# hw 2 - Version 209 hw02 McCord(53580 This print-out should...

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Version 209 – hw02 – McCord – (53580) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Calculate the molality oF sucrose in a solution composed oF 13 . 34 g oF sucrose (C 12 H 22 O 11 ) dissolved in 633 mL oF water. Correct answer: 0 . 0615672 m . Explanation: m C 12 H 22 O 11 = 13 . 34 g V H 2 O = 633 mL = 0 . 633 L MW C 12 H 22 O 11 = 342 . 296 g / mol m H 2 O = (0 . 633 L) p 1 kg 1 L P = 0 . 633 kg Thus the molality is m C 12 H 22 O 11 = moles solute kg solvent = p 13 . 34 g sucrose 342 . 296 g / mol sucrose P 0 . 633 kg H 2 O = 0 . 0615672 m 002 10.0 points How many grams oF methanol (CH 3 OH) is required to prepare a 0 . 234 m solution in 396 g oF water? Correct answer: 2 . 96525 g. Explanation: m CH 3 COOH = 0 . 234 m m H 2 O = 396 g m methanol (g) = ? mol CH 3 OH = 0 . 234 mol kg H 2 O × 0 . 396 kg H 2 O = 0 . 092664 mol CH 3 OH 0 . 092664 mol CH 3 OH × 32 g CH 3 OH 1 mol CH 3 OH = 2 . 96525 g CH 3 OH 003 10.0 points A solution is 40.0% silver nitrate (AgNO 3 ) by mass. The density oF this solution is 1.48 grams/mL. The Formula weight oF AgNO 3 is 170 grams/mol. Calculate the molality oF AgNO 3 in this solution. Correct answer: 3 . 92157 m . Explanation: density = 1.48 g/mL ±W AgNO 3 = 170 g/mol % AgNO 3 = 40% by mass molality = n AgNO 3 kg water In 100 g oF 40.0% AgNO 3 (aq) there are 40.0 g AgNO 3 and 60.0 g water. n AgNO 3 = 40 . 0 g AgNO 3 × ± 1 . 0 mol AgNO 3 170 g AgNO 3 ² = 0 . 235294 mol AgNO 3 m = 0 . 235 mol AgNO 3 0 . 060 kg water = 3 . 92157 m 004 10.0 points You’ll need to use the Clausius-Clapeyron Equation (twice) on this one. See page 808 in your book. I lectured on this on ±riday, 1/18. Liquid metallic zinc (Zn) has a vapor pres- sure oF 40 torr at 673 C and 100 torr at 736 C. Predict the normal boiling point oF zinc. 1. 1252 K 2. 1452 K 3. 906 K 4. 1076 K 5. 1183 K correct Explanation: T 1 = 673 C + 273 = 946 K P 1 = 40 torr T 2 = 736 C + 273 = 1009 K P 2 = 100 torr We must frst calculate Δ H vap , Zn : ln p P 2 P 1 P = Δ H vap , Zn R p 1 T 1 - 1 T 2 P

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Version 209 – hw02 – McCord – (53580) 2 Δ H vap , Zn = R ln p P 2 P 1 P 1 T 1 - 1 T 2 = p 8 . 314 J mol · K P ln p 100 torr 40 torr P 1 946 K - 1 1009 K = 115421 J / mol The normal boiling point is at 760 torr, standard atmospheric pressure. When the vapor pressure of Zn equals 760 torr, Zn will boil. We need to calculate the temperature at which this occurs: P 1 = 100 torr P 2 = 760 torr T 1 = 736 C + 273 = 1009 K T 2 = ? 1 T 1 - 1 T 2 = R Δ H vap , Zn ln p P 2 P 1 P 1 T 2 = 1 T 1 - R Δ H vap , Zn ln p P 2 P 1 P = 1 1009 K - 8 . 314 J mol · K 115421 J / mol ln p 760 torr 100 torr P = 0 . 000844989 K 1 Thus T 2 = 1183 . 45 K 005 10.0 points The vapor pressure of water at 37 C is 47.1 torr and its enthalpy of vaporization is 44.0 kJ · mol 1 . Estimate the vapor pressure of wa- ter at 87 C. Assume the enthalpy of vaporiza- tion of water is independent of temperature. 1. 713 torr 2. 504 torr correct 3. 256 torr 4. 52 torr 5. 112 torr Explanation: 006 10.0 points The phase diagram for a pure substance is given below. Solid Liquid Vapor 50 100 150 200 250 300 100 200 300 400 Temperature, K Pressure, atm What is the critical temperature?
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hw 2 - Version 209 hw02 McCord(53580 This print-out should...

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