properties of solutions

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PRACTICE PROBLEMS WITH DETAILED SOLUTIONS CHAPTER 13 - 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 (INCLUDING COLLOIDS)

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 = iXsolventPo

solvent iXsolvent = insolvent/(insolvent + insolute) iXA + iXB = 1 PTotal = PA + PB = iXAPo

A + iXBPoB

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 Tf 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 162-2010 EXAM I Chapter 12 – Properties of Solutions Solution concentrations concepts (molarity, mole fraction, etc.) 7. A solution can be carefully prepared to hold more solute than the maximum solubility. Such a solution is called A. supersaturated B. subsaturated C. saturated D. isotonic E. hypertonic A. True. A supersaturated solution is one that contains more solute than is present in a saturated solution. B. False. A subsaturated solution is one that contains less solute than is present in a saturated solution. C. False. A saturated solution is one that contains the maximum amount of solute, while at equilibrium with undissolved solute. D. False. An isotonic solution is one that has the same osmotic pressure as body fluids. E. False. A hypertonic solution is one that has an osmotic pressure greater than that of body fluids. CHEM 162-2009 EXAM I CHAPTER 12 – PROPERTIES OF SOLUTIONS Solution concentrations concepts (Molarity, mole fraction, etc.) 21. Which of the following statements is true regarding molality and molarity? A. Neither molality nor molarity depend on temperature. B. Both molality and molarity depend on temperature. C. Molarity depends on temperature but molality does not. D. Molality depends on temperature but molarity does not. E. Temperature dependence of molality and molarity depends on the material. Molarity is moles of solute per liter of solution. If the solution is heated, the density decreases, resulting in the same number of moles but an increase in volume of the solution. That is, if the temperature is increased, the concentration of solute in solution (i.e., molarity) decreases. Therefore, molarity is dependent on temperature. The purpose of the development of the molality concentration, was to have a concentration which is temperature independent. That is, since molality is moles of solute per kg solvent, and neither moles nor kg is affected by temperature, then molality is independent of temperature change. A. False. Molality doesn’t depend on temperature, but molarity does.

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B. False. Molality doesn’t depend on temperature, but molarity does. C. True. Molarity depends on temperature but molality does not. D. False. Molarity depends on temperature but molality does not. E. False. Although temperature dependence of molarity depends on the material, molality, regardless of the material, is not dependent on temperature.

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SOLUTION CONCENTRATIONS CALCULATIONS

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1 Chem 162-2011 Final exam Chapter 12 - Properties of Solutio–s Solution Concentrations Calculations (Molarity, Mole Fraction, Etc.) A solution consists of 0.10 moles of NaCl in 1000 g of water. If we add 50.0 g of ethanol, C2H5OH(l) which one of the following quantities will NOT change?

A. Molality of NaCl. B. Molarity of NaCl. C. Mass % of NaCl. D. Mole fraction of NaCl. E. Vapor pressure of water above the solution. ET note: The addition of ethanol might cause confusion in this problem because it is not clear whether ethanol should be considered to be a solute or a solvent. Ethanol might be considered to be a solute because it is a smaller quantity than the water (which is clearly the solvent). On the other hand, ethanol might be considered to be a solvent (i.e., a “co-solvent”) rather than a solute because a solvent is defined as a substance having the same state as the final solution, and both ethanol and the final solution are liquids. If ethanol was considered to be a solvent rather than a solute, then this problem would have no solution. This problem would have been better written if it used a solid, such as glucose, sucrose or urea instead of ethanol. Use of a solid would have avoided the ambiguity of whether to call the liquid ethanol a solute or a solvent.

A. Molality = moles solute/1 kg solvent. In this case it is moles NaCl/1 kg H2O 0.10 moles NaCl/1kg H2O = 0.10 molal 50.0g C2H5OH(l) is added. 50.0g/46.08gmol-1 = 1.09 mol C2H5OH. The moles of NaCl/1 kg of H2O is still 0.10 molal; i.e., the moles of C2H5OH is irrelevant. This is no change. B. Molarity = moles solute/1 liter solution 0.10 moles NaCl/1kg H2O → mol NaCl/1000 mL solution Numerator: Done Denominator: 1000 g H2O → mL solution 0.10 mol NaCl x 58.45g/mol = 5.845g NaCl Therefore, total solution = 1000 g + 5.845g = 1005.845g Assume the density of such a weak solution is 1.0g/mL Therefore, 1005.845g x 1.0g/mL = 1005.845mL Numerator/Denominator: 0.10molNaCl/1.005845Lsolution = 0.099 molar solution If we assume that C2H5OH has a density of 1.0 (it really has a density of approximately 0.8g/mL) then 50.0 g would be 50.0 mL = 0.050L Therefore, the new molarity would be 0.10molNaCl/(1.005845L + 0.050L) = 0.0947M, a change. C. Mass percent = grams of NaCl/100 g solution 0.10 moles NaCl/1kg H2O solvent → g NaCl/100 g H2O solution Numerator: 0.10 mol x 58.45g/mol = 5.845 g NaCl Denominator: gsolute + gsolvent = gsolution gsolution = 5.845gsolute + 1000gsolvent = 1005.845gsolution Numerator/Denominator = (5.845gNaClsolute/1005.845gsolution) x 100 = 0.58 mass percent Add 50.0g C2H5OH to the solution. (5.845gNaClsolute/(1005.845gsolution + 50.0gsolue)) x 100 = 0.55 mass percent, a change. D. Mole fraction = mol NaCl/molTotal 0.10mol NaCl/1000gH2O → mol NaCl/(molNaCl + molH2O) Numerator: OK Denominator: 1000gH2O x (1mol/18.02g) = 55.55 mol H2O Numerator/Denominator = 0.10mol/(0.10molNaCl + 55.55 mol H2O) = 0.00180 mol fraction Add 50.0g C2H5OH. mol = 50.0g/46.08gmol-1 = 1.09 mol C2H5OH. Mole fraction = 0.10molNaCl/(0.10molNaCl + 55.55molH2O + 1.09mol C2H5OH) = 0.00176 mol fraction, a change. E. Raoult’s law: Psoln = iXsolventAPo

solventA + iXsolventBPosolventB

XH2O = 55.55 mol H2O/(55.55 mol H2O + (2 x 0.10molNaCl)) = 0.9964 x Po Add 1.09mol C2H5OH: XH2O = 55.55 mol H2O/(55.55 mol H2O + (2 x 0.10molNaCl) + 1.09mol C2H5OH) = 0.9773 x Po, a change.

A

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26 Chem 162-2011 Final exam

Chapter 12 - Properties of Solutio–s Solution concentrations calculations (molarity, mole fraction, etc.) A soft drink contains 10.5% sucrose by mass and has a density of 1.04g/mL. What volume of this soft drink would contain 78.5g sucrose? A. 747 mL B. 812 mL C. 719 mL D. 655 mL E. 911 mL 10.5g sucrose/100gsolution 10.5gsucrose/100gsolution = 78.5gsucrose/Xgsolution 78.5g sucrose would be in 747.6gsolution 747.6gsolution x (1mL/1.04g) = 719 mL

C


Chapter 12 - Properties of Solutio–s Solution concentrations calculations (molarity, mole fraction, etc.) An aqueous solution contains 0.72% Al2(SO4)3 by mass and has a density of 1.05 g/cm3. What is the molarity of Al2(SO4)3? Molar mass of Al2(SO4)3 = 342.12 g/mol. A. 0.0111M B. 0.0221M C. 0.0884M D. 0.0442 E. 0.0307 0.72gAl2(SO4)3/100gsolution → mol Al2(SO4)3/Lsolution Numerator: 0.72g Al2(SO4)3/(342.12gmol-1) = 0.0021045mol Al2(SO4)3 Denominator: 100gsolution x (1cm3/1.05g) = 95.24cm3 = 95.24mL solution = 0.09524Lsolution Numerator/Denominator: 0.0021045mol Al2(SO4)3/0.09524Lsolution = 0.0221 mol/L= 0.0221M

B

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Chem 162-2010 Final Exam + Answers Chapter 12 – Physical Properties of Solutions Solution concentrations calculations

42. What is the molality of 2.158M C6H12O6(aq) whose density is 1.1434g/cm3? A. 1.777m

B. 2.860m C. 2.158m D. 1.887m E. 2.467m 2.158 mol C6H12O6/Lsoln → ?mol C6H12O6/kgsolv First do numerator, then denominator. Numerator is already done: 2.158 mol → 2.158 mol Denominator: Plan: Lsoln → gsoln → gsolv → kgsolv 1L solution x 1.1434g/0.001L = 1143.4g solution gsolute + gsolvent = gsolution gsolute = 2.158mol x 180.18g/mol = 388.83 g 388.83gsolute + gsolvent = 1143.4gsolution gsolvent = 754.57g = 0.75457kg Combine numerator with denonimator: 2.158mol/0.75457kg = 2.86 molsolute/kgsolvent = 2.86molal

CHEM 162-2010 EXAM I Chapter 12 - Properties of Solutio–s Solution concentrations calculations (molarity, mole fraction, etc.) 19. What is the mole fraction Cl- in a 1.43m MgCl2(aq) solution? A. .333 B. 0.0249 C. 0.0478 D. 0.00143

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E. 1.43 MgCl2 → Mg2+ + 2Cl- Hence, a 1.43 m solution of MgCl2 is equivalent to a 2.86 m solution of Cl-. 2.86mol Cl-solute/kg water → ?mol Cl-solute/(mol Cl- solute + mol H2O solvent) Numerator: 2.86 mol Cl- solute → mol Cl- solute Done. Denominator: 1 kg water X 1000g/kg x 1/(18.02g/mol-1) = 55.5 mol water solvent Numerator/Denominator: 2.86 mol Cl-solute/(2.86 mol Cl- solute + 55.5 mol H2O solvent) = 0.0490 CHEM 162-2010 EXAM I Chapter 12 - Properties of Solutio–s Solution concentrations calculations (molarity, mole fraction, etc.) 23. An aqueous solution of H2O2 is 30.0% by mass. What is the molality of the solution? A. 9.70 m B. 0.185 m C. 11.4 m D. 12.6 m E. 8.82 m 30.0gH2O2/100.0 g solution → mol H2O2/kg solvent Numerator: 30.0g /34gmol-1 = 0.882 mol H2O2 Denominator: kg solute + kg solvent = kg solution kg solvent = kg solution – kg solute kg solvent = 0.100kg solution – 0.0300kg solute = 0.070 kg solvent Numerator/Denominator: 0.882 mol H2O2/0.070 kg solvent = 12.6 mol H2O2/kg solvent = 12.6m CHEM 162-2010 EXAM I Chapter 12 - Properties of Solution– Solution concentrations calculations (molarity, mole fraction, etc.) 3. What is the molarity of 2.67m CuSO4(aq) solution with a density of 1.67g/cm3? A. 2.15M B. 3.13M C. 1.60M D. 4.46M

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E. 1.00M 2.67moles CuSO4/kg H2O → ?moles CuSO4/L solution Numerator: moles = moles; done Denominator: gsolute + gsolvent = gsolution 2.67mol CuSO4 x 159.55gmol-1 = 426.0 g CuSO4 426.0g CuSO4 + 1000gH2O = 1426.0gsolution 1426.0gsolution x (1 mL/1.67g) = 853.9 mL solution = 0.8539 L solution Numerator/Denominator: 2.67molCuSO4/0.8539Lsolution = 3.13M CHEM 162-2009 EXAM I CHAPTER 12 - PROPERTIES OF SOLUTION– Solution concentrations calculations (Molarity, mole fraction, etc.) 8. What is the molarity of 13.7m formic acid solution (HCOOH), which has a density of 1.129g/cm3? A. 27.5M B. 9.49M C. 15.5M D. 12.1M E. 13.7M 13.7molHCOOH/1kg solvent → molHCOOH/L solution Numerator: 13.7 mol HCOOH → 13.7 mol HCOOH; Done Denominator: g solute + g solvent = g solution 13.7 mol HCOOH x 46.03gmol-1 = 630.6 g HCOOH (630.6g HCOOH + 1000g solvent) x (1mL/1.129g) x (1L/1000mL) = 1.444 L solution Numerator/Denominator = 13.7 mol HCOOH/1.444L solution = 9.486 M solution CHEM 162-2009 FINAL EXAM CHAPTER 12 - PROPERTIES OF SOLUTION– SOLUTION CONCENTRATIONS CALCULATIONS (MOLARITY, MOLE FRACTION, ETC.) 4. Procaine hydrochloride (M = 272.77g/mol) is used as a local anesthetic. Calculate the molarity of a 4.666 m solution which has a density of 1.1066 g/mL. A. 4.056 M B. 2.272 M C. 4.216 M D. 4.666 M

