Exam 1 - Version 268 – Exam 1 – mccord –(51620 1 This...

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Unformatted text preview: Version 268 – Exam 1 – mccord – (51620) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. McCord CH302 1pm class only - 51620 For water: k f = 1 . 86 ◦ C / m k b = 0 . 51 ◦ C / m C s , ice = 2 . 