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Unformatted text preview: Howard (clh2528) H03: Colligative Properties mccord (51620) 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points We dissolve 7.5 grams of urea (a nonelec trolyte with MW 60 g/mol) in 500 grams of water. At what temperature would the solu tion boil? 1. 99.54 C 2. 0.13 C 3. 100.13 C correct 4. 99.87 C 5. 100.46 C Explanation: m urea = 7.5 g m H 2 O = 500 g T b = K b m = K b mol urea kg water = (0 . 515 C / m) 7 . 5 60 mol urea . 5000 kg water = 0 . 12 C T b = T b + T = 100 . 12 C 002 10.0 points 30 grams of NaNO 3 is dissolved in 754 grams of water. What is the boiling point elevation in degrees Celsius? Note that K b for water is 0.512 C /m . Assume complete dissociation of the salt and ideal behavior of the solution. Correct answer: 0 . 479326 C. Explanation: m NaNO 3 = 30 g m water = 754 g K b water = 0 . 512 C /m NaNO 3 Na + + NO 3 i = 2 m = (30 g NaNO 3 ) parenleftBig 1 mol NaNO 3 85 . 0 g NaOH parenrightBig = 0 . 352941 mol NaNO 3 T b = K b m i = parenleftbigg . 512 C m parenrightbiggparenleftbigg . 352941 . 754 m parenrightbigg (2) = 0 . 479326 C 003 10.0 points Which of the following aqueous solutions would exhibit the highest boiling point? 1. . 1 m urea 2. . 1 m CaCl 2 correct 3. . 1 m NaCl Explanation: Here the solution with the highest effec tive molality would have the highest boiling point. Urea does not ionize in water, so its stated molality and effective molality should be the same. When NaCl dissolves, 2 ions are formed, but when CaCl 2 dissolves, 3 ions are formed. The effective molality for CaCl 2 would then be (approximately) 3 times the stated molality, while for NaCl the effective molality would only be (approximately) twice the stated molality. Thus, we would expect that CaCl 2 would exhibit the highest boiling point. 004 10.0 points The freezing point of an aqueous solution con taining a nonelectrolyte dissolved in 305 g of water is 1 . 8 C. How many moles of solute are present? Given that k f is 1 . 86 K kg / mol. Correct answer: 0 . 295161 mol. Explanation: k f = 1 . 86 K kg / mol T f = 1 . 8 C = 1 . 8 K m solvent = 305 g = 0 . 305 kg Howard (clh2528) H03: Colligative Properties mccord (51620) 2 T f = k f m = k f n solute m solvent n solute = T f m solvent k f = (1 . 8 K) (0 . 305 kg) 1 . 86 K kg / mol...
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This note was uploaded on 03/22/2011 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.
 Spring '07
 Holcombe
 Chemistry

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