Exam 1-solutions[1] - Version 287 Exam 1 vanden bout(51640 This print-out should have 31 questions Multiple-choice questions may continue on the

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Version 287 – Exam 1 – vanden bout – (51640) 1 This print-out should have 31 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 0.0 points An animal cell assumes its normal volume when it is placed in a solution with a total solute molarity oF 0.3 M. IF the cell is placed in a solution with a total solute molarity oF 0.1 M, 1. the escaping tendency oF water in the cell increases. 2. water leaves the cell, causing contrac- tion. 3. no movement oF water takes place. 4. water enters the cell, causing expansion. correct Explanation: 002 0.0 points Which oF the Following statements about col- ligative properties oF solutions is ±ALSE? 1. Colligative properties assume ideal solu- tions 2. The higher the concentration oF solute in the solution, the lower the vapor pressure oF the solvent. 3. Osmosis is a colligative property. 4. Colligative properties are identical For all solvents correct 5. Colligative properties arise From the con- centration oF the solute but not the inter- molecular Forces oF the solute Explanation: Colligative properties, which include osmo- sis, vapor pressure lowering, melting and boil- ing point elevations, depend only on the num- ber oF solute particles present in solution, not on their properties. However, di²erent solvents have very diF- Ferent colligative properties 003 0.0 points Given that the Freezing point depression con- stant For water is 1.86 K m 1 , what is the Freezing point oF a solution that con- tains 0.5 moles KNO 3 and 1 mole oF sucrose (C 12 H 22 O 11 ) in 500 g oF water? 1. +2.79 C 2. -5.58 C 3. -3.72 C 4. -7.44 C correct 5. -1.86 C 6. -2.79 C 7. -37.2 C Explanation: The van’t Ho² Factor For KNO 3 is 2. So 0.5 moles is e²ectively 1 mole oF ions. There is another mole oF sucrose. This is 2 total moles in 500 g making a 4 molal solution oF all solutes. Δ T f = - mK f = - (4)(1 . 86) = - 7 . 44 004 0.0 points Consider the reaction PCl 5 (g) PCl 3 (g) + Cl 2 (g) . At a certain temperature, iF the initial concen- tration oF PCl 5 (g) is 2.0 M, at equilibrium the concentration oF Cl 2 (g) is 0.30 M. Calculate the value oF K c at this temperature. 1. 0.045 2. 19 3. 0.064 4. 0.090
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Version 287 – Exam 1 – vanden bout – (51640) 2 5. 0.053 correct Explanation: 005 0.0 points A particular small molecule drug works by binding to the active site in a given enzyme Drug ( aq ) + Enzyme ( aq ) BoundComplex ( aq ) If the equilibrium constant for this reaction is 10 10 , at what concentration of free drug is there 1000 times more bound enzyme (com- plex) than unbound enzyme? 1. 1 M 2. 10 10 M 3. 10 7 M correct 4. 10 13 M 5. 10 3 M Explanation: K = [complex]/[drug][enzyme] [drug] = [complex]/[enzyme] x 1/K = 1000/10 10 = 10 7 M 006 0.0 points If more points are awarded on this assign- ment, would you like them added to your score?
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This note was uploaded on 03/27/2011 for the course CHEM 302 taught by Professor Mccord during the Spring '10 term at University of Texas at Austin.

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Exam 1-solutions[1] - Version 287 Exam 1 vanden bout(51640 This print-out should have 31 questions Multiple-choice questions may continue on the

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