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E. 1.106 M 4.666 moles P/1000 g solvent → moles P/L solution Numerator: moles → moles; Done Denominator: g solution = g solute + g solvent g P = 4.666 moles P x 272.77 g/mol = 1273 g P = 1273 g solute 1000 g solvent + 1273 g solute = 2273 g solution 2273 g solution x (1 mL/1.1066 g) = 2054 mL solution = 2.054 L solution Combining numerator and denominator: 4.666 mol P/2.054 L solution = 2.27 mol P/L solution = 2.27M B CHEM 162-2009 FINAL EXAM CHAPTER 12 - PROPERTIES OF SOLUTION– SOLUTION CONCENTRATIONS CALCULATIONS (MOLARITY, MOLE FRACTION, ETC.) 45. An aqueous solution of H2O2 is 30.0% by mass and has a density of 1.10 g/cm3. What is the molarity of the solution? A. 9.70M B. 12.6M C. 4.80M D. 1.45M E. 7.93M 30.0gsolute/100.0gsolution → molsolute/Lsolution Numerator: 30.0gsolute/34.02gmol-1 = 0.8818molsolute Denominator: 100.0gsolution x (0.001L/1.10g) = 0.0909Lsolution Numerator/Denominator = 0.8818molsolute/0.0909Lsolution = 9.70mol/L = 9.70 M A

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50 Chem 162-2008 Final Exam + Answers Chapter 12 – Physical 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. 4.35m B. 3.37m C. 3.75m D. 4.61m E. 3.05m

3.75 mol/Lsoln → ?mol/kgsolv First do numerator, then denominator. Numerator is already done: 3.75 mol → 3.75 mol Denominator: 1L solution x 1.230g/0.001L = 1230g solution gsolute + gsolvent = g solution (3.75mol x 98g/mol) + gsolvent = 1230gsolution gsolvent = 862.5g = 0.865kg Combine numerator with denonimator: 3.75mol/0.865kg = 4.34 molsolute/kgsolvent = 4.34molal

A

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8 CHEM 162-2008 EXAM I CHAPTER 12 - PROPERTIES OF SOLUTION– SOLUTION CONCENTRATIONS CALCULATIONS (Molarity, mole fraction, etc.)

A 1.211m solution of NaNO3(aq) has a density of 1.065g/cm3. What is its molarity of this solution?

A. 1.169M B.1.259M C. 1.137M D. 1.290M E. 1.211M 1.211 molessolute/kgsolvent → molsolute/Lsolution Numerator: OK = 1.211 molsolute Denominator: MW NaNO3 = 22.99 + 14.01 + 48.00 = 85g/mol 1.211 mol x 85g/mol = 102.94g NaNO3 = solute gsolute + gsolvent = gsolution 102.94gNaNO3 + 1000gsolvent = 1102.94gsolution 1102.94gsolution x 1mL/1.065g x (1L/1000mL) = 1.036Lsolution Combining numerator and denominator: 1.211 molsolute/1.036Lsolution = 1.169molsolute/Lsolution = 1.169M

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36 Chem 162-2007 Final exam + answers Chapter 12 – Properties of Solutions Solution Concentrations Calculations gsolute/gsolution → M An automobile antifreeze contains equal volumes of ethylene glycol (d = 1.114 g/mL); FW = 62.07) and water (d = 1.00 g/mL) at 25 °C. The solution has a density of 1.06 g/mL. What is the molarity of ethylene glycol in the solution? [Hint: the volumes are NOT additive!]

A. 9.00 B. 5.68 C. 4.50 D. 0.323 E. 8.97

VEG = VH2O DEG = 1.114g/mL DH2O = 1.00g/mL Molarity = moles ethylene glycol/1000 mL solution Let’s say that there is 1 mL each of EG and H2O present. Therefore, there is 1.114 g EG and 1.00 g H2O present. 1.114gEG/1.00gH2O 1.114gEG/2.114gsolution Convert gEG/gsolution into moleEG/Lsoln. First do numerator, then denom. 1.114gEG/62.07gmol-1 = 0.0179475molEG 2.114gsolution x (1.000L/1060g) = 1.99434 x 10-3 L 0.0179475molEG/(1.99434 x 10-3L) = 8.9992 molEG/L = 9.00M Note: Having 9.00 and 8.97 as options is anal retentive. With a very small rounding-off error, one can easily get an answer closer to 8.97.

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6. CHEM 162-2007 EXAM I + ANSWERS CHAPTER 12 - PROPERTIES OF SOLUTION– SOLUTION CONCENTRATIONS CALCULATIONS (MOLE FRACTION) What mass of sucrose, C12H22O11, must be dissolved in 1 L of water (d = 0.998 g/ml) to obtain a solution with 0.025 mole fraction C12H22O11? A) 165

B) 486 C) 362 D) 246 E) 452 mole fraction = molesolute/molesolution 0.025mol sucrose/mol solution → gsucrose/1Lsolvent Numerator: 0.025 mol sucrose x 342.34gmol-1 = 8.5585g sucrose Denominator: molsolute + molsolvent = molsolution molsolvent = molsolution – molsolute molsolvent = 1molsolution – 0.025molsucrose = 0.975molsolvent 0.975molsolvent x 18.02g/mol = 17.5695 g solvent 17.5695gH2O x (1 mL/0.998g) = 17.60 mL solvent = 0.01760Lsolvent numerator/denominator: 8.5585gsucrose/0.01760Lsolvent = 486.3gsucrose/Lsolvent

B

2. CHEM 162-2007 EXAM I + ANSWERS

CHAPTER 12 - PROPERTIES OF SOLUTION– SOLUTION CONCENTRATIONS CALCULATIONS (Molarity) The density of a solution that is 20.0% HClO4 by mass is 1.138 g/mL. Calculate the molarity of the HClO4. A) 1.75

B) 2.26 C) 1.99 D) 3.45 E) 0.442 Percent by mass = (gsolute/gsolution) x 100 Molarity = molsolute/Lsolution 20.0 g HClO4/100gsolution → molHClO4/Lsolution Numerator: 20gHClO4/100.46gmol-1 = 0.199 mol HClO4 Denominator: 100g solution x (1mL/1.138g) = 87.87 mL solution Numerator/Denominator: (0.199 mol HClO4/87.87 mL solution) x (1000 mL/L) = 2.26 mol/L = 2.26M

B

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ENERGETICS OF SOLUTIONS AND SOLUBILITY CONCEPTS

4 Chem 162-2011 Exam I Hill & Petrucci et al. 4th edition Chapter 12 – Properties of Solutions Energetics of solutions and solubility concepts Which of the following would result in an ideal solution?

A. The enthalpy of solvation is greater in magnitude than the sum of the enthalpies of expansion of the solute and of the solvent. B. The enthalpy of solvation is equal in magnitude to the sum of the enthalpies of expansion of the solute and of the solvent. C. The enthalpy of solvation is less in magnitude than the sum of the enthalpies of expansion of the solute and of the solvent. D. The enthalpy of expansion of the solute is large. E. The enthalpy of expansion of the solute is small.

Comment: Terminology to characterize solute-solute bond breakage or solvent-solvent bond breakage is “breakage”, “separation” and “expansion”. Since “expansion” is neither used by this professor nor the H&P textbook, I think that its usage in this exam problem was not the best choice. An ideal solution is one in which the sum of all three enthalpies (solute-solute bond breaking, solvent-solvent bond breaking, solute-solvent bond formation) equals zero. Another way of saying this is that an ideal solution is one in which the enthalpy of the solute-solvent bond formation can be calculated from the solute-solute bond breakage and the solvent-solvent bond breakage. Examples of ideal solutions are pentane-hexane, and benzene-toluene. A. False. This is not an ideal solution. It is a negative deviation from an ideal solution. B. True C. False. This is not an ideal solution. It is a positive deviation from an ideal solution. D. False. This doesn’t determine an ideal solution. E. False. Same answer as D.

B

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8 Chem 162-2011 Exam I Hill & Petrucci et al. 4th edition Chapter 12 – Properties of Solutions Energetics of solutions and solubility concepts

When an ionic salt dissolves in water, the strongest solute-solvent interaction is

A. hydrogen bonding. B. London forces. C. ion-dipole. D. ion-ion forces. E. dipole-dipole. Clarification regarding water. Water is a dipole. So when water reacts with another dipole it is a dipole-dipole interaction. When the H atom in water bonds to a F, O or N atom of another molecule, this is a particularly strong dipole-dipole interaction, and is given the special name of hydrogen bonding. When an ionic salt, e.g., NaCl dissolves in water, the strongest solute-solvent bond is between these ions and water. The Na+ H2O bond is an ion-dipole bond; the Cl- H2O bond is an ion-dipole bond. An ion-dipole bond is stronger than hydrogen bonding. This question might have been written better if it said, “When an ionic salt dissolves in water, the solute-solvent interaction is”.

A

CHEM 162-2010 FINAL EXAM Chapter 12 - Properties of Solution– Energetics of solutions and solubility concepts 23. Which compound below would be expected to be the least soluble in water? A. CH3CH2CH2I B. CH3CH2CH2NH2 C. CH3CH2-O-CH3 D. CH3CH2COOH E. CH3CH2COOCH3 The molecule that would be the most soluble would be the one that forms the strongest bonds with water. B, C, D and E can all form hydrogen bonds with water. However, A is not capable of forming hydrogen bonds with water. It forms a slightly weaker dipole-dipole bond with water, and is therefore the least soluble.

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CHEM 162-2010 EXAM I Chapter 12 - Properties of Solution– Energetics of solutions and solubility concepts 6. Non-polar materials won’t dissolve in water because A. the enthalpy of hydration is much larger than the sum of enthalpies to separate the solute particles and to separate the solvent particles. B. the enthalpy of hydration is endothermic. C. the enthalpy of hydration is exothermic. D. the enthalpy of hydration is much smaller than the sum of the enthalpies to separate the solute particles and to separate the solvent particles. E. the enthalpy of hydration is equal to the sum of the enthalpies to separate the solute particles and to separate the solvent particles. There is confusion in this problem of what is meant by “larger” or “smaller”. Generally, if something is “larger”, this means it is more positive. If something is “smaller” it means it is more negative. Hence, +6 is larger than +4; +6 is larger than -2; +6 is larger than -8. If something is smaller, this means it is more negative. Hence, -8 is smaller than +6. I believe that in this problem larger or smaller refers to the absolute number; hence, -8 is larger than +6. Another unclear part of this problem is the meaning of the term “enthalpy of hydration”. In Zumdahl and Zumdahl enthalpy of hydration (∆Hhydration) is defined as the sum of ∆H of solvent-solvent bond breakage and ∆H of solvent-solute bond formation. However, based on the context of this problem, ∆Hhydration seems to simply mean ∆H of solvent-solute bond formation. Let B = the non-polar solute, e.g., benzene, and let W = the polar water Therefore, ∆HBBbondbreakage + ∆HWWbondbrreakage + ∆HBWbondformation + ∆HBWbondformation = ∆Hsoln Code: ∆HBBbondbreakage + ∆HWWbondbrreakage = ∆HBBparticleseparation + ∆HWWparticleseparation ∆HBWbondformation = ∆Hhydration Substances will dissolve if the ∆Hsolution is either slightly positive, neutral, or negative. Hence, add up the enthalpies of bond breakage with the enthalpies of bond formation. Facts: Bond breakage of non-polar molecules is small and positive; bond breakage of polar molecules is large and positive; bond-formation of polar-nonpolar molecules is small and negative.

∆HBBbondbreakage + ∆HWWbondbrreakage + ∆HBWbondformation + ∆HBWbondformation = ∆Hsoln

+ + - - = +

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A. False: ∆HBBbondbreakage + ∆HWWbondbrreakage + ∆HBWbondformation + ∆HBWbondformation = ∆Hsoln

+ + - - = -

Based on the provided information, ∆Hsoln calculates to be negative, which would mean that the non-polar benzene will dissolve in the water. The reason this is incorrect is because the negative value of delta enthalpy of solution is based on a very large negative value of the enthalpy of bond formation. In reality, the enthalpy of bond formation is a small negative, due to the very weak London forces between benzene and water. B. False: ∆HBBbondbreakage + ∆HWWbondbrreakage + ∆HBWbondformation + ∆HBWbondformation = ∆Hsoln

+ + + + = +

Clearly this reaction will not go because delta H of solution is strongly endothermic. But what is more important is that it doesn’t make sense that the ∆HBWbondformation is endothermic, since bond formation is always exothermic. C. True? False?:


+ + -? - ? -? -? = +? -?

