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Unformatted text preview: Version 352 – Quiz 1 – laude – (51635) 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Constants R = 8 . 314 J · K − 1 · mol − 1 Equations ln parenleftbigg P 2 P 1 parenrightbigg = Δ H R parenleftbigg 1 T 1- 1 T 2 parenrightbigg Π = MRT Δ T b = iK b m Δ T f = iK f m P i = P ◦ i χ i P = summationdisplay i P i q = mc Δ T q = m Δ H 001 5.0 points Consider a membrane, permeable to water but not sugar, which can withstand an os- motic pressure of 5 atm. If the osmotic pres- sure is currently just below the membrane’s breaking point, and you measure a sugar con- centration of 0.15 M on the left side, what is the concentration on the right side? (Assume RT = 25 L · atm · mol − 1 for the purposes of this question.) 1. 0.25 M sugar 2. 0.35 M sugar correct 3. 0 M sugar (pure water) 4. 0.2 M sugar 5. 0.5 M sugar Explanation: The difference in the molarities of the sugar solution must be large enough to produce 5 atm of pressure, which would be produced by a 0.2 M concentration gradient, so the other side needs to be at least 0.35 M. 002 5.0 points If acetic acid has a pure vapor pressure of 20 torr at 30 ◦ C and acetaldehye has a pure vapor pressure of 1000 torr at 30 ◦ C, a mixture of 24 moles of acetic acid and 16 moles of acetaldehyde would have what total vapor pressure?...
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This note was uploaded on 09/17/2011 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.
- Spring '07