University of Texas CH 302 McCord H03 - Colligative Properties

# University of Texas CH 302 McCord H03 - Colligative Properties

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Unformatted text preview: byrne (cmb3744) – 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. 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.

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University of Texas CH 302 McCord H03 - Colligative Properties

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