This might be true or false. There is not enough information to be certain. The information just says that the enthalpy of hydration is exothermic. What it fails to say is how exothermic it is as compared to the endothermic bond breaking. If it is weakly exothermic (which is true) then ∆Hsoln would be moderately endothermic, and the reaction won’t go. If it is strongly exothermic (not true) then ∆Hsoln would be moderately exothermic and the reaction will go. D. True


+ + - - = + Unlike answer option “C”, this answer option does say how exothermic the bond formation step is as compared to the endothermic bond breaking. Since it is weakly exothermic then ∆Hsoln would be moderately endothermic, and the reaction won’t go. E. False

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+ + - - = 0 Unlike answer option “C”, this answer option does say how exothermic the bond formation step is as compared to the endothermic bond breaking. It says (incorrectly) that magnitude of the negative ∆HBWbondformation is equal to the sum of the positive ∆Hbondbreakage. Therefore, the ∆Hsoln is zero. If the ∆Hsoln is zero, then the reaction would go, contrary to the observation that the non-polar material is insoluble in water. The error lies in the magnitude of the ∆HBWbondformation. ∆HBWbondformation is very small negative (see option D), not moderate negative (as in option E). CHEM 162-2010 EXAM I Chapter 12 - Properties of Solution– Energetics of solutions and solubility concepts 8. Which compound below would be expected to be the least soluble in water? A. CH3CH2CH2Br B. CH3CH2CH2NH2 C. CH3CH2-O-CH3 D. CH3CH2COOH E. CH3CH2NHCH3 The molecule that would be the most soluble would be the one that foms the strongest bonds with water. B, C, D and E can all form hydrogen bonds with water. However, A is not capable of forming hydrogen bonds with water. It forms a slightly weaker dipole-dipole bond with water, and is therefore the least soluble.

CHEM 162-2009 EXAM I CHAPTER 12 - PROPERTIES OF SOLUTION– Energetics of solutions and solubility concepts 18. In an ideal solution: A. ∆H of hydration > 0 B. ∆H of hydration = 0 C. ∆H of hydration + ∆H of expansion of the solvent + ∆H of expansion of the solute > 0 D. ∆H of hydration + ∆H of expansion of the solvent + ∆H of expansion of the solute < 0 E. ∆H of hydration + ∆H of expansion of the solvent + ∆H of expansion of the solute = 0 A. False. In an ideal solution, ∆H of hydration is <0, because bonds are being formed, which is always exothermic.

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B. False. In an ideal solution, ∆H of hydration is <0, because bonds are being formed, which is always exothermic. C. False. In an ideal solution, ∆H of hydration is exactly equal to and opposite in sign to the sum of the ∆H of expansion of the solvent and the ∆H of expansion of the solute. Therefore, the total of the three ∆H’s = 0 D. False. In an ideal solution, ∆H of hydration is exactly equal to and opposite in sign to the sum of the ∆H of expansion of the solvent and the ∆H of expansion of the solute. Therefore, the total of the three ∆H’s = 0 E. True. In an ideal solution, ∆H of hydration is exactly equal to and opposite in sign to the sum of the ∆H of expansion of the solvent and the ∆H of expansion of the solute. Therefore, the total of the three ∆H’s = 0

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5 CHEM 162-2008 EXAM I CHAPTER 12 - PROPERTIES OF SOLUTION– ENERGETICS OF SOLUTIONS AND SOLUBILITY CONCEPTS Which condition results in an ideal solution?

A. Enthalpy of mixing the solute and solvent is very large and exothermic.

B. Enthalpy of mixing the solute and solvent is approximately equal in magnitude to the sum of the enthalpies of separating of the solute molecules and to separate the solvent molecules. C. The enthalpy to separate the solvent molecules is very small. D. The enthalpy to separate of the solute molecules is very small. E. The enthalpy mixing solute and solvent is much smaller in magnitude than the enthalpies of separating solute from solute and solvent from solvent.

ΔHsoln = ΔHsolute-solutebondbreaking + ΔHsolvent-solventbondbreaking + ΔHsolute-solventbondforming An ideal solution is one in which the molecules of solute and the molecules of solvent are so similar that there is no significant change in energy whether a solute molecule is bonded to a solute molecule, or a solute molecule is bonded to a solvent molecule. That is, solute-solute, solvent-solvent, and solute-solvent bond formation would all be approximately the same, because the molecules are so similar, e.g., pentane and hexane, or methanol and ethanol. The positive ∆H of solute-solute bond breaking would be equal to the positive ∆H of solvent-solvent bond breaking, which would be equal to the negative ∆H of solute-solvent bond formation, resulting in a ∆H of solution being zero. A. False. This could be an ideal solution. If the solute-solute bond breaking is very large and endothermic, and solvent-solvent bond breaking is very large and endothermic, then in an ideal solution the solute solvent bond formation would be very large and exothermic. However, very large and exothermic energy can also occur in a non-ideal solution, such as acetone forming a solution with chloroform. B. True. Consider pentane mixing with hexane as an example of an ideal solution. Let’s arbitrarily assume that pentane is the solute and hexane is the solvent. The energy to separate the solute molecules is small and positive, since a pentane-pentane bond is held together by Lewis forces. The energy to separate the solvent-solvent molecules is small and positive, since a hexane-hexane bond is held together by Lewis forces. The energy to form bonds between the solute and the solvent molecules (pentane and hexane) would be small and negative, since the solute and solvent molecules are held together by Lewis forces. However, the fact that the enthalpy of mixing the solute and solvent is equal in magnitude to the sum of the enthalpies of solute-solute separation and solvent-solvent separation doesn’t prove that this is an ideal solution, because it is possible that the solute is held together by strong electrostatic forces (ion-ion bonding) (e.g., NaCl) and that the solvent is held together by more moderate dipole-dipole bonding forces (e.g., acetone-acetone), and that the solute and solvent would come together to form ion-dipole forces (Na+ - acetone), which is roughly an average of the ion-ion bonding and dipole-dipole bonding. This would not be an ideal solution. Nevertheless, this option is the best answer. C. False. If the enthalpy to separate the solvent-solvent molecules is small, but the energy to separate the solute-solute molecules is large, then this would not be an ideal solution., because the energy to break the solvent-solvent molecules and solute-solute molecules are different. That is, the solvent and solute molecules are not behaving the same with regard to bond breaking and making. D. False. If the enthalpy to separate the solute-solute molecules is small, but the energy to separate the solvent-solvent molecules is large, then this would not be an ideal solution., because the energy to break the solute-solute molecules and solvent-solvent molecules are different. That is, the solute and solvent molecules would not be behaving the same with regard to bond breaking and making. E. False. I don’t know if it’s possible to have a situation in which the enthalpies of separating solute from solute and solvent from solvent would be large, and the enthalpy of mixing solute and solvent would be much smaller. In any case, the sum of the enthalpies wouldn’t add up to zero, which means that this would not be an ideal solution.

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42 Chem 162-2007 Final exam + answers Chapter 12 – Properties of Solutions Energetics of solutions and solubility concepts When an ionic salt dissolves in water, the strongest solute-solvent interaction is

A. ion-dipole. B. hydrogen bonding. C. London forces. D. ion-ion forces. E. dipole-dipole.

Use a typical salt, such as NaCl NaCl → Na+ + Cl- A. True. Na+ (or Cl-) is an ion; H2O is polar (i.e., it’s a dipole). The interaction is an ion-dipole interaction. B. False. There is no hydrogen bonding. There would have been if the solute contained a F, N or O atom. C. False. There is London forces interaction. However, that interaction is very weak compared to the ion-dipole interaction. D. False. H2O is not an ion, so it cannot be ion-ion forces. E. False. Na+ and Cl- are ions, not a dipole. Hence, it cannot be dipole-dipole interaction.

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1. CHEM 162-2007 EXAM I + ANSWERS CHAPTER 12 - PROPERTIES OF SOLUTION– ENERGETICS OF SOLUTIONS AND SOLUBILITY CONCEPTS Which of the following hypothetical steps is predicted to be exothermic? A) pure solvent → separated solvent molecules B) pure solute → separated solute molecules

C) separated solvent and separated solute molecules → solution D) pure solvent and pure solute → heterogeneous mixture E) not enough information to answer this question (A) Breaking a solvent-solvent bond requires energy input. Therefore it is endothermic. (B) Breaking a solute-solute bond requires energy input. Therefore it is endothermic. (C) Forming a solution from separated solvent and separated solute molecules means solute-solvent bond formation. This gives off energy. It is exothermic. (D) If pure solute (e.g. sucrose[s]) and pure solvent (e.g., hexane[l]) form a heterogeneous mixture, then they didn’t react with each other, suggesting that the reaction is endothermic. (E) There is enough information to answer this question.

C

ENERGETICS OF SOLUTIONS AND SOLUBILITY

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

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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

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VAPOR PRESSURE OF SOLUTIONS CONCEPTS


Chapter 12 – Properties of Solutions Vapor pressure of solutions concepts (Henry’s law, Raoult’s law) What amount of sucrose must be dissolved in 5.00 moles of water in order to reduce the vapor pressure of water by 10.0%?

A. 0.315 moles B. 0.748 moles C. 0.556 moles D. 0.772 moles E. 0.280 moles

Psoln = iXsolventPosolvent

Let X = Vapor pressure of pure H2O. Let 0.9X =Vapor pressure of aqueous solution in which the vapor pressure was reduced by 10%. Mole fraction of water = 5.00molH2O/(5.00molH2O + Ymolsucrose) 0.9X = (5.00/(5.00+Y)) x X (0.9X/X) = (5.00/(5.00+Y)) 0.9 = (5.00/(5.00+Y)) Y = 0.556 mol sucrose

C

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3 Chem 162-2011 Exam I Hill & Petrucci et al. 4th edition Chapter 12 – Properties of Solutions Vapor pressure of solutions concepts (Henry’s Law, 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. X and Z only D. Y and Z only E. Z only X. True: An example of a negative deviaion from Raoult’s law would be a mixture of acetone and chloroform. An acetone-acetone mixture is held together by dipole-dipole interaction. A chloroform-chloroform mixture is held together by dipole-dipole interaction. An acetone-chloroform mixture is held together by hydrogen bonding, stronger interaction than dipole-dipole interaction. This stronger interaction results in lower vapor pressure than predicted by Raoult’s law. Y. False: The stronger bonding results in an exothermic reaction; hence, a negative ΔHsol’n. Z. True: See X above.

E

CHEM 162-2010 EXAM I Chapter 12 – Properties of Solutions Vapor pressure of solutions concepts (Henry’s law, Raoult’s law) 11. Non-ideal solutions vary from Raoult’s law. Which of the following would likely cause negative deviations from (vapor pressure lower than predicted by) Raoult’s Law? A. A substance that won’t dissolve. B. Repulsion between the solute and solvent. C. Endothermic enthalpy of solution D. Extremely volatile solute E. Strong attraction between solute and solvent

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Negative deviations from Raoult’s law occur when the bonding between the solute-solvent bonding is greater than the average of the solute-solute bonding and the solvent-solvent bonding. In this way, since the bonding between the solute and solvent is greater than expected, then the vapor pressure is lower than expected. Hence, this is a negative deviation from Raoult’s law. An example of this is chloroform mixed with acetone. Chloroform-chloroform bonding is dipole-dipole interaction; acetone-acetone bonding is dipole-dipole interaction; but chloroform-acetone bonding is the stronger hydrogen bonding. CHEM 162-2010 EXAM I Chapter 12 – Properties of Solutions Vapor pressure of solutions concepts (Henry’s law, Raoult’s law) 14. An open bottle of Coke in the refrigerator will still fizz when brought up to room temperature. Why? A. CO2 moves more slowly at low temperature. B. CO2 is more soluble in water at high temperature than low. C. CO2 reacts with air at room temperature. D. The partial pressure of CO2 is higher at room temperature. E. CO2 is more soluble in water at low temperature than high. The fizz is the gas escaping from the solution. A. False. Whereas CO2 indeed moves more slowly at low temperature, the speed of the movement has nothing to do with its status at equilibrium, which is that at room temperature gases are less soluble than at lower temperatures, so the gases will escape (fizz) at room temperature. B. False. Gases are less soluble at high temperature than at low temperature. C. False. CO2 does not react with air. D. False. CO2 is less soluble in water at room temperature than at refrigerated temperature. According to Henry’s law, since it is less soluble it will have a lower vapor pressure. E. True. Whereas solid and liquid solutes are more soluble in solvents at high temperatures, gases are more soluble in solvents at low temperatures. This can be explained with entropy considerations. CHEM 162-2010 EXAM I Chapter 12 – Properties of Solutions Vapor pressure of solutions concepts (Henry’s law, Raoult’s law) 16. Distillation works because of Raoult’s Law. How does distillation allow for compounds to be separated? A. The mole fraction in the gas of the more volatile (low boiling point) compound is higher than in the starting solution. B. The mole fraction in the gas of the more volatile (low boiling point) compound is lower than in the starting solution. C. Only the more volatile component vaporizes. D. Only the less volatile component vaporizes. E. The solution boils at a higher temperature, which allows the compounds to separate.

