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Properties of Solutions Chapter 17 Web-site:

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Properties of Solutions – ch Predict the relative solubility of the following: a. O 2 in H 2 0 vs. O 2 in CCl 4 b. CH 3 OH in H 2 O vs. CH 3 CH 2 CH 2 CH 2 OH in H 2 O c. AgCl (s) in H 2 O at 25°C vs. AgCl (s) in H 2 O at 45°C d. CO 2(g) in C 8 H 18 at 25°C vs. CO 2(g) in C 8 H 18 at 45°C e. N 2(g) in C 6 H 6(g) at 1 atm vs. N 2(g) in C 6 H 6(g) at 8 atm

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Properties of Solutions – ch If the solubility of CO in water is 0.08 M at 25 °C and 0.75 atm, what is the solubility of CO in water at 25 °C and 2.6 atm?

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Properties of Solutions – ch Calculate the molality and mole fraction of an aqueous solution that is 8 M NaCl, the density of the solution is 1.18 g/ml. Concentrations Molarity => M = n solute /L solution Molality => m = n solute /kg solvent Mole fraction => X a = n a /n total

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Properties of Solutions – ch Calculate the heat of hydration for the following ionic solids. a. KF b. RbF c. Compare the ion-dipole forces for K + vs. Rb + in H 2 O CompoundLattice EnergyHeat of Solution KF-804 kJ/mol-15 kJ/mol RbF-768 kJ/mol-24 kJ/mol

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Properties of Solutions – ch Rank the following aqueous solutions by their boiling points, freezing points, vapor pressure and osmotic pressure. a. 0.1 M C 6 H 12 O 6 b. 0.1 M KBr c M Na 2 SO 4 d M CH 3 COOH

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Properties of Solutions – ch. 17 Colligative properties => the properties of a solution with a non-volatile solute relative to a pure solvent As the concentration of solute particles ↑ vapor pressure ↓ “depression” => P a = X a P a ° freezing point ↓ “depression” => ΔT f = - k f i m boiling point ↑ “elevation” => ΔT b = k b i m osmotic pressure ↑ => π = iMRT

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Properties of Solutions – ch A salt solution sits in an open beaker. Assuming constant temperature, the vapor pressure of the solution a. increases over time b. decreases over time c. stays the same over time d. We need to know which salt is in the solution to answer this e. We need to know the temperature and pressure to answer this

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Properties of Solutions – ch You place a beaker with 300 mL of salt water and a beaker with 300 ml of pure water under a bell jar. What will you observe over time? Salt water Pure

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Properties of Solutions – ch Calculate the boiling point and freezing point of a solution made by dissolving 110 g of K 3 PO 4 (212.3g/mol) in 800 mL of water at 1 atm. For water k b = 0.51 °Ckg/mol and k f = 1.86 °Ckg/mol.

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Properties of Solutions – ch Calculate the vapor pressure of a solution (in torr) made by dissolving 159 g of ethylene glycol (HOCH 2 CH 2 OH – 62.08g/mol) in 500 g of water at 27 °C. At 27 °C the vapor pressure of pure water is 26.7 torr.

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Properties of Solutions – ch Benzene (C 6 H 6 – g/mol) and toluene (C 7 H 8 – g/mol) form an ideal solution. What is the vapor pressure of a solution prepared by mixing 40 g of toluene with 15 g of benzene at 25 °C? At 25 °C the vapor pressures of pure toluene and pure benzene are 28 and 95 torr respectively. What is the mole fraction of the benzene in the vapor above the solution?

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Properties of Solutions – ch Pentane and hexane form an ideal solution. What composition of a pentane and hexane solution at 25 °C would give a vapor pressure of 350 torr? At 25 °C the vapor pressures of pure pentane and hexane are 511 torr and 150 torr respectively.

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Properties of Solutions – ch Draw a vapor pressure curve for the following solutions at 25 °C. a. Ethanol (CH 3 CH 2 OH) and methanol (CH 3 OH), heat of solution is 0 kJ b. Methanol and hexane (C 6 H 14 ), solution feels cooler upon mixing c. Ethanol and water, solution feels warmer upon mixing

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Properties of Solutions – ch. 17 Ideal vs. Non-Ideal Solutions a. ΔH sol’n ~ 0 => Ideal solution – the actual vapor pressures will agree with Raoult’s law b. ΔH sol’n > 0 => Non-Ideal solution – the forces in the solution are weaker than in the pure substances resulting in higher VPs than expected from Raoult’s Law c. ΔH sol’n Non-Ideal solution – the forces in the solution are stronger than in the pure substances resulting in lower VPs than expected from Raoult’s Law

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Properties of Solutions – ch The vapor pressures of several solutions of water and butanol were determined at various compositions and the data is given below: a. Are the solutions of water and butanol ideal? b. Which of the above solutions would have the lowest boiling point? XH2OXH2O VP (torr)

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Properties of Solutions – ch Calculate the osmotic pressure of a solution made by dissolving 83 g of glucose (C 6 H 12 O 6 ) in 100 mL of water at 30 °C.

