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solution3_pdf - dang(rtd356 – H03 Colligative Properties...

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dang (rtd356) – H03: Colligative Properties – Mccord – (52450) 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. 100.46 C 2. 99.54 C 3. 99.87 C 4. 0.13 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 0 . 5000 kg water = 0 . 12 C T b = T 0 b + Δ T = 100 . 12 C 002 10.0 points 50 grams of NaNO 3 is dissolved in 538 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: 1 . 11962 C. Explanation: m NaNO 3 = 50 g m water = 538 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 0 . 512 C m parenrightbigg parenleftbigg 0 . 588235 0 . 538 m parenrightbigg (2) = 1 . 11962 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 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 . 6 C. How many moles of solute are present? Given that k f is 1 . 86 K · kg / mol. Correct answer: 0 . 206452 mol. Explanation: k f = 1 . 86 K · kg / mol Δ T f = - 1 . 6 C = 1 . 6 K m solvent = 240 g = 0 . 24 kg
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dang (rtd356) – H03: Colligative Properties – Mccord – (52450) 2 Δ T f = k f m = k f n solute m solvent n solute = Δ T f m solvent k f = (1 . 6 K) (0 . 24 kg) 1 . 86 K · kg / mol = 0 . 206452 mol .
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