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exam1 - Version 058 – Exam 1 – Fakhreddine –(52395...

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Version 058 – Exam 1 – Fakhreddine – (52395) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Which of the following would you expect to have the highest heat of vaporization? 1. C 8 H 18 2. NaF correct 3. CH 4 4. NH 3 5. BCl 3 Explanation: 002 10.0 points What is the molal concentration m of NaCl, a strong electrolyte in water, if the ob- served boiling temperature of the solution is 100 . 361 C? ( K b = 0 . 515 C/m for water.) 1. 0.350 m correct 2. 0.175 m 3. 0.701 m 4. 0.186 m 5. 0.372 m Explanation: T f = 100 C T 0 f = 100 . 361 C Δ T = 100 . 361 C - 100 C = 0 . 361 C Δ T = K b m m total = Δ T K b = 0 . 361 C 0 . 515 C /m = 0 . 700971 m m total = m Na + m Cl - , so m NaCl = 1 2 m total = 0 . 350485 m 003 10.0 points Consider 0.01 m aqueous solutions of each of the following. a) NaI; b) CaCl 2 ; c) K 3 PO 4 ; and d) C 6 H 12 O 6 (glucose) Arrange the solutions in order of freezing point from lowest to highest. Assume that each compound behaves ideally. 1. d, c, b, a 2. c, b, d, a 3. c, d, a, b 4. a, b, c, d 5. a, d, b, c 6. b, a, d, c 7. c, b, a, d correct 8. d, a, b, c 9. None of these Explanation: solvent is H 2 O m = 0.01 m solutes are NaI, CaCl 2 , K 3 PO 4 , C 6 H 12 O 6 FP depression of each solution = ? Δ t f = K f m a) For the solute NaI, NaI(s) -→ Na + + I (aq) each formula unit of NaI yields two ions in solution: Δ t f = - 1 . 86 C · kg H 2 O mol ions · 0 . 01 mol NaI kg H 2 O · 2 mol ions mol NaI = - 0 . 04 C b) For the solute CaCl 2 , CaCl 2 (s) -→ Ca 2+ + 2 Cl (aq)
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Version 058 – Exam 1 – Fakhreddine – (52395) 2 each formula unit of CaCl 2 yields three ions in solution: Δ t f = - 1 . 86 C · kg H 2 O mol ions · 0 . 01 mol CaCl 2 kg H 2 O · 3 mol ions mol CaCl 2 = - 0 . 06 C c) For the solute K 3 PO 4 , K 3 PO 4 (s) -→ 3 K + + PO 3 4 (aq) each formula unit of K 3 PO 4 yields four ions in solution: Δ t f = - 1 . 86 C · kg H 2 O mol ions · 0 . 01 mol K 3 PO 4 kg H 2 O · 4 mol ions mol K 3 PO 4 = - 0 . 07 C d) For the solute C 6 H 12 O 6 (glucose), Δ t f = - 1 . 86 C · kg H 2 O mol ions · 0 . 01 mol C 6 H 12 O 6 kg H 2 O · 1 mol ions mol C 6 H 12 O 6 = - 0 . 02 C Arranging these from lowest to highest, - 0 . 07 C < - 0 . 06 C > - 0 . 04 C < - 0 . 02 C , so the order of increasing Δ t f is c, b, a, d. 004 10.0 points Consider two liquids A and B. The vapor pressure of pure A (molecular weight = 50 g/mol) is 225 torr at 25 C and the vapor pressure of pure B (molecular weight = 75 g/mol) is 90 torr at the same temperature. What is the total vapor pressure at 25 C of a solution that is 70% A and 30% B by weight? 1. 335 torr 2. 203 torr 3. 135 torr 4. 124 torr 5. 76 torr 6. 195 torr correct 7. 108 torr 8. 115 torr 9. 225 torr Explanation: For A, P 0 = 255 torr MW = 50 g / mol For B, P 0 = 90 torr MW = 75 g / mol The mole fractions are 7 9 for A and 2 9 for B. parenleftBig 7 9 parenrightBig (225) + parenleftBig 2 9 parenrightBig (90) = 175 + 20 = 195 torr 005 (part 1 of 3) 5.0 points Please disregard the shading in the graph as it may get you confused!
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