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Properties of Solutions – ch A solution contains 3.75 g of a nonvolatile hydrocarbon in 95 g of acetone. The boiling points of pure acetone and the solution are 55.9 °C and 56.5 °C respectively. What is the molar mass of the hydrocarbon? For acetone the K b = 1.71 °CKg/mol.

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Properties of Solutions – ch A solution that contains 29.4 g of non-volatile/non-ionizing solute in g of water has a vapor pressure of torr at 27 °C. What is the molar mass of the solute? The vapor pressure of water at 27 °C is torr.

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Properties of Solutions – Answers 1. Predict the relative solubility of the following: a. O 2 in H 2 0 vs. O 2 in CCl 4 b. CH 3 OH in H 2 O vs. CH 3 CH 2 CH 2 CH 2 OH in H 2 O c. AgCl (s) in H 2 O at 25°C vs. AgCl (s) in H 2 O at 45°C d. CO 2(g) in C 8 H 18 at 25°C vs. CO 2(g) in C 8 H 18 at 45°C e. N 2(g) in C 6 H 6(g) at 1 atm vs. N 2(g) in C 6 H 6(g) at 8 atm 2. If the solubility of CO in water is 0.08 M at 25 °C and 0.75 atm, what is the solubility of CO in water at 25 °C and 2.6 atm? C 1 /P 1 = C 2 P 2 (0.08M/0.75atm) = C 2 /2.6atm C 2 = 0.28M

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Properties of Solutions – Answers 3. The term proof is defined as twice the percent by volume of ethanol (CH 3 CH 2 OH) in solution. Thus, a solution that is 95% ethanol by volume is 190 proof. What is the molarity, molality and mole fraction of ethanol in a 92 proof ethanol/water solution? The density of pure ethanol is 0.8 g/mL. 92 proof => 46 % ethanol by volume So if we had 1L of solution => 0.46 L of ethanol and 0.54 L of water (460 mL ethanol x 0.8g/mL)/(46 g/mol) = 8 mol ethanol M ethanol = 8 mol/1L = 8M Since the density of water is 1Kg/L => 0.54 Kg of water m ethanol = 8 mol/0.54Kg = 14.8m 540g/18g/mol = 30 mol of water X ethanol = 8mol/38mol = 0.21

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Properties of Solutions – Answers 4. Calculate the heat of hydration for the following ionic solids. a. KF b. RbF c. Compare the ion-dipole forces for K + vs. Rb + in H 2 O ΔH hydration = ΔH LE + ΔH sol’n KF => ΔH hydration = (-804kJ/mol) + (-15kJ/mol) = -819kJ/mol RbF => ΔH hydration = (-768kJ/mol) + (-24kJ/mol) = -792kJ/mol Since both species have fluoride we can compare the ion-dipole forces of K + vs Rb + => the more negative the heat of hydration the stronger the force => K + has the stronger ion-dipole force => this should make sense since K + is smaller and can form shorter bonds with water

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Properties of Solutions – Answers 5. Rank the following aqueous solutions by their boiling points, freezing points, vapor pressure and osmotic pressure. a. 0.1 M C 6 H 12 O 6 => conc. of solute particles = 0.1M b. 0.1 M KBr => conc. of solute particles = 0.2M c M Na 2 SO 4 => conc. of solute particles = 0.15M d M CH 3 COOH => conc. of solute particles = 0.05M (slightly higher than 0.05 M b/c it will ionize roughly 2%) Relative FP or VP => b < c < a < d Relative BP or OP => b > c > a > d

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Properties of Solutions – Answers 6. What will happen to the vapor pressure of a beaker of water if it’s left out in the sun? VP decreases over time - as the water evaporates the concentration of the salt increases making it harder to evaporate 7. You place a beaker with 300 mL of salt water and a beaker with 300 ml of pure water under a bell jar. What will you observe over time? Eventually all of the pure water will be gone and the volume of the salt water solution will double

