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Unformatted text preview: echols (ce6563) – H03: Col. Prop. / Eq Condition – mccord – (51180) 1 This printout should have 23 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Calculate the osmotic pressure at 25 ◦ C of a 2 . 9 m solution with a solution density of 1 . 9 g/mL. The solute is nonionizing and has a MW of 220 g / mol. Correct answer: 82 . 2995 atm. Explanation: density = 1 . 9 g/mL MW = 220 g / mol m = 2 m T = 25 ◦ C + 273 = 298 K (2 . 9 mol) · (220 g / mol) = 638 g solute 1000 g solvent + 638 g solute = 1638 g soln (1638 g soln) · parenleftbigg 1 mL 1 . 9 g parenrightbigg · parenleftbigg 1 L 1000 mL parenrightbigg = 0 . 862105 L soln 2 . 9mol . 862105 L soln = 3 . 36386 M π = M RT = (3 . 36386 M)(0 . 0821)(298 K) = 82 . 2995 atm 002 (part 1 of 3) 10.0 points A 1% by mass NaCl(aq) solution has a freezing point of . 593 ◦ C. What is the van’t Hoff i factor? k f = 1 . 86 K · kg / mol. Correct answer: 1 . 84454. Explanation: Δ T f = . 593 ◦ C = 0 . 593 K MW = 58 . 44 g / mol k f = 1 . 86 K · kg / mol A 1% aqueous solution of NaCl will contain 1 g of NaCl for 99 g of water. The molality of the solution is m = parenleftbigg 1 g 58 . 44 g / mol parenrightbigg . 099 kg = 0 . 172844 mol / kg . The van’t Hoff i factor is calculated from Δ T f = ik f m i = Δ T f k f m = . 593 K (1 . 86 K · kg / mol) (0 . 172844 mol / kg) = 1 . 84454 . 003 (part 2 of 3) 10.0 points What is the total molality of all solute species? Correct answer: 0 . 318817 mol / kg. Explanation: The total molality of all solute species (NaCl(aq) + Na + (aq) + Cl − (aq)) is im = (1 . 84454)(0 . 172844 mol / kg) = 0 . 318817 mol / kg . 004 (part 3 of 3) 10.0 points The molality calculated from the freezing point depression is the sum of the molalities of the undissociated ion pairs, the Na + ions, and the Cl − ions.) What is the percentage dissociation of NaCl in this solution? Correct answer: 84 . 4536%. Explanation: If all of the NaCl has dissociated, the total molality in solution would have been (0 . 172844 mol / kg)(1 . 84454) = 0 . 345688 mol / kg having an i value of 2. If no dissociation had taken place, the molality in solution would have equaled 0 . 172844 mol / kg. NaCl → Na + + Cl − . 172844 mol / kg x x x . 318817 mol / kg = 0 . 172844 mol / kg x + x + x x = 0 . 318817 mol / kg . 172844 mol / kg = 0 . 145973 mol / kg . echols (ce6563) – H03: Col. Prop. / Eq Condition – mccord – (51180) 2 The percentage dissociation of NaCl in this solution is P = x m = . 145973 mol / kg . 172844 mol / kg × 100% = 84 . 4536% . 005 10.0 points What is the percent ionization of a solution made by adding 6.00 grams of CH 3 COOH to 100 g of water ( K b = 0 . 512 ◦ C/m) if the boiling point of the solution is 100.712 ◦ C?...
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This note was uploaded on 02/13/2012 for the course CH 301 taught by Professor Fakhreddine/lyon during the Spring '07 term at University of Texas at Austin.
 Spring '07
 Fakhreddine/Lyon
 Chemistry

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