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Exam 1-solutions[1]

# Exam 1-solutions[1] - Version 287 Exam 1 vanden bout(51640...

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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 – find all choices before answering. 001 0.0points 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 0.0points Which of the following statements about col- ligative properties of solutions is FALSE? 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, different solvents have very dif- ferent colligative properties 003 0.0points 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 Hoff factor for KNO 3 is 2. So 0.5 moles is effectively 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 0.0points 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 0.0points 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 0.0points If more points are awarded on this assign- ment, would you like them added to your score? 1. YES, I would like the points and the higher score. correct 2. NO, leave my score alone, I prefer the lower score Explanation: This should be a no-brainer. Most students want higher scores. If you picked yes, you got credit for the question and you got the extra points you asked for (if they were granted by your instructor). If you answered NO, you also got what you wanted...
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