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Properties of Solutions – Answers 8. Calculate the boiling point and freezing point of a solution made by dissolving 110 g of K 3 PO 4 (212.3g/mol) in 800 mL of water at 1 atm. For water k b = 0.51 °Ckg/mol and k f = 1.86 °Ckg/mol. (110 g K 3 PO 4 )/(212.3g/mol) = 0.52 mol K 3 PO 4 m = 0.52 mol/0.8 kg = 0.65m ΔT f = -(0.65 mol/kg)(4)(1.86 °Ckg/mol) = -4.8°C ΔT b = (0.65 mol/kg)(4)(0.51 °Ckg/mol) = 1.3°C Since pure water has a FP = 0°C and BP = 100°C then the solution will have a FP = -4.8°C and BP = 101.3°C

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Properties of Solutions – Answers 9. Calculate the vapor pressure of a solution (in torr) made by dissolving 159 g of ethylene glycol (HOCH 2 CH 2 OH – 62.08g/mol) in 500 g of water at 27 °C. At 27 °C the vapor pressure of pure water is 26.7 torr. (159g)/(62.08g/mol) = 2.6 mol HOCH 2 CH 2 OH (500g)/(18.02g/mol) = 27.7 mol water P water = (27.7mol/30.3mol)(26.7torr) = 24.4torr

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Properties of Solutions – Answers 10. Toluene and benzene form an ideal solution. What is the vapor pressure of a solution prepared by mixing 40 g of toluene with 15 g of benzene at 25 °C? At 25 °C the vapor pressures of pure toluene and pure benzene are 28 and 95 torr respectively. (40g)/(92.15g/mol) = 0.43 mol toluene (15g)/(78.12g/mol) = 0.19 mol benzene P toluene = (0.43 mol/0.62 mol)(28 torr) = 19.6 torr P benzene = (0.19 mol/0.62 mol)(95 torr) = 29.1 torr P solution = 48.7 torr

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Properties of Solutions – Answers 11. Pentane and hexane form an ideal solution. What composition of a pentane and hexane solution at 25 °C would give a vapor pressure of 350 torr? At 25 °C the vapor pressures of pure pentane and hexane are 511 torr and 150 torr respectively. P total = P pentane + P hexane P total = X p P p ° + X h P h ° Since X p + X h = 1 => X p = 1 – X h P total = (1 – X h ) P p ° + X h P h ° 350 torr = (1 – X h )(511 torr) + X h (150 torr) X h = 0.45 => X p = 0.55

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Properties of Solutions – Answers 12. Draw a vapor pressure curve for the following solutions at 25 °C. a. Ethanol (CH 3 CH 2 OH) and methanol (CH 3 OH) b. Methanol and hexane (C 6 H 14 ), solution feels cooler upon mixing c. Ethanol and water, solution feels warmer upon mixing

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Properties of Solutions – Answers 13. The vapor pressures of several solutions of water and butanol were determined at various compositions and the data is given below: a. are the solutions of water and butanol ideal? No – since the VP’s of some of the solutions are outside of the range of pure water and pure butanol b. which of the above solutions would have the lowest boiling point? Highest VP will result in the lowest BP => X H 2 O = 0.58 XH2OXH2O VP (torr)

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Properties of Solutions – Answers 14. Calculate the osmotic pressure of a solution made by dissolving 83 g of glucose (C 6 H 12 O 6 ) in 100 mL of water at 30 °C. π = iMRT π = (1)((83g/180g/mol)/(01.L))( atmL/molK)(303K) π = 115 atm

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Properties of Solutions – Answers 15. A solution contains 3.75 g of a nonvolatile hydrocarbon in 95 g of acetone. The boiling points of pure acetone and the solution are 55.9 °C and 56.5 °C respectively. What is the molar mass of the hydrocarbon? For acetone the K b = 1.71 °CKg/mol. molar mass = g/mol => given grams and you can get moles from ΔT b = k b i m => mol = ΔT b kg solvent / k b i mol = (56.5 °C – 55.9 °C)(0.095kg)/(1.71°CKg/mol)(1) mol = molar mass = 3.75g/0.056mol = 66.8g/mol

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Properties of Solutions – Answers 16. A solution that contains 29.4 g of non-volatile/non-ionizing solute in g of water has a vapor pressure of torr at 27 °C. What is the molar mass of the solute? The vapor pressure of water at 27 °C is torr. molar mass = g/mol => given grams and you can get moles from P water = X water P water ° => n solute = (n w P w °/P w ) – (n w ) => n solute = (100.8g/18g/mol)(26.74torr)/(25.81torr) – (100.8g/18g/mol) n solute = mole molar mass = 29.4g/0.202mol = 146g/mol

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