# hw2 - Version 147 – hw02 – McCord –(53580 1 This...

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Unformatted text preview: Version 147 – hw02 – McCord – (53580) 1 This print-out should have 24 questions. Multiple-choice 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 10 . 75 g of sucrose (C 12 H 22 O 11 ) dissolved in 646 mL of water. Correct answer: 0 . 0486154 m . Explanation: m C 12 H 22 O 11 = 10 . 75 g V H 2 O = 646 mL = 0 . 646 L MW C 12 H 22 O 11 = 342 . 296 g / mol m H 2 O = (0 . 646 L) parenleftbigg 1 kg 1 L parenrightbigg = 0 . 646 kg Thus the molality is m C 12 H 22 O 11 = moles solute kg solvent = parenleftbigg 10 . 75 g sucrose 342 . 296 g / mol sucrose parenrightbigg . 646 kg H 2 O = 0 . 0486154 m 002 10.0 points How many grams of methanol (CH 3 OH) is required to prepare a 0 . 13 m solution in 473 g of water? Correct answer: 1 . 96768 g. Explanation: m CH 3 COOH = 0 . 13 m m H 2 O = 473 g m methanol (g) = ? mol CH 3 OH = 0 . 13 mol kg H 2 O × . 473 kg H 2 O = 0 . 06149 mol CH 3 OH . 06149 mol CH 3 OH × 32 g CH 3 OH 1 mol CH 3 OH = 1 . 96768 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 Clausius-Clapeyron 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. 1183 K correct 2. 1452 K 3. 906 K 4. 1076 K 5. 1252 K 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 147 – 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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hw2 - Version 147 – hw02 – McCord –(53580 1 This...

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