter 11 BOILING POINT ELEVATION

A solution of an electrolyte is prepared from 0.168 mol of the electrolyte and 500. g of water. The boiling point of this solution is 100.291ºC. Given that Kb for water is 0.512 ºC/m, what is the value of i for the electrolyte in this solution?

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A. choose this choice if none of the others is correct. B. 3.38

C. 3.00 D. 2.00

E. 1.69

ΔT = Tf - Ti = Kbimsolute (100.291 - 100.000) = 0.512 x i x ((0.168 mol/500 g) x (1000 g/kg)) i = 1.692



Colligative Properties (van’t Hoff factor) Which one of the following solutions has the lowest freezing point? Assume all salts are 100% ionized. A. 0.10 mol Al(NO3)3 in 200 g water

B. 2.0 mol C12H22O11 in 1000 g water C. 0.10 mol Fe2(SO4)3 in 250 g water D. 0.40 mol BaCl2 in 500 g water E. 0.50 mol NaCl in 500 g water

Determine the moles of particles per liter of water (A). Al(NO3)3 → Al3+ + 3NO3- i = 4

0.10 mol in 200 g H2O = 0.50 mol in 1.00 L water 0.50 x 4 = 2.00 moles of particles (B) C12H22O11 → No reaction i = 1

2.0 mol in 1000 g H2O = 2.0 mol in 1.00 L water 2.0 x 1 = 2.0 moles of particles (C) Fe2(SO4)3 → 2Fe3+ + 3SO42-

i = 5 0.10 mol in 250 g H2O = 0.40 mol in 1.00 L water 0.40 x 5 = 2.0 moles of particles (D) BaCl2 → Ba2+ + 2Cl-

i = 3 0.40 mol in 500 g H2O = 0.80 mol in 1.00 L water 0.80 x 3 = 2.40 moles of particles (E) NaCl → Na+ + Cl-

i = 2 0.50 mol in 500 g H2O = 1.00 mol in 1.00 L water 1.00 x 2 = 2.00 moles of particles BaCl2 has the greatest number of moles of particles and therefore the lowest freezing point.

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Colligative Properties (van’t Hoff factor)

A solution of 1.570 g of a certain compound dissolved in 16.08 g of camphor freezes at a

temperature 15.2 o C below the normal freezing point of pure camphor. Determine the molar mass a of the compound. Kf = 40.0 K kg/mol for camphor.

A. 128 g

mol

B. 64.0 g

mol

C. 257 g

mol

D. 96.0 g

mol

E. 192 g

mol

Freezing point depression: ΔT = Tf -Ti = -Kfimsolute

-15.2 = -40.0 x (mol compound/0.01608 kg) -15.2 = -40.0 x ((1.570g/MW)/0.01608 kg)

X = 256.9 g/mol = molecular weight = molar mass


CHAPTER 11 - PROPERTIES OF SOLUTIONS BOILING POINT ELEVATION CALCULATION A 2.00-g sample of an unknown nonelectrolyte is dissolved in 15.0 g of CCl4. The

boiling point of the solution is 77.85 C. The boiling point constant for CCl4 is

5.03 K/m and the boiling point of pure CCl4 is 76.50 C. The molar mass of the

unknown is

A. 37,300 g/mol

B. 154 g/mol C. 24,900 g/mol D. 249 g/mol

E. 497 g/mol

Boiling-point elevation: ΔT = Tf - Ti = Kbimsolute m = molsolute/kgsolvent = (gsolute/MWsolute)/kgsolvent

(77.85 - 76.50) = 5.03 x 1 x ((2.00 g/MW)/0.015) MW = 496.79 g/mol

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OSMOTIC PRESSURE CALCULATION The mass of glucose (MW = 180.2) required to prepare 2.00 L of solution with an

osmotic pressure of 1.50 atm at 25 C is

A. 263 g

B. 1.46 g C. 44.6 g

D. 132 g

E. 22.1 g

Osmotic pressure: πV = inRT 1.50 atm x 2.00 L = 1 x (g/180.2gmol-1) x 0.08205 latmdeg-1mol-1 x 298K

X = 22.10 g

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Colligative Properties Calculation (freezing point depression)

What is the freezing point of a solution prepared by dissolving 1.50 g of naphthalene,

C10H8, in 20.0 g of benzene? The freezing point of pure benzene is 5.50 o C, and Kf = 5.12 for benzene.

A. 3.00 o C

B. 2.50 o C

C. 2.00 o C

D. 1.50 o C

E. 1.00 o C

I disagree with the textbook’s formula of ΔT = +Kfmsolute. The result of using this formula provides an absolute change in temperature, and therefore doesn’t tell whether the final

temperature is lower or higher than the original. By inserting a “-“ in front of K permits use of the formula ΔT = Tf - Ti, and gives the correct answer every time. ΔT = Tf - Ti = -Kfmsolute

Tf - Ti = -Kfmsolute Tf - 5.50 = (-5.12) x ((1.50/128)/0.020)

X = 2.50oC = Tf

MISCELLANEOUS

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27 CHEM 162-2006 FINAL EXAM + ANSWERS CHAPTER 11 – PROPERTIES OF SOLUTIONS MISCELLANEOUS

Which one of the following is an example of a colloidal suspension?

A. mayonnaise

B. steel C. filtered sea water

D. sand E. diamond

A colloid is a suspension (not a solution) of particles in a medium. These particles can be precipitated under certain conditions.

A. Mayonnaise is a type of colloid called an emulsion, in which two insoluble liquids (in this case oil in water) are mixed and one of the liquids remains suspended in the other.

B. Steel is a solid solution of various metals in iron. C. Filtered sea water is a solution of salts in water.

D. Sand is a solid solution of various sodium silicates. E. Diamond is a form of an element (carbon). It consists of only one component, carbon.


CHAPTER 11 - PROPERTIES OF SOLUTIONS MISCELLANEOUS CONCEPT

A colloid is stabilized by

A. Electrostatic repulsions

B. Lattice energy

C. Energy of solution D. Electrostatic attractions

E. Kinetic Energy Page 543 of Zumdahl and Zumdahl:

“What stabilizes a colloid? Why do the particles remain suspended rather than forming larger aggregates and precipitating out? The answer is complicated but the main factor

seems to be electrostatic repulsion. The center of a colloidal particle (a tiny ionic crystal, a group of molecules, or a single large molecule) attracts from the medium a layer of ions, all of the same charge. This group of ions, in turn, attracts another layer of

oppositely charged ions . . .Because the colloidal particles all have an outer layer of ions with the same charge, they repel each other and do not easily aggregate to form particles

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that are large enough to precipitate.”

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