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A. True. The more volatile component preferentially goes into the vapor, leaving the less volatile component in the residue. When the vapor is condensed and re-distilled, it will become even richer in the more volatile component. By repeating this process numerous times, eventually one will get a very pure substance in the vapor phase. B. False. The opposite is true. The mole fraction of the gas of the more volatile compound is higher than its mole fraction in the starting solution. C. False. Both components vaporize, but more of the more volatile substance vaporizes. D. False. Both components vaporize. E. False. This is a meaningless statement. CHEM 162-2009 EXAM I CHAPTER 12 – PROPERTIES OF SOLUTIONS Vapor pressure of solutions concepts (Henry’s Law, Raoult’s Law) 23. In a non-ideal solution, which statements are true? X. The total vapor pressure may be greater than that predicted by Raoult’s law Y. The vapor pressure may be less than that predicted by Raoult’s Law. Z. A volatile solute becomes non-volatile. A. X only B. X and Y only C. X and Z only D. Y and Z only E. X, Y, and Z X. True. The total vapor pressure may be greater than that predicted by Raoult’s law. An example is a mixture of acetone and carbon disulfide. Y. True. The vapor pressure may be less than predicted by Raoult’s Law. (This should say “total vapor pressure”.) An example is a mixture of acetone and chloroform. Z. False. In the above cited cases of acetone, carbon disulfide and chloroform, all are non-volatile solutes, and none have become non-volatile. I suppose that there may be some case in which the volatile solute is adsorbed so strongly to the solvent that it becomes non-volatile, but this would be the exception, not the rule. CHEM 162-2009 EXAM I CHAPTER 12 - PROPERTIES OF SOLUTIO–S Vapor pressure of solutions concepts (Henry’s Law, Raoult’s Law) 25*. Which of the following will not affect solubility, expressed in moles/liter, of a solid or gas dissolving in a solvent? A. Temperature of the solution B. Partial pressure of the gas C. Relative polarity of solute and solvent

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D. Volume E. Lattice energy of the solid. *Option D would have been better stated as, “Volume of solution”. A. Will affect solubility. If the temperature is increased, the solubility of solid will increase in the solution, while the solubility of gas will decrease. B. Will affect the solubility. According to Henry’s law, as the partial pressure of a gas is increased above a solution, the solubility of the gas in the solution will increase. C. Will affect the solubility. A polar solute will be highly soluble in a polar solvent, but somewhat insoluble in a non-polar solvent. D. Will not affect the solubility. Does this mean volume of solution? If the volume of the solution increases, then there is no affect on solubility. The term solubility here is expressed as moles/liter. Whether the solution is 10 mL or 1000 mL the solubility of a substance, in moles per liter, is not affected. Does “volume” mean volume of solvent? If the volume of the solvent increases, more solute will dissolve, but the concentration of soluble solute won’t have changed. Hence, the solubility wouldn’t have changed if the volume of solvent was increased. We have to distinguish between “solubility” and “concentration”. If we just increase the volume of the solvent, then the concentration of solute will change, but the solubility of the solute won’t change. E. Will affect the solubility. Take the solubility of diamond in pentane, as an example. The lattice energy of diamond is very large. Therefore, the energy of expanding the C-C bonds in diamond is very large. This energy is not likely to be made up for by the weak exothermicity of a pentane-carbon bond formation (London forces). Hence, diamond won’t be soluble in pentane. On the other hand, let’s consider the solubility of a crystal of C20H42 (assuming that that forms a crystal at room temperature) in hexane. This large non-polar hydrocarbon is held together by weak London forces. Therefore, the energy of expanding C20H42-C20H42 bonds is very small. This energy is likely to be made up for by the weak exothermicity of a pentane-C20H42 bond formation. Hence, C20H42 would be expected to be soluble in pentane. CHEM 162-2009 FINAL EXAM CHAPTER 12 - PROPERTIES OF SOLUTIO–S VAPOR PRESSURE OF SOLUTIONS CONCEPTS (HENRY’S LAW, RAOULT’S LAW) 15. The solubility of O2(g) in pure water is 4.4 mg O2 in 100.0 mL of water at 20oC and 1.0 atm pressure. At what pressure of O2 would its solubility be 0.010 M at 20oC? A. 1.8 atm B. 2.4 atm C. 7.3 atm D. 3.8 atm E. 5.4 atm Henry’s law is that the ratio of the pressure of a gas above a solvent, to its concentration in that solvent, is a constant. 0.0044gO2/(32g/mol) = 0.0001375 mol O2

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0.0001375 mol O2/0.100L = 0.001375 mol O2/L 1 atm/1.375 x 10-3M = X atm/0.010M X = 7.27 atm This problem can also be solved more formally using the Henry’s law formula: SA = kPA, where SA is the solubility of dissolved gas A in the solution. k = SA/PA k = 0.001375M/1atm = 0.001375 k = SB/PB 0.001375 = 0.010/PB PB = 7.27 atm C

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10 CHEM 162-2008 EXAM I CHAPTER 12 - PROPERTIES OF SOLUTIO–S VAPOR PRESSURE OF SOLUTIONS CONCEPTS (Henry’s Law, Raoult’s Law) Negative deviations from Raoult’s law (lower vapor pressure than predicted by Raoult’s law) is caused by:

A. Increasing the pressure on the system. B. Greater repulsion between the solute and solvent and solute-solute or solvent-solvent interaction. C. The solute being non-volatile.

D. Greater attraction between the solute and solvent than solute-solute and solvent solvent interactions E. The solvent being non-volatile

This problem is related to the problem on ideal solutions. Negative deviations from Raoult’s law occur when the solute and solvent molecules don’t form an ideal solution, in particular when the reaction between the solute and solvent is greater than the reaction between the solute and solute, and solvent and solvent. This results in a lower vapor pressure than what would be predicted from an ideal solution, such as mixing pentane and hexane. An example of a negative deviation from Raoult’s law is the mixing of (CH3)2CO (acetone) and CHCl3 (chloroform). Acetone mixed with acetone forms a dipole-dipole intermolecular bond; chloroform mixed with chloroform forms a dipole-dipole intermolecular bond. But acetone mixed with chloroform forms a much stronger hydrogen-bond. This stronger bond results in the acetone-chloroform combination wanting to stay in the liquid phase where it has this strong bond, and not go into the gas phase in which the bond would be broken (ideal gas molecules have no intermolecular bonds). Hence the vapor pressure would be lower than predicted based on acetone alone or chloroform alone. A. False. Increasing the pressure on the system does not come into play with Raoult’s law. B. False. This could have been worded better. It could have said, “Greater repulsion between the solute and solvent than between the combination of solute-solute and solvent-solvent.” In any case, “greater repulsion between solute and solvent” means a positive deviation from Raoult’s law. Formation of an acetone and carbon disulfide solution is an example of this. Acetone is polar and carbon disulfide is non-polar. Acetone-acetone is a dipole-dipole interaction, which has moderate bond formation energy. Carbon disulfide-carbon disulfide is a London forces interaction, which has weak bond formation energy. The combination of acetone and carbon disulfide forms London forces, which is weaker than the average expected based on acetone-acetone and carbon disulfide-carbon disulfide bonds, and therefore forces more vapor into the vapor phase where there is no bonding between the molecules. This is a positive deviation from Raoult’s law. C. False. Deviations from Raoult’s law only occurs with two volatile substances. A possible explanation of this is that non-volatile substances usually are non-volatile because they have a strong solute-solute interaction with each other. If they have a stronger solute-solute interaction with each other than with the solvent, then they wouldn’t be likely to affect the solvents volatility. D. True. As in the case described above with acetone and chloroform, the relatively strong interaction between acetone and chloroform, as compared to acetone-acetone or chloroform-chloroform, results in a negative deviation from Raoult’s law. E. False. Deviations from Raoult’s law only occurs with two volatile substances. A possible explanation of this is that non-volatile substances usually are non-volatile because they have a strong solvent-solvent interaction with each other. If they have a stronger solvent-solvent interaction with each other than with the solute, then they wouldn’t be likely to affect the solutes volatility.

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VAPOR PRESSURE OF SOLUTIONS CALCULATIONS 2 Chem 162-2011 Exam I

Hill & Petrucci et al. 4th edition Chapter 12 – Properties of Solutions Vapor pressure of solutions calculations The vapor pressure of pure benzene (C6H6) and toluene (C7H8) at 25°C are 95.1 and 28.4 Torr, respectively. A solution is prepared with a mole fraction of toluene of 0.750. Determine the mole fraction of toluene in the gas phase. Assume the solution to be ideal.

A) 0.750 B) 0.250 C) 0.213

D) 0.473 E) 0.0280 Psoln = iXsolventAPo


Xbenzene = 1 - 0.750 = 0.250 Xtoluen– = 0.750 Po

benzene = 95.1 Torr Po

toluene = 28.4 Torr i = 1 for benzene and toluene Pbsoln = (0.250 x 95.1) = 23.78 mm Ptsoln = (0.750 x 28.4) = 21.30 mm The moles are directly related to the vapor pressure, according to Dalton’s Law. Therefore: Xtolueneinvapor = 21.30/(21.30 + 23.78) = 0.472

D

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9 Chem 162-2011 Exam I Hill & Petrucci et al. 4th edition Chapter 12 – Properties of Solutions Vapor pressure of solutions calculations The mass of naphthalene (C10H8) that must be dissolved in 175 g of heptane (C7H16 ) to lower its vapor pressure by 25.0% is A. 701 g B. 525 g C. 175 g D. 74.6 g E. 101 g Let y = vapor pressure of pure heptane. Since the vapor pressure was lowered by 25.0%, then the vapor pressure of heptane in solution is 0.75y Psoln = iXsolventPo

solvent Xh = nh/(nh + nn) = (gh/MWh)/((gh/MWh) + (gn/MWn)) MWheptane = 84.07 + 16.16 = 100.23gmol-1 MWnaphthalene = 120.10 + 8.08 = 128.18gmol-1 Xh = (175g/100.23gmol-1)/((175g/100.23gmol-1) + (g/128.18gmol-1)) Psoln = iXsolventPo

solvent 0.75y = (((175/100.23)/((175/100.23) + (g/128.18))) x 1.00y) The y’s cancel out. g = 74.6g naphthalene

D

16 Chem 162-2011 Exam I

Hill & Petrucci et al. 4th edition Chapter 12 – Properties of Solutions Vapor pressure of solutions calculations (Henry’s Law, Raoult’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

A

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C. 1.8×109M/atm D. 4.6×10-3M/atm E. 2.5×10-6M/atm Henry’s Law: SA=kPA The unit for Henry’s law constant given in the answer options is “M”. Hence, change the concentration of Helium in water into molarity. 1.83 x 10-6g/L x (1mol/4.00g) = 4.575 x 10-7mol/L = 4.575 x 10-7 M Now solve for k. 4.575x10-7M = k x (1.2x10-3atm) k = 3.81 x 10-4 M/atm

CHEM 162-2010 EXAM I Chapter 12 - Properties of Solutio–s Vapor pressure of solutions calculations (Henry’s law; Raoult’s law) 17. What is the mole fraction of pentane in the gas phase of a mixture of 20.0 g pentane (C5H12) with 40.0 g of heptane (C7H16) at 20.0oC. The vapor pressures of pure pentane and heptane at 20.0oC are 420 Torr and 36.0 Torr, respectively. A. 0.11 B. 0.41 C. 0.59 D. 0.89 E. 0.50 This is a typical Raoult’s law problem. Psoln = XsolventPo

solvent XP = nP/(nP + nH) XP = (20g/72.17gmol-1)/((20g/72.17gmol-1) + (40g/100.23gmol-1)) = 0.4098 PP = XPinsolnPo

solvent PP = 0.4098 x 420 = 172.1 Torr Psoln = XsolventPo

solvent XH = nH/(nH + nP) XH = (40g/100.23gmol-1)/((40g/100.23gmol-1) + (20g/72.17gmol-1)) = 0.5902 PH = XHinsolnPo

solvent PH = 0.5902 x 36.0 = 21.25 Torr

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According to Dalton, moles are directly related to pressure. Hence, the mole fraction of pentane in the vapor phase is equal to the pressure fraction of pentane in the vapor phase. XP = PP/(PP + PH) = 172.1Torr/(172.1Torr + 21.25Torr) = 0.89 CHEM 162-2010 EXAM I Chapter 12 - Properties of Solutio–s Vapor pressure of solutions calculations (Henry’s law, Raoult’s law) 12. At 25oC the partial pressure of carbon dioxide in a can of soda is 5.00 atm. The Henry’s Law constant for carbon dioxide in soda is 3.3 x 10-2 mole L-1atm-1. Calculate the mass of carbon dioxide in a 375 mL can at 25C. A. 0.165g B. 2.72g C. 0.0619g D. 7.26g E. 1.11g PA = 5.00 atm k = 3.3 x 10-2 mole L-1atm-1 V = 0.375L Henry’s Law: SA=kPA S = 3.3 x 10-2 mole L-1atm-1 x 5 atm = 0.165M 0.165 mol/L x 0.375L x 44.01g/mol = 2.72g CO2 CHEM 162-2009 EXAM I CHAPTER 12 - PROPERTIES OF SOLUTIO–S Vapor pressure of solutions calculations 3. What mass of sodium sulfate must be added to 150.0 g of water at 22oC to reduce the vapor pressure of the resulting solution to 20.0 Torr? The vapor pressure of pure water at 22oC is 23.0 Torr. Assume the sodium sulfate is completely dissociated in solution. A. 18.5g B. 200g C. 417g* D. 139g E. 0.416g *Note: Answer options were replaced just prior to the exam. The correct answer option is C, 59.1g.

