# HW3 - rote (cmr2495) H03: Colligative Properties mccord...

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rote (cmr2495) – H03: Colligative Properties – mccord – (51620) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – fnd 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. 100.46 C 2. 99.87 C 3. 100.13 C correct 4. 0.13 C 5. 99.54 C 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 0 . 5000 kg water = 0 . 12 C T b = T 0 b + Δ T = 100 . 12 C 002 10.0 points 10 grams oF NaNO 3 is dissolved in 899 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 . 134005 C. Explanation: m NaNO 3 = 10 g m water = 899 g K b water = 0 . 512 C /m NaNO 3 Na + + NO 3 i = 2 m = (10 g NaNO 3 ) p 1 mol NaNO 3 85 . 0 g NaOH P = 0 . 117647 mol NaNO 3 Δ T b = K b · m · i = ± 0 . 512 C m ²± 0 . 117647 0 . 899 m ² (2) = 0 . 134005 C 003 10.0 points Which oF the Following aqueous solutions would exhibit the highest boiling point? 1. 0 . 1 m urea 2. 0 . 1 m CaCl 2 correct 3. 0 . 1 m NaCl Explanation: Here the solution with the highest efec- tive molality would have the highest boiling point. Urea does not ionize in water, so its stated molality and e±ective molality should be the same. When NaCl dissolves, 2 ions are Formed, but when CaCl 2 dissolves, 3 ions are Formed. The e±ective molality For CaCl 2 would then be (approximately) 3 times the stated molality, while For NaCl the e±ective 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 195 g oF water is - 1 . 9 C. How many moles oF solute are present? Given that k f is 1 . 86 K · kg / mol. Correct answer: 0 . 199194 mol. Explanation: k f = 1 . 86 K · kg / mol Δ T f = - 1 . 9 C = 1 . 9 K m solvent = 195 g = 0 . 195 kg

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rote (cmr2495) – 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 . 9 K) (0 . 195 kg) 1 . 86 K · kg / mol = 0 . 199194 mol .
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## This note was uploaded on 04/22/2011 for the course CHEM 302 taught by Professor Mccord during the Spring '10 term at University of Texas.

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HW3 - rote (cmr2495) H03: Colligative Properties mccord...

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