# CH3 - Howard (clh2528) H03: Colligative Properties mccord...

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Unformatted text preview: Howard (clh2528) H03: Colligative Properties mccord (51620) 1 This print-out should have 16 questions. Multiple-choice 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.

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CH3 - Howard (clh2528) H03: Colligative Properties mccord...

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