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Name: _______________________________________________ Date: ______________________ CHEM 1C Wednesday, May 1, 2019 – Chapter 17 Part 2 – Colligative Properties - Key I. Colligative Properties the properties of a solution with a non - volatile solute relative to a pure solvent II. Raoult’s Law P a = X a P a ° 1. A salt solution sits in an open beaker. Assuming constant temperature, the vapor pressure of the solution will… 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 2. Rank the following aqueous solutions in order of increasing vapor pressure: a. 0.1 M C 6 H 12 O 6 b. 0.1 M KBr c. 0.05 M Na 2 SO 4 d. 0.05 M CH 3 COOH 3. 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. For non - volatile soluted ⇒ P sol’n = X solvent P° solvent 159 g HOCH 2 CH 2 OH x 1 ௠௢௟� 62 . 08 � = 2.56 mole HOCH 2 CH 2 OH and 500 g H 2 O x 1 ௠௢௟� 18 � = 27.8 mol H 2 O X H 2 O = ௡ ಹ 2 � ௡ ಹ 2 � + ே �ಸ ⇒ X H 2 O = 27 . 8 ௠௢௟ � 2 ை 27 . 8 ௠௢௟ � 2 ை + 2 . 56 ௠௢௟� �ை�� 2 �� 2 ை� = 0.916 P sol’n = (0.916)(26.7 torr) = 24.4 torr 4. Benzene (C 6 H 6 – 78.12 g/mol) and toluene (C 7 H 8 – 92.15 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? For solutions with 2 volatile species ⇒ P sol’n = X a P a °+ X b P b ° 40 g toluene x 1 ௠௢௟� 92 . 15 � = 0.434 mole toluene 15 g benzene x 1 ௠௢௟� 78 . 12 � = 0.192 mole benzene X toluene = 0 . 434 ௠௢௟� 0 . 434 ௠௢௟� + 0 . 192 ௠௢௟� = 0.693 X benzene = 0 . 192 ௠௢௟� 0 . 434 ௠௢௟� + 0 . 192 ௠௢௟� = 0.307 P sol’n = (0.693)(28 torr) + (0.307)(95 torr) = 48.6 torr mole fraction of a gas or vapor = ௉ � ௉ �� ��೗ ⇒ ݉݋݈݁ ݂�ܽ��݋݊ ݋݂ ܾ݁݊�݁݊݁ �ܽ݌݋� = 29 . 2 ௧௢௥௥ 48 . 6 ௧௢௥௥ = 0.6

2 as conc x i ↑ vapor pressure ↓
as conc x i ↑ vapor pressure ↓ b d vapor pressure ↓ “depression” ⇒ P a = X a P a ° freezing point ↓ “depression” ⇒ Δ T f = - k f i m boiling point ↑ “elevation” ⇒ Δ T b = k b i m osmotic pressure ↑ ⇒ π = iMRT i = vant Hoff factor = moles of solute particles moles of solute III. Ideal vs. Non - Ideal Solutions 9 . 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. For soluti ons with 2 volatile species ⇒ P sol’n = X a P a °+ X b P b ° since X hexane + X pentane = 1 ⇒ P sol’n = X hexane P hexane °+ (1 - X hexane )P pentane ° 350 torr = (X hexane )(150 torr) + (1 - X hexane )(511 torr) ⇒ X hexane = 0.446 ⇒ X pentane = 0.554 10. After substa nce A is mixed with substance B the solution feels hotter than before they were mixed. What deviation from Raoult’s Law (if any) would be expected for this solution? a. no deviation b. positive deviation – A and B form stronger forces than pure A and B c. positive deviation – A and B form weaker forces than pure A and B d. negative deviation – A and B form stronger forces than pure A and B e. negative deviation – A and B form weaker forces than pure A and B 11. Predict if the following solutions will be ideal or non - ideal. What type of deviation from Raoult’s law would you expect for the non - ideal solutions? a. carbon tetrachloride (CCl 4 ) and dichloromethane (CH 2 Cl 2 ) non - ideal polar and non - polar give po sitive deviations b. water and acetone (CH 3 COCH 3 ) non - ideal polar H - bonding with polar with O give negative deviations c. carbon tetrachloride and benzene (C 6 H 6 ) non - polar and non - polar are ideal 12. 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 – data is illustrating positive deviations b. Which of the above solutions would have the lowest boiling point? Hi ghest VP is lowest BP IV. Freezing Point Depression & Boiling Point Elevation 13. Rank the following aqueous solutions by their boiling points, and freezing points. a. 0.1 M CH 2 O as concentration x i ↑ FP ↓ BP ↑ b. 0.1 M LiF FP ⇒ b c a c. 0.05 M (NH 4 ) 2 SO 4 BP ⇒ a c b 14. 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. ΔT f = – k f im and ΔT b = k b im 110 g K 3 PO 4 x 1 ௠௢

3 ௟� 212 . 3 �
௟� 212 . 3 � = 0.518 mole K 3 PO 4 ⇒ m = 0 . 518 ௠௢௟� 0 . 8 ௞� = 0.648 mole/kg ΔT f = – (1.86 °Ckg/mol)(4)( 0.648 mole/kg) = 4.8 °C ⇒ since the FP of pure water is 0 °C the FP of this solution is – 4 °C ΔT b = k b im = (0.51 °Ckg/mol)(4)(0.648 mol/kg) = 1.32 °C ⇒ since the BP of pure water is 100 °C the BP of this solution is 101.32 °C 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 Δ T b = k b i m =� mol = � � � ( ௞� ௦௢௟��௡௧ ) ௞ � � mol = ( 56 . 5 ° � – 55 . 9 ° � ) ( 0 . 095 ௞� ) ( 1 . 71° �௄� / ௠௢௟ ) ( 1 ) mol = 0.033 molar mass = 3.75g/0.056mol = 66.8g/mol IV. Osmotic Pressure 16. 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 = ( 83 � � 1 ೘�೗� 180 � 0 . 1 ௅ ) (0.08206atmL/molK)(303K) π = 115 atm 17. A solution that contains 29 g of non - volatile/non - ionizing solute in 126 g of water has a vapor pressure of 723.4 torr at 100 °C. What is the molar mass of the solute? 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 = (126g/18g/mol)(760torr)/(723.4torr) – (126g/18g/mol) n solute = 0.354 moles molar mass = 29g/0.354mol = 82 g/mol 18. A solid mixture contains MgCl 2 (molar mass= 95.218 ݃ ݉݋݈ ) and NaCl (molar mass = 58.443 ݃ ݉݋݈ ). When 0.500 g of this solid i s dissolved in enough water to form 1.000 L of solution, the osmotic pressure at 25.0˚C is observed to be 0.3950 atm. What is the mass percent of MgCl 2 in the solid? (Assume ideal behavior for the solution.) If you have X g of MgCl 2 then there’s 0.50 g – X g of NaCl moles of MgCl 2 = X/95.218 = 0.0105X moles of NaCl = (0.5 – X)/58.443 = 0.00855 – 0.0171X π = iMRT = (iM (for MgCl 2 )+iM (for NaCl))RT 0.3950 = (3(0.0105X/1) + 2((0.00855 – 0.0171X)/1)(0.08206)(298) X = 0.352 g % MgCl 2 = (0.352g/0.5g)(100) = 70.4% Osmosis : The flow of solvent from a lower concentration solution through a semipermeable membrane to a higher concentration solution Osmotic Pressure : The pressure needed to stop the flow of solvent through a semipermeable membran