H03- Col. Prop. _ Eq Condition-solutions

H03 Col Prop_ - echols(ce6563 – H03 Col Prop Eq Condition – mccord –(51180 1 This print-out should have 23 questions Multiple-choice

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: echols (ce6563) – H03: Col. Prop. / Eq Condition – mccord – (51180) 1 This print-out should have 23 questions. Multiple-choice 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 non-ionizing 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?...
View Full Document

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.

Page1 / 7

H03 Col Prop_ - echols(ce6563 – H03 Col Prop Eq Condition – mccord –(51180 1 This print-out should have 23 questions Multiple-choice

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online