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Unformatted text preview: Version 209 – hw02 – McCord – (53580) 1 This printout should have 24 questions. Multiplechoice questions may continue on the next column or page – find 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) parenleftbigg 1 kg 1 L parenrightbigg = 0 . 633 kg Thus the molality is m C 12 H 22 O 11 = moles solute kg solvent = parenleftbigg 13 . 34 g sucrose 342 . 296 g / mol sucrose parenrightbigg . 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 × . 396 kg H 2 O = 0 . 092664 mol CH 3 OH . 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 FW 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 × parenleftBig 1 . 0 mol AgNO 3 170 g AgNO 3 parenrightBig = 0 . 235294 mol AgNO 3 m = . 235 mol AgNO 3 . 060 kg water = 3 . 92157 m 004 10.0 points You’ll need to use the ClausiusClapeyron Equation (twice) on this one. See page 808 in your book. I lectured on this on Friday, 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 first calculate Δ H vap , Zn : ln parenleftbigg P 2 P 1 parenrightbigg = Δ H vap , Zn R parenleftbigg 1 T 1 1 T 2 parenrightbigg Version 209 – hw02 – McCord – (53580) 2 Δ H vap , Zn = R ln parenleftbigg P 2 P 1 parenrightbigg 1 T 1 1 T 2 = parenleftbigg 8 . 314 J mol · K parenrightbigg ln parenleftbigg 100 torr 40 torr parenrightbigg 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 = ?...
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This note was uploaded on 03/23/2008 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.
 Spring '07
 Holcombe

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