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FW Na2SO4 = (22.99x2 + 32.07 + 16.00x4) = 142.05g/mol Na2SO4 → 2Na+ + SO4

2- i = 3 150.0 g water 150.0gwater/18.02gmol-1 = 8.32 mol H2O Psoln = 20.0 Torr Po

solvent = 23.0 Torr ?gNa2SO4 Psoln = iXsolventPo

solvent iXsolvent = ((i x nsolvent)/((i x nsolvent) + (i x nsolute))) Psoln = ((i x nsolvent)/((i x nsolvent) + (i x nsolute))) x Po

solvent (20.0Torr) = ((1x8.32molH2O)/((1x8.32molH2O) + (3 x (gNa2SO4/142.05gmol-1))) x 23.0 Torr X = 59.09g CHEM 162-2009 EXAM I CHAPTER 12 - PROPERTIES OF SOLUTIO–S Vapor pressure of solutions calculations 11*. The solubility of oxygen in water is 4.00 ppm at 25oC. What is Henry’s Law constant for water at this temperature? [The mole fraction of oxygen in air with a total pressure of 1 atm is about 0.21 and that the density of water is 1.00g/cm3] *This problem probably could have been stated better. It should have said, “The solubility of oxygen in water is 4.00 ppm at 25oC when the air pressure is 1 atm.” A. 5.95 x 10-4 Matm-1 B. 5.95 x 10-7 Matm-1 C. 1.90 x 10-5 Matm-1 D. 37.2 Matm-1 E. 1.90 x 10-2 Matm-1 S = 4gO2/1000000g solution Pair = 1 atm Mole fraction of O2 in air = 0.21 Henry’s Law: SA = kPA Answer option units suggests that the S should be in units of M. S = 4gO2/1000000gsoln → molO2/Lsoln Numerator: 4gO2/32gmol-1 = 0.125 mol O2 Denominator: 1000000g solution x (1cm3/1.00g) x (1 mL/cm3) x (1L/1000 mL) = 1000 Lsolution Numerator/Denominator = 0.125molO2/1000Lsolution = 0.000125MO2 solution Since, according to Dalton’s law, moles are directly related to pressure, then: 0.21 mol/1.00 mol total = 0.21 atm/1.00 atm total

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0.000125MO2 = k x 0.21atm O2 k = 5.95 x 10-4Matm-1 CHEM 162-2009 EXAM I CHAPTER 12 - PROPERTIES OF SOLUTIO–S Vapor pressure of solutions calculations 24. At 313 K the vapor pressure of pure methanol is 303 Torr and the vapor pressure of pure propanol is 44.6 Torr. A solution of these two substances is prepared and it has a vapor pressure of 174 Torr at 313 K. The mole fraction of methanol in the solution is A. 0.501 B. 0.151 C. 0.574 D. 0.871 E. 0.750 Po

methanol = 303 Torr Po

propanol = 44.6 Torr Psoln = 174 Torr Psoln = iXsolventAPo


Let X = mole fraction of methanol, and 1-X = mole fraction of propanol 174 Torr = (X x 303 Torr) + ((1-X) x 44.6) Torr X = 0.501

CHEM 162-2009 FINAL EXAM CHAPTER 12 - PROPERTIES OF SOLUTIO–S VAPOR PRESSURE OF SOLUTIONS CALCULATIONS 26. How many moles of benzene, C6H6, must be added to 2.00 moles of carbon tetrachloride, CCl4, in order to reduce the vapor pressure of carbon tetrachloride by 20%. A. 0.25 mol B. 0.50 mol C. 1.00 mol D. 1.50 mol E. 1.75 mol Psoln = iXsolventPo

solvent Psoln = (insolv/(insolv + insolute)) x Po

solvent CCl4 = solvent For pure CCl4, let Psoln = 1 atm, and Po

solvent = 1 atm (The exact value of the pressure is irrelevant.): For the 20% decrease in the vapor pressure above the solution:

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0.80 = (2.00/(2.00 + X)) x 1 X = 0.50 mol B

27 Chem 162-2008 Final Exam + Answers Chapter 12 – Physical Properties of Solutions Vapor pressure of solutions calculations

Henry’s law constant for O2(g) is 2.10×10-4 atmwaterg

Og 2 . Fish need an O2(aq)

concentration of 4.0 ppm (by mass) to survive. What is the minimum pressure of O2(g) to maintain this concentration if the density of water is 1.000g/mL?

A. 0.019 Torr B. 14.5 Torr C. 0.609 Torr D. 462.7 Torr E. 152.0 Torr Convert k into the same units as the concentration (ppm). k = (2.10x10-4 gO2/gwateratm) x ((1x106)/(1x106)) = 210gO2/106gwateratm SA = kPA (4.0gO2/106gwater) = (210gO2/106gwateratm) x PA PA = 0.019 atm 0.019atm x (760 torr/atm) = 14.4 Torr

B

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6 CHEM 162-2008 EXAM I CHAPTER 12 – PROPERTIES OF SOLUTIONS VAPOR PRESSURE OF SOLUTIONS CALCULATIONS (Henry’s Law, Raoult’s Law) An aqueous solution of O2 and water has an [O2] concentration of 4.57×10-4M. What is the partial pressure of O2 over this solution? kH for O2 = 2.18×10-3 M/atm

A. 760 Torr B. 7.57×10-4 Torr C. 0.210 Torr D. 4.77 Torr

E. 159 Torr

Henry’s Law: SA = kPA 4.57 x 10-4M = 2.18 x 10-3M/atm x Pox Pox = 0.2096 atm 0.2096 atm x 760 Torr/1 atm = 159 Torr

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18 CHEM 162-2008 EXAM I CHAPTER 12 – PROPERTIES OF SOLUTIONS VAPOR PRESSURE OF SOLUTIONS CALCULATIONS (Henry’s Law, Raoult’s Law) What is the mole fraction of acetone (C3H6O) in the gas phase resulting from a solution of 25.0 g acetone and 25.0 g chloroform (CHCl3) in the liquid phase? The vapor pressure of C3H6O is 216 Torr and CHCl3 is 157 Torr.

A. 0.401 B. 0.673

C. 0.739 D. 0.327 E. 0.500

For two volatile components: Psoln = iXsolventAPosolventA + iXsolventBPo

solventB iXA = inA/(inA + inB) mol = g/MW molacetone = 25.0g/58.09gmol-1 = 0.43037 mol acetone molchloroform = 25.0g/119.37gmol-1 = 0.2094 mol chloroform iXacetone = (1 x 0.43037)/((1 x 0.43037) + (1 x 0.2094)) = 0.6727 iXchloroform = (1 x 0.2094)/((1 x 0.2094) + (1 x 0.43037)) = 0.3273 Pacetone = iXacetonePo

acetone Pacetone = 0.6727 x 216 Torr = 145.3 Torr Pchloroform = 0.3273 x 157 Torr = 51.4 Torr Combination gas law: P1V1/n1T1 = P2V2/n2T2 At constant V and T, P1/P2 = n1/n2 Therefore, there is a direct relationship between pressure and moles. This is also known as Dalton’t law. Xacetonegas = 145.3Torr/(145.3Torr + 51.4Torr) = 0.739

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14 Chem 162-2007 Final exam + answers Chapter 12 – Properties of Solutions Vapor Pressure of Solutions Calculations Raoult’s Law The vapor pressure of pure benzene (C6H6) and toluene (C7H8) at 25 °C are 95.1 and 28.4 Torr respectively. A solution is prepared with a mole fraction of toluene of 0.750. Assuming the solution to be ideal, determine the mole fraction of toluene in the vapor phase.

A. 0.473 B. 45.1 C. 0.527 D. 0.896 E. 21.3

Po

B = 95.1 Torr Po

T = 28.4 Torr XT = 0.750 Psoln = PB + PT = iXsolventAPo


XB = 1.000 – 0.750 = 0.250 PB = iXsolventBPo

solventB = 0.250 x 95.1 = 23.78 Torr PT = iXsolventTPo

solventT = 0.750 x 28.4 = 21.30 Torr Mole fraction of gases is proportional to their vapor pressures. XB = 23.78/(23.78 + 21.30) = 0.528 XT = 21.30/(21.30 + 23.78) = 0.472

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4. CHEM 162-2007 EXAM I + ANSWERS CHAPTER 12 – PROPERTIES OF SOLUTIONS VAPOR PRESSURE OF SOLUTIONS CALCULATIONS (RAOULT’S LAW) The vapor pressure of pure benzene (C6H6) and toluene (C7H8) at 25°C are 95.1 and 28.4 mm Hg respectively. A solution is prepared with a mole fraction of toluene of 0.750. Determine the total pressure above the solution, in mm Hg. Assume the solution to be ideal. A) 61.8 B) 66.7 C) 123.5 D) 45.1 E) 78.4 Raoult’s law: iXsolventAPosolventA + iXsolventBPosolventB Pobenzene = 95.1 mm Potoluene = 28.4 mm Xtoluene = 0.750 Psoln = iXsolventAPosolventA + iXsolventBPosolventB Since the van’t Hoff factor for benzene and toluene are both 1, then the mole fraction may be used rather than the particle fraction. Xtoluene = 0.750 Xbenzene = 1- 0.750 = 0.250 Psoln = (0.250 x 95.1) + (0.750 x 28.4) = 45.08 mm

D

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3. CHEM 162-2007 EXAM I + ANSWERS CHAPTER 12 - PROPERTIES OF SOLUTIONS VAP–R PRESSURE OF SOLUTIONS CALCULATIONS (HENRY’S LAW) The solubility of O2(g), in water at 25°C is 3.16 mL O2/100 g H2O when the gas pressure is maintained at 1 atm. What pressure of O2 (g) would be required to produce a saturated solution containing 10.9 mL O2/100 g H2O? A) 4.45 atm

B) 3.45 atm C) 0.290 atm D) 3.16 atm E) 10.9 atm S1 = 3.16 mL/100gH2O P1 = 1 atm ?P2 @ 10.9 mL/100 g H2O Henry’s Law = S = kP 3.16 mL/100 g = k x 1 atm k = 0.0316mL/g S2 = kP2 (10.9mL/100g) = 0.0316mL/g x P2 P2 = 3.45 atm This problem may also be solved with a simple proportion. (3.16 mL O2)/1 atm = (10.9 mL O2)/X atm X = 3.45 atm

B

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COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE CHANGES) CONCEPTS 12 Chem 162-2011 Exam I

Hill & Petrucci et al. 4th edition Chapter 12 – Properties of Solutions Colligative properties (boiling point, freezing point, osmotic pressure changes) concepts All colligative properties A. depend on the properties of both the solute and the solvent. B. depend only on the properties of the solvent and the amount of the solute. C. depend only on the properties of the solute and the amount of the solvent. D. depend only on the amounts of solute and solvent. E. depend only on the temperature of the solution. A. False. If one looks at the formula, ΔT = -Kfimsolute, one can say that msolute depends on the concentration of the solute, not the property of the solute. Therefore, the colligative property of freezing point depression doesn’t depend on the property of the solute. However, one can also argue that option “A” is true. “i” depends on the dissociative property of the solute. Some solutes have an i of 1, 2, 3, etc. Kf is a property of the solvent. Hence, freezing point depression does depend on the properties of both the solute and the solvent. B. True. If one looks at the formula, ΔT = -Kfimsolute, one can say that Kf depends on the property of the solvent, and msolutedepends on the moles (i.e., the amount) of solute. However, one can also argue against this option being true. (a) “i” depends on the dissociative property of the solute. Some solutes have an i of 1, 2, 3, etc. This is a property of the solute. The property of the solute is not listed in option B. (b) The “amount” of the solute is an incorrect term. Molality is not amount; it is concentration of the solute. (c) One can argue that it is not only the amount of the solute, but it is the amount of the solute and the amount of the solvent (actually the concentration of the solute and the relative concentration of the solvent). Bottom line is that the choice of “B” as the answer is quite weak. C. False. It depends on the properties of the solute, but not the amount of solvent. D. False. It also depends on the properties of the solute. E. False. It doesn’t depend on the temperature of the solution.

