CH 302 HW 3

# CH 302 HW 3 - wei (cyw224) H03: Colligative Properties...

This preview shows pages 1–3. Sign up to view the full content.

wei (cyw224) – H03: Colligative Properties – Mccord – (52450 )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. 100.13 C correct 3. 0.13 C 4. 99.87 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 40 grams oF NaNO 3 is dissolved in 556 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 . 866695 C. Explanation: m NaNO 3 = 40 g m water = 556 g K b water = 0 . 512 C /m NaNO 3 Na + + NO - 3 i =2 m = (40 g NaNO 3 ) ± 1 mol NaNO 3 85 . 0 g NaOH ² . 470588 mol NaNO 3 Δ T b = K b · m · i = ³ 0 . 512 C m ´³ 0 . 470588 0 . 556 m ´ (2) . 866695 C 003 10.0 points Which oF the Following aqueous solutions would exhibit the highest boiling point? 1. 0 . 1 m NaCl 2. 0 . 1 m CaCl 2 correct 3. 0 . 1 m urea 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 205 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 . 121237 mol. Explanation: k f =1 . 86 K · kg / mol Δ T f = - 1 . 1 C = 1 . 1K m solvent = 205 g = 0 . 205 kg

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
wei (cyw224) – 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 . 1 K) (0 . 205 kg) 1 . 86 K · kg / mol =0 . 121237 mol .
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 03/22/2010 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.

### Page1 / 6

CH 302 HW 3 - wei (cyw224) H03: Colligative Properties...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online