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Unformatted text preview: byrne (cmb3744) – 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. 0.13 ◦ C 2. 99.54 ◦ C 3. 99.87 ◦ C 4. 100.46 ◦ C 5. 100.13 ◦ C correct Explanation: m urea = 7.5 g m H 2 O = 500 g Δ 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 + Δ T = 100 . 12 ◦ C 002 10.0 points 50 grams of NaNO 3 is dissolved in 786 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 . 766352 ◦ C. Explanation: m NaNO 3 = 50 g m water = 786 g K b water = 0 . 512 ◦ C /m NaNO 3 → Na + + NO − 3 i = 2 m = (50 g NaNO 3 ) parenleftBig 1 mol NaNO 3 85 . 0 g NaOH parenrightBig = 0 . 588235 mol NaNO 3 Δ T b = K b · m · i = parenleftbigg . 512 ◦ C m parenrightbiggparenleftbigg . 588235 . 786 m parenrightbigg (2) = 0 . 766352 ◦ C 003 10.0 points Which of the following aqueous solutions would exhibit the highest boiling point? 1. . 1 m urea 2. . 1 m NaCl 3. . 1 m CaCl 2 correct 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 240 g of water is 1 . 1 ◦ C. How many moles of solute are present? Given that k f is 1 . 86 K · kg / mol. Correct answer: 0 . 141935 mol. Explanation: k f = 1 . 86 K · kg / mol Δ T f = 1 . 1 ◦ C = 1 . 1 K m solvent = 240 g = 0 . 24 kg byrne (cmb3744) – H03: Colligative Properties – mccord – (51620) 2 Δ T f = k f m = k f n solute m solvent n solute = Δ T f m solvent k f = (1 . 1 K) (0 . 24 kg) 1 . 86 K · kg / mol = 0...
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This note was uploaded on 05/08/2011 for the course CHEM 302 taught by Professor Mccord during the Spring '10 term at University of Texas.
 Spring '10
 McCord

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