B

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14 Chem 162-2011 Exam I Hill & Petrucci et al. 4th edition Chapter 12 – Properties of Solutions Colligative properties (boiling point, freezing point, osmotic pressure changes) concepts Which of the following aqueous solutions has the lowest freezing point?

A. 0.1m glucose B. 0.1m NaCl

C. 0.1m Na2SO4 D. 0.1m NaHSO4 E. 0.1m CH3COOH

Freezing-point depression: ΔT = Tf - Ti = -Kfimsolute Since K is a constant, and m is a consta–t, the one with the lowest freezing point is the one with the largest “i” value. A. Glucose → Glucose Organic molecules, other than organic acids, don’t dissociate. i =1 B. NaCl → Na+ + Cl- i = 2 C. Na2SO4 → 2Na+ + SO4

2- i = 3 D. NaHSO4 → Na+ + HSO4

- HSO4- is a very weak acid, and dissociates very

slightly i = 2 (perhaps 2.01? 2.02?) E. CH3COOH → CH3COO- + H+ Organic acids dissociate very weakly. i ≈ 1.01

A

CHEM 162-2010 EXAM I Chapter 12 - Properties of Solutions Colligative properties (boiling point– freezing point, osmotic pressure changes) concepts 1. Rank the following 0.30 m aqueous solutions from lowest to highest freezing points. NaCl, C6H5OH, CaCl2, CH3COOH A. NaCl, C6H5OH, CaCl2, CH3COOH B. CaCl2, NaCl, CH3COOH, C6H5OH C. C6H5OH, CH3COOH, NaCl, CaCl2 D. CH3COOH, CaCl2, NaCl, C6H5OH E. NaCl, CH3COOH, C6H5OH, CaCl2 NaCl → Na+ + Cl- i = 2

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C6H5OH is an organic molecule (not an organic acid), so doesn’t dissociate; i = 1. (Actually, phenol does dissociate, but less than the weak acid CH3COOH, so if CH3COOH is 1.01, then let’s call phenol approximately 1.0001.) CaCl2 → Ca2+ + 2Cl- i = 3 CH3COOH CH3COO- + H+ i is a little more than 1, perhaps 1.01, since weak acids dissociate slightly. H2O, itself, would have the highest freezing point. The substance with the greatest i would have the lowest freezing point. Therefore, CaCl2 < NaCl < CH3COOH < C6H5OH CHEM 162-2010 EXAM I Chapter 12 – Properties of Solutions Colligative properties (boiling point, freezing point, osmotic pressure changes) concepts 4. The value of the boiling point elevation constant, Kb, depends on: A. The identity of the solvent B. The identity of the solute C. The identities of both the solvent and the solute D. The concentration of the solution E. The van’t Hoff factor, i. The value of Kb depends solely on the identity of the solvent. For example, H2O has a Kb of 0.512; benzene has a Kb of 2.53. CHEM 162-2009 EXAM I CHAPTER 12 - PROPERTIES OF SOLUTIONS C–lligative properties (boiling point, freezing point, osmotic pressure changes) concepts 9. Assume you had 0.30m aqueous solutions of the compounds listed below. Which of the compounds below would be most effective in removing ice from a sidewalk? A. CaCl2 B. C6H12O6 C. KNO3 D. NaBr E. C12H22O11 The one that would be the most effective is the one that dissociates into the largest number of particles, that is, the one with the largest van’t Hoff value (i). A. CaCl2 → Ca2+ + 2Cl- i = 3 B. C6H12O6 → NR i = 1

48

C. KNO3 → K+ + NO3- i = 2

D. NaBr → Na+ + Br- i = 2 E. C12H22O11 → NR i = 1 CHEM 162-2009 EXAM I CHAPTER 12 - PROPERTIES OF SOLUTIONS C–lligative properties (boiling point, freezing point, osmotic pressure changes) concepts 10. All colligative properties A. depend on the properties of both the solute and the solvent. B. depend on the properties of the solute. C. depend on the amount of the solute. D. depend on the amounts of solvent. E. depend only on the temperature of the solution. There is a difference of opinion between the Course Coordinator, and me on this choice. Option C (The Course Coordinator’s choice): The Course Coordinator says that C is the correct answer because if you change the amount of solute it will correspondingly change the colligative properties. For example, if you increase the moles of solute then the boiling point elevation will correspondingly increase. My response to that is that if you increase the amount of solute, and correspondingly increase the amount of solvent, then the colligative property won’t change. That is, it is not the amount of solute that is important, but it is the concentration of solute that is important. Note that all four colligative properties contain concentration terms, not mole terms. One can argue, however, that the C answer option is acceptable, if you assume that there are no other changes, i.e., changing the moles of solute, and making no other changes, is equivalent to changing the concentration of solute. Option D: Changing the concentration of the solute corresponds to changing the concentration of the solvent. If one is increased, the other is decreased. If one is decreased, the other is increased. Since the model for all colligative properties is related to the solvent concentration, then if the amounts (moles) of solvent is equivalent to the concentration of solvent, then D must be considered as an acceptable answer, also. Option B (my choice). The Course Coordinator said that it is the number of particles, not the properties of the particles, that is important. He says that that is why option B is not an acceptable answer. I agree that it is the number of particles, not the properties of the particles, that is important. But the problem is not referring to the properties of the particles. It is referring to the properties of the solute. Dissociation of a solute into its particles in water is a chemical property. The chemical properties of urea, NaCl and CaCl2 are different, in that urea doesn't dissociate, NaCl dissoci’tes into two ions, and CaCl2 dissociates into three ions. The greater the degree of dissociation, the greater the effect on colligative properties. Hence, the chemical properties of the solute do affect the colligative properties of the solution. Option A. I’m not sure that option A is not acceptable. Assuming I am correct with regard to option B, then accepting option A means the colligative properties also depend on the properties of the solvent. Let’s consider boiling point elevation and freezing point depression. Every solvent has a different Kb,Kf and Po. Doesn’t that mean that the physical properties of these solvents are different, and that these

49

properties affect the boiling point elevation, freezing point depression, and vapor pressure? I think so. (I’m not certain of how the solvent affects the osmotic pressure.) Option E is clearly not acceptable. Although Raoult's law and osmotic pressure’depend on the temperature of the solution, they also dependent on other things. For example, all colligative properties depend on the concentration of solute.

3 CHEM 162-2008 EXAM I CHAPTER 12 - PROPERTIES OF SOLUTIONS C–LLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE CHANGES) CONCEPTS Osmometry uses osmotic pressure to measure the molar mass of large compounds. Why is this technique generally preferred over either freezing point depression or boiling point elevation?

A. The van’t Hoff factor for dissociation of ions plays a more important role in freezing point depression than it does in osmotic pressure. B. The Kf and Kb of the solvent is not known for enough materials.

C. The temperature change from freezing and boiling of solutions is not large enough for an accurate measurement. D. Freezing and boiling take too long to measure. E. Freezing point and boiling point measurements require ionic materials, but osmotic pressure can work with any material.

A. False. The van’t Hoff factor for dissociation of ions plays an equal role in freezing point depression and osmotic pressure. B. False. The Kf and Kb of the solvent is known for a huge number of materials. C. True. The temperature change from freezing and boiling of solutions is small. Since it depends on molality, and molality will be very small for a small quantity of a large molecular weight compound, then the temperature changes in freezing point depression and boiling point elevations will be very small, perhaps too small to be measurable. However, the change in osmotic pressure during osmotic pressure measurements is very large. So even if the molality of the unknown substance is small, the osmotic pressure change will be large, and easy to measure. D. False. Freezing and boiling point changes take minutes to measure. Even if it took long, the time of analyses that provides critical information is not a factor in using these tools. E. False. Freezing and boiling point measurements work with any materials, just as osmotic pressure measurements work with any materials.

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26 Chem 162-2007 Final exam + answers

Chapter 12 – Properties of Solutions Colligative Properties Concepts Freezing point

Assume you had 0.30 m aqueous solutions of the compounds listed below. Which

of these solutions would be most effective in removing ice from a sidewalk?

A. CaCl2

B. C6H12O6 C. KNO3 D. NaBr

E. C12H22O11 The substance that is the most effective in removing ice from a sidewalk would be the one with the lowest freezing temperature. Freezing-point depression: ΔT = Tf - Ti = -Kfimsolute Sinc– Kf is the same for each (because the solvent, H2O, is the same for each), and m is the same for each (0.30m), and Ti is the same for each (0oC), then the only parameter that is different, which could affect the value of Tf, is i, the van’t Hoff factor. The substance with the largest van’t Hoff factor is the substance that would provide the lowest freezing temperature.

A. CaCl2 → Ca2+ + 2Cl- i = 3 B. C6H12O6 → C6H12O6 i = 1 C. KNO3 → K+ + NO3

- i = 2 D. NaBr → Na+ + Br- i = 2 E. C12H22O11 → C12H22O11 i = 1

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8. CHEM 162-2007 EXAM I + ANSWERS CHAPTER 12 - PROPERTIES OF SOLUTIONS C–LLIGATIVE PROPERTIES CONCEPTS (VAN’T HOFF) Which of the following aqueous solutions has the lowest freezing point? A) 0.1 m glucose B) 0.1 m NaCl

C) 0.1 m Na2SO4 D) 0.1 m NaHSO4 E) 0.1 m CH3COOH Freezing-point depression: ΔT = Tf - Ti = -Kfimsolute Since Ti– Kf and molality are all the same, the greatest ∆T (i.e., the lowest freezing point [Tf]) will be the largest “i”. (A) glucose: an organic molecule; it doesn’t dissociate; i = 1 (B) NaCl: NaCl → Na+ + Cl-; i = 2 (C) Na2SO4: Na2SO4 → 2Na+ + SO4

2-; i = 3 (D) NaHSO4: NaHSO4 → Na+ + HSO4

-; i = 2

(E) CH3COOH: CH3COOH ←→ CH3COO- + H+; CH3COOH dissociates only about 1%;

hence i ≈ 1.01.

C

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COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE CHANGES) CALCULATIONS 5 Chem 162-2011 Exam I

Hill & Petrucci et al. 4th edition Chapter 12 – Properties of Solutions Colligative properties (boiling point, freezing point, osmotic pressure changes) calculations 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 Tf - Ti = Kbimsolute Tf - Ti = Kbi x (molesolute/kgsolvent) Tf - Ti = Kbi((masssolute/MWsolute)/kgsolvent) (77.85oC–– 76.50oC) = (5.03K/m) x ((2.00g/X)/0.015kg) X = 496.8g/mol

E

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11 Chem 162-2011 Exam I Hill & Petrucci et al. 4th edition Chapter 12 – Properties of Solutions Colligative properties (boiling point, freezing point, osmotic pressure changes) calculations What mass of glucose (C6H12O6) in grams, must be added to 125 g H2O to lower the freezing point to −2.45°C? (Kf = 1.86°C m−1)

A. 180 g B. 23.7 g C. 237 g

D. 29.7 g E. 1.32 g

Freezing-point depression: ΔT = Tf - Ti = -Kfimsolute Tf - Ti = -Kfi(molesolute/kgs–lvent) Tf - Ti = -Kbi((ma–ssolute/MWsolute)/kgsolvent) MW C6H1–O6 = 156.18 (-2.45oC – 0oC) = -(1.86oC/m-1) x 1 x ((Xg/180.18gmol-1)/0.125kg) X = 29.7g

D

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25 Chem 162-2011 Exam I Hill & Petrucci et al. 4th edition Chapter 12 – Properties of Solutions Colligative properties (boiling point, freezing point, osmotic pressure changes) calculations 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. 22.1 g D. 44.6 g E. 132 g Osmotic pressure: πV = inRT πV = i(g/MW)RT 1.50 atm x 2.00L = 1 x (g/180.2gmol-1) x 0.08206Latmdeg-1mol-1 x 298K X = 22.1g

E

CHEM 162-2010 FINAL EXAM CHAPTER 12 - PROPERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES (BO–LING POINT, FREEZING POINT, OSMOTIC PRESSURE CHANGES) CALCULATIONS 32. When 0.64g of the heart stimulant epinephrine was dissolved in 36.00g of benzene, the normal freezing point of the resulting solution was found to be 5.03oC. What is the approximate molar mass of epinephrine? Tf of benzene = 5.53oC; Kf for benzene = 5.12oC-kg/mol A. 180 g/mol B. 970 g/mol C. 510 g/mol D. 390 g/mol E. 96 g/mol Freezing-point depression: ΔT = Tf - Ti = -Kfimsolute Tf - Ti = -Kfimolsolute/kgsol–ent Tf - Ti = -Kfi((mass–olute)/(MWsolute))/kgsolvent 5.03–– 5.53 = -5.12 x 1 x ((0.64/MW)/0.036) MW = 182g/mol CHEM 162-2010 EXAM I Chapter 12 - Properties of Solutions

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Colligative properties (bo–ling point, freezing point, osmotic pressure changes) calculations 5. If a 1.00x10-6 g of a protein is dissolved in 1.000 μL of water and exerts an osmotic pressure of 0.372 Torr at a temperature of 25.0oC, what is the molar mass of the protein? A. 5.52x106 g/mol B. 5.52 g/mol C. 65.8 g/mol D. 5.00x104 g/mol E. 6.58x104 g/mol 1.00x10-6 g of protein 1.000 μL of water x (10-6 L/μL) = 10-6 L 0.372Torr 298K ?MW πV = inRT πV = i(g/MW)RT MW = igRT/πV MW = 1 x 1.00x10-6 g x 0.08206Latmdeg-1mol-1 x 298K/0.372Torr x (1atm/760Torr) x 1x10-6L = 5.00x104 g/mol CHEM 162-2010 EXAM I Chapter 12 - Properties of Solutions Colligative properties (bo–ling point, freezing point, osmotic pressure changes) calculations 22. Reverse osmosis can be used to purify drinking water. What is the minimum pressure that must be applied to a 0.13M KCl solution at 25oC to cause solvent to be removed from the mixture? A. 2420 Torr B. 6.36 Torr C. 4830 Torr D. 644 Torr E. 322 Torr 0.13M KCl 298K

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The minimum pressure would be the amount of pressure to counteract the osmotic pressure. Hence, the minimum pressure equals the osmotic pressure. πV = inRT π = i x (n/V)RT π = i x MRT i for KCl = 2 π = 2 x 0.13M x 0.08206Latmdeg-1mol-1 x 298K = 6.36 atm 6.36 atm x 760 Torr/atm = 4832 Torr CHEM 162-2009 EXAM I CHAPTER 12 - PROPERTIES OF SOLUTIONS Colligative properties (bo–ling point, freezing point, osmotic pressure changes) calculations 2. What is the osmotic pressure of 0.010M CaCl2(aq) solution at 25oC, assuming ideal behavior? A. 558 Torr B. 186 Torr C. 5.65x104 Torr D. 1.88x104 Torr E. 0.734 Torr 0.010M CaCl2 298K Osmotic pressure: πV = inRT π = (n/V)iRT π = iMsoluteRT CaCl2 → Ca2+ + 2Cl- i = 3 π = 3 x 0.010M x 0.08206Latmdeg-1mol-1 x 298K = 0.734 atm 0.734 atm x 760 Torr/atm = 558 Torr CHEM 162-2009 EXAM I CHAPTER 12 - PROPERTIES OF SOLUTIONS Colligative properties (bo–ling point, freezing point, osmotic pressure changes) calculations 5. A 2.50m solution of a certain HClO2* boils at 101.5oC. What fraction of the acid molecules have dissociated [f = ((i – 1)/(imax – 1)), Kb = 0.51oC/m] *A better way of writing this is to say, “a certain aqueous solution of HClO2.” A. 1.18 B. 0.50 C. 0.080 D. 1.00

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E. 0.18 2.50 mol HClO2/kg water BP = 101.5oC Kb = 0.51oC/m ?fraction of acid dissociated Boiling-point elevation: ΔT = Tf - Ti = Kbimsolute HClO2 + H2O ←→ H–O+ + ClO2

- 101.5oC – 100.0oC = 0.51oC/m x i x 2.50m i = 1.176 imax = 2.0 f = ((i – 1)/(imax – 1)) f = ((1.176 – 1)/(2 – 1)) = 0.176 CHEM 162-2009 EXAM I CHAPTER 12 – PROPERTIES OF SOLUTIONS Colligative properties (boiling point, freezing point, osmotic pressure changes) calculations 14*. A of a solution of toluene in cyclohexane if it freezes at 4.65oC? What is the boiling point of this solution? *The question was reworded in an errata sheet prior to the exam. The rewording was something like, “If a solution of toluene in cyclohexane freezes at 4.65oC, what is the boiling point of this solution?”

Data for Cyclohexane Tf 6.55oC Kf 20.0oC/m Tb 80.7oC Kb 2.79oC/m

A. 80.0oC B. 81.3oC C. 100.6oC D. 81.0oC E. 94.3oC Freezing-point depression: ΔT = Tf – Ti = -Kfimsolute 4.65oC – 6.55oC = -20.0oC/m x 1 x msolute msolute = 0.095 Boiling-point elevation: ΔT = Tf – Ti = Kbimsolute Tf – 80.7oC = 2.79oC/m x 1 x 0.095m Tf = 80.07oC

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CHEM 162-2009 EXAM I CHAPTER 12 – PROPERTIES OF SOLUTIONS Colligative properties (boiling point, freezing point, osmotic pressure changes) calculations 19. What is the freezing point of a CsCl(aq) solution that is 1.50% by mass? The Kf for water is 1.86oC/m. Assume ideal conditions. A. -0.17oC B. -55.8oC C. -0.34oC D. -0.55oC E. -0.65oC ΔT = Tf – Ti = -Kfimsolute 1.50gCsCl/100gsoln → molCsCl/kgsolvent MW CsCl = 132.91g + 35.45g = 168.36g Numerator: 1.50gCsCl/168.36gmol-1 = 8.909 x 10-3 mol Denominator: gsolute + gsolvent = gsolution gsolvent = gsolution – gsolute gsolvent = 100gsolution – 1.50gCsCl = 98.50gsolvent 98.50g x 1kg/1000g = 0.0985 kg solvent Numerator/Denominator: (8.909 x 10-3 mol)/0.0985 kg solvent = 0.09045 m ΔT = Tf – Ti = -Kfimsolute Tf – 0oC = -1.86oC/m x 2 x 0.09045m Tf = -0.336oC CHEM 162-2009 FINAL EXAM CHAPTER 12 – PROPERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE CHANGES) CALCULATIONS 3. When 0.200 mol of substance X is dissolved in 500.0 g of water, the freezing point of the resulting solution is 270.92K. Which one of the following substances could be X? Kf = 1.86oC/m and the freezing point of water is 273.15K. A. C6H12O6 B. NaCl C. Na3PO4 D. CaCl2 E. Mg3(PO4)2 Identification of the substance can be determined by the value of the van’t Hoff factor, “i”. 0.400 mol dissolved in 1 Kg = 0.400 molal ΔT = Tf – Ti = -Kfimsolute (270.92 – 273.15) = (-1.86 x i x 0.400) i = 2.997 = CaCl2 = D

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CHEM 162-2009 FINAL EXAM CHAPTER 12 – PROPERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE CHANGES) CALCULATIONS 14. Dichloracetic acid, Cl2CHCOOH (molar mass 129) partially ionizes into H+ and Cl2CHCOO- ions in water solution. When 6.5 grams of this acid is dissolved in 100 grams of water, the freezing point of the resultant solution is 272.01 K. The freezing point of pure water is 273.15 K and Kf = 1.86 K/m for water. Calculate the percent ionization of dichloroacetic acid. A. 60.8% B. 43.2% C. 32.4% D. 10.2% E. 21.6% Cl2CHCOOH + H2O ←→ H3O+ + Cl2CHCOO- (6.5g/(129 g/mol)) = 0.05039 mol/100g = 0.5039m ∆T = Tf – Ti = -Kfim (272.01 – 273.15) = -1.86 x i x 0.5039m i = 1.216 Formula in back of exam: % dissociation = (i – 1)/(imax – 1) x 100% = ((1.216-1)/(2-1)) x 100 = 21.6% E CHEM 162-2009 FINAL EXAM CHAPTER 12 – PROPERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE CHANGES) CALCULATIONS 31. Dimethylglyoxime, DMG, is an organic compound used to test for aqueous nickel(II) ions. A solution prepared by dissolving 65.0 g of DMG in 375 g of ethanol boils at 80.3oC. What is the molar mass of DMG? Kb for ethanol is 1.22oC/m, boiling point of pure ethanol = 78.5oC A. 44.1 g/mol B. 65.8 g/mol C. 131.6 g/mol D. 117 g/mol E. 553 g/mol Tf – Ti = Kbim (80.3 – 78.5) = +1.22 x 1 x m m = 1.48 (65g DMG/375gEtOH) = (XgDMG/1000gEtOH) X = 173.3 g DMG moles = g/MW 1.48 mol = 173.3 g DMG/MW

60

MW = 117.1g/mol D

7 CHEM 162-2008 EXAM I CHAPTER 12 – PROPERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE CHANGES) CALCULATIONS The solution formed when 1.00 μg sample of a molecular compound is dissolved in 0.1000μL of water exerts an osmotic pressure of 5.8 Torr at 25 °C. What is the molar mass of the compound?

A. 420 g/mol B. 350 g/mol C. 4200 g/mol D. 550000 g/mol

E. 32000 g/mol

πV = inRT πV = igRT/MW (5.8 Torr x (1 atm/760 Torr) x 0.1000μL x (1L/106μL)) = (1 x (1.00μg x 1g/106μg) x 0.08205Latmdeg-1mol-1 x 298K)/(MW) MW = 32,039 g/mol

61

23 CHEM 162-2008 EXAM I CHAPTER 12 – PROPERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE CHANGES) CALCULATIONS A 0.0300m solution of PbCl2( aq) freezes at -0.14 °C. What fraction of the PbCl2 is dissociated? Kf for water is 1.86 °C/m.

A. 0.75 B. 0.40 C. 0.66 D. 2.50 E. 1.50

Freezing-point depression: ΔT = Tf - Ti = -Kfimsolute -0.14 - 0 = -–.86 x i x 0.0300 i = 2.51 Pb–l2 ←→ Pb2+ + 2Cl- PbCl2 ←→ Pb2+ + 2Cl- Initial 1.000 0 0 Change -X +X +2X Equilibrium 1.000 – X +X +2X

The total particles = 2.51. (1.000 – X) + X + 2X = 2.51 X = 0.755; therefore, 75.5% of the PbCl2 dissociated. Therefore, PbCl2 = 0.245; Pb2+ = 0.755; Cl- = 1.51 Total particles = 2.51 or use the following formula from the exam formula sheet: %Dissociation = ((i - 1)/(imax -1)) x 100% ((2.51 – 1)/(3 – 1)) x 100% = 75.5% dissociation

62

1 CHEM 162-2008 EXAM I CHAPTER 12 – PROPERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE CHANGES) CALCULATIONS What is the boiling point of a solution of CHCl3 in C6H6 that freezes at 5.22 °C? Data for C6H6: Kf = 5.12 °C/m, Kb = 2.53 °C/m Tf = 5.53 °C, Tb = 80.10 °C

A. 79.95 °C B. 80.15 °C C. 82.68 °C

D. 80.25 °C E. 77.52 °C

Boiling-point elevation: ΔT = Tf – Ti = Kbimsolute Freezing-point depression: ΔT = Tf – Ti = -Kfimsolute Plan: You’re not given enough information to find the Tfinal for the boiling point, because you don’t know the molality of the solute. But you have enough information to find the molality of the solute in the freezing point depression formula. Therefore, find the molality of the solute using the freezing point formula, then use that value of the molality in the boiling point formula to find the boiling point. Tf – Ti = -Kfimsolute 5.22o – 5.53o = -5.12oC/m x 1 x msolute msolute = 0.0605 Tf – Ti = Kbimsolute Tf – 80.10o = 2.53oC/m x 1 x 0.0605 Tf = 80.25o

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27 Chem 162-2007 Final exam + answers Chapter 12 – Properties of Solutions Colligative Properties Calculations Osmotic Pressure A new, molecular compound produces an osmotic pressure of 0.0964 atm when 3.15 g of the compound is dissolved in enough water to make 150. mL at 25 °C. What is the molecular mass of the compound?

A. 447g/mol B. 5330 g/mol C. 45,300 g/mol D. 49.5 g/mol E. 1690 g/mol

0.0964 atm 3.15 g in 0.150L 298K ?MW Osmotic pressure: πV = inRT πV = igRT/MW MW = igRT/πV MW = (1 x 3.15g x 0.08205Latmdeg-1mol-1 x 298K)/(0.0964atm x 0.150L) = 5326 g/mol

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16 Chem 162-2007 Final exam + answers Chapter 12 – Properties of Solutions Colligative Properties Calculations Freezing Point What is the freezing point of a 0.100 M aqueous solution of a weak acid HX with a Ka of 1.25×10-4? Kf of water = 1.86 °C kg/mol and assume the solution has a density of 1.00 g/mL. [Hint i is not 1 in this problem.]

A. –0.186 °C B. –0.372 °C C. –0.192 °C D. –0.198 °C E. +0.192 °C

0.10M HX Ka = 1.25 x 10-4 Kf = 1.86oCkg/mol D = 1.00 g/mL ?Tf ΔT = Tf – Ti = -Kfimsolute Calculate i based on dissociation of HX

HX + H2O ←→ H3O+ + X-

HX + H2O ←→ H3O+ + X-

Initial 0.100M 0 0 Change -X +X +X Equilibrium 0.100-X +X +X

([H3O+][X-])/[HX] = 1.25 x 10-4 ([X][X])/[0.10-X] = (1.25 x 10-4) Simplify: ([X][X])/[0.10] = (1.25 x 10-4) X = 0.00353 This is a 3.53% dissociation of HX. Therefore, i = 1 + 0.0353 = 1.0353. Find m. 0.10mol/1000 mL solution = molarity Molality = mol/1000 g solvent 0.10mol/1000 mL solution x (1 mL/g) = 0.10mol/1000gsolution Not enough information provided to calculate molality, so assume that because the density of the solution is 1.00 g/mL, the solution is dilute enough so that molarity ≈ molality. (Note: The concentration of the aqueous solution should have been given in molality, not molarity, so that the student would not have to make this assumption.) Tf – 0oC = -1.86oCkg/mol x 1.0353 x 0.10 = -0.193 Tf = -0.193oC Note that if the van’t Hoff factor wasn’t considered: Tf – 0oC = -1.86oCkg/mol x 1 x 0.10 = -0.186

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CHAPTER 12 – PROPERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES CALCULATIONS (VAN’T HOFF) A 0.10 m aqueous solution of a strong electrolyte solute has a freezing point of -0.744oC. Which of the following is the solute? The Kf for water is 1.86oC/m. A. CaCl2 B. Al2(SO4)3 C. NaCl

D. FeCl3 E. Na2(SO4) ΔT = Tf - Ti–= -Kfimsolute m = 0.10 Tf = -7.44oC (-0.744 – 0.00) = -1.86oC/m x i x 0.10m i = 4.00 (A) CaCl2 → Ca2+ + 2Cl- i = 3 (B) Al2(SO4)3 → 2Al3+ + 3SO4

2- i = 5 (C) NaCl → Na+ + Cl- i = 2 (D) FeCl3 → Fe3+ + 3Cl- i = 4 (E) Na2(SO4) → 2Na+ + SO4

2- i = 3

D

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10. CHEM 162-2007 EXAM I + ANSWERS CHAPTER 12 - PR–PERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES CALCULATIONS (OSMOTIC PRESSURE) 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 298K. What is the molar mass of hemoglobin?

A) 6.7 × 104 B) 3.3 × 103 C) 3.3 × 104 D) 3.8 × 103 E) 5.6 × 102

2.05g hemoglobin/100.0 mL solution π = 5.69 mm @ 298K ?MW π = inRT/V π = i(g/MW)RT/V MW = (i x g x R x T)/(π x V) MW = (1 x 2.05g x 0.08205Latmdeg-1mol-1 x 298K)/(5.69 mm x (1atm/760 mm) x 0.100L) = 6.69 x 104 g/mol

A

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9. CHEM 162-2007 EXAM I + ANSWERS CHAPTER 12 - PR–PERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES CALCULATIONS (OSMOTIC PRESSURE) The freezing point of an aqueous solution of glucose (molar mass = 180 g/mol) was measured to be −1.52°C. Calculate the osmotic pressure of 100. mL of this solution at 25.0°C? (Kfwater = 1.86°C/m; density of the solution is 1.15g/mL) A) 2.00 atm B) 0.168 atm

C) 20.0 atm D) 16.8 atm E) 69.1 atm ΔT = Tf - Ti–= -Kfimsolute (-1.52 – 0.00) = -1.86°C/m x 1 x msolute msolute = 0.8172 m = 0.8172 mol glucose/kgsolvent Dsolution = 1.15g/mL Osmotic pressure: πV = inRT V = 100mL; therefore, find moles of solute in 100 mL solution 0.8172molglucose/1kgsolvent → molglucose/0.100Lsolution Numerator: OK; 0.8172 mol Denominator: gsolute + gsolvent = gsolution 0.8172molglucose x (180.18g/mol) = 147.24 g glucose 147.24g glucose + 1000gH2O = 1147.24gsolution 1147.24gsolution x (1mL/1.15g) = 997.6 mL solution 0.8172 mol glucose/0.9976 L solution = 0.8192 mol glucose/L solution = 0.08192 mol glucose/0.100L solution πV = inRT π = inRT/V π = (1 x 0.08192 mol x 0.08205Latmdeg-1mol-1 x 298K)/0.100L = 20.03 atm

C

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7. CHEM 162-2007 EXAM I + ANSWERS CHAPTER 12 - PR–PERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES CALCULATIONS (FREEZING POINT) An aqueous solution of a non volatile, non electrolyte solute boils at 100.32°C. What is its expected freezing point? (Kb = 0.512°C/m, Kf H2O = 1.86°C/m) A) 0oC

B) −1.2oC C) −0.32oC D) −0.83oC E) -2.1oC Boiling-point elevation: ΔT = Tf - Ti–= Kbimsolute (100.32o – 100.00o) = 0.512oC/m x 1 x msolute msolute = 0.625 Freezing-point depression: ΔT = Tf - Ti–= -Kfimsolute Tf – 0.00o = -1.86oC/m x 1 x 0.625m Tf = -1.16oC

B


CHAPTER 12 - PR–PERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES CALCULATIONS (FREEZING POINT DEPRESSION) What mass of glucose (C6H12O6) in grams, must be added to 125 g H2O to lower the freezing point to −2.45°C? (Kf = 1.86°C m−1) A) 180 B) 23.7 C) 237

D) 29.7 E) 1.32 ∆T = Tf – Ti = -Kf x i x msolute ∆T = -2.45 – 0.00 = -1.86oCm-1 x 1 x (mol/0.125kg) ∆T = -2.45 – 0.00 = -1.86oCm-1 x 1 x ((g/MW)/0.125kg) ∆T = -2.45 – 0.00 = -1.86oCm-1 x 1 x ((g/180.18)/0.125kg) g = 29.67

D

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41 Chem 162-2008 Final Exam + Answers Chapter 12 – Physical Properties of Solutions Colligative properties (Boiling point, freezing point, osmotic pressure changes) calculations A 0.150m aqueous solution of some weak acid boils at 100.077 °C. What is the % ionization of this acid? The boiling point elevation constant, Kb, for water is 0.512 °C/m A. 1.0026% B. 2.6×10-3% C. 1.0×10-7% D. 0.26% E. 0.0510% ∆T = Tf – Ti = Kbimsolute 100.077 – 100.000 = 0.512 x i x 0.150 i = 1.0026 (Example: If i = 1.03 then there are 0.97 particles of HA, 0.03 particles of H3O+ and 0.03 particles of A- at equilibrium, or a total of 1.03 particles. This is (0.03 x 100 =) 3% dissociation of the original 1 M solution.) 0.0026 x 100 = 0.26% dissociation of the weak acid.

D

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39 Chem 162-2008 Final Exam + Answers Chapter 12 – Physical Properties of Solutions Colligative properties (boiling point, freezing point, osmotic pressure changes) calculations How many liters of ethylene glycol antifreeze (C2H6O2) should be added to a car radiator containing 15.0 L of water, if the engine needs to be protected from freezing to −17.8 oC. Kf for water is 1.86 oC/m Density of ethylene glycol is 1.1 g/mL Density of water is 1.0 g/mL A. 15.0 L B. 10.5 L C. 8.1 L D. 20.1 L E. 5.5 L ∆T = Tf – Ti = -Kfimsolute -17.8 = -1.86 x 1 x mEG mEG = 9.570molEG/1000gH2O 9.570molEG/1000gH2O → ?LEG/15000mLH2O 9.570molEG x 62g/mol x (1mL/1.1g) x 1L/1000mL) = 0.539LEG 1000gH2O x 1mL/1.0g = 1000 mL H2O numerator/denominator = 0.539LEG/1000 mL H2O 0.539LEG/1000mLH2O = ?LEG/15000mLH2O 0.539LEG/1000mLH2O = 8.09LEG/15000mLH2O

C

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25 Chem 162-2008 Final Exam + Answers Chapter 12 – Physical Properties of Solutions Colligative properties (boiling point, freezing point, osmotic pressure changes) calculations Calculate the osmotic pressure of 0.010 M calcium chloride (CaCl2) solution at 25 oC. Assume an ideal solution.

A. 0.28 atm B. 0.12 atm C. 0.49 atm D. 0.98 atm E. 0.73 atm CaCl2 ←→ Ca2+ + 2Cl- i = 3 πV = inRT π = inRT/V π = i(MV)RT/V π = iMRT π = 3 x 0.010mol/L x 0.08205Latm/degmol x 298K = 0.734 atm

E

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MISCELLANEOUS (E.G., COLLOIDS) 15 Chem 162-2011 Exam I

Hill & Petrucci et al. 4th edition Chapter 12 – 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 Note: I think the word “suspension” is confusing, since mayonnaise is a “dispersion”, and not a “suspension”. However, the term “colloidal suspension” is frequently used to distinguish a colloid from a typical suspension. A typical suspension would be a mixture of sand and water immediately after shaking. A. Mayonnaise is a colloidal dispersion (aka colloidal suspension). B. Steel is a solid solution. C. Filtered sea water is a solution. D. Sand is a mixture. E. Diamond is an allotropic form of an element.

A

CHEM 162-2010 EXAM I Chapter 12 - Properties of Solutions Miscellaneous

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2. Colloids are A. solutions with solid solutes and liquid solvents B. any heterogeneous mixture C. mixtures with suspended particles small enough to scatter light D. Colored solutions E. solutions in which water is the solute instead of the solvent There are three options when a solute and solvent are mixed. One is solution, one is suspension and the last is precipitation. If the resulting mixture is homogeneous, and the resulting particles are so small that they do not scatter light, it is a solution. If the resulting mixture is homogeneous or heterogeneous, and the resulting particles are large enough to scatter light (Tyndall effect), it is a suspension, also called a colloidal dispersion; if the resulting mixture precipitates, then we are left with a solution or suspension in combination with a precipitate. Hence, a colloid is a suspension which lies between a solution and a precipitate. A. False. Solutions are solutions, not colloids. Solutions don’t scatter light. B. False. Rocks and water is a heterogeneous mixture, not a colloidal suspension. C. True. A suspension that has particles small enough to scatter light is a colloid. D. False. A colored solution is a solution. The particles are too small to scatter light. E. False. A solution is a solution, not a colloidal suspension. The particles are too small to scatter light.

CHEM 162-2009 EXAM I CHAPTER 12 - PROPERTIES OF SOLUTIONS MISCELLANEOUS 22. Which of the following is not a colloid? A. Smog B. Salt water C. mayonnaise D. Milk E. Gelatin Smog, mayonnaise, milk and gelatin are all colloids. Salt water is not.

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4 CHEM 162-2008 EXAM I CHAPTER 12 - PROPERTIES OF SOLUTIONS MISCELLANEOUS Which of the following is best described as a colloid? A. Sand B. Ice cream C. Brass D. Filtered sea water E. Glass 4. A colloid is a mixture of substances that lies between a true solution and a suspension. In a true solution the molecules remain in the solution indefinitely. In a suspension, the substances remain in the solution momentarily, and then precipitate out (e.g., a mixture of sand and water). The particles in a colloid are larger than in a solution (i.e., larger than indivdual molecules), but smaller than in a suspension. Whereas the particles in a solution are molecules, the particles in colloids and suspensions are much larger than molecules. The particles in a colloid stay suspended indefinitely, like molecules in a solution, but scatter light (Tyndall effect), when light is shined on it. Colloids are frequently cloudy. A colloid can be considered as a stable suspension. A. False. Sand doesn’t fit the definition. Sand is not a mixture. It is simply sodium silicate, with impurities. B. True. Ice cream is a colloid in that some particles, such as fat, are suspended as particles, rather than dissolved as molecules. C. False. Brass is an alloy of copper and zinc. An alloy is a solid solution. Hence, all of the zinc suspended in the copper or copper suspended in the zinc is in molecular form. D. False. Filtered sea water is a true solution of molecules, such as sodium chloride, dissolved (not suspended) in water. E. False. Glass is a solid solution containing silicates, boric oxide, aluminum oxide, and other substances, all dissolved in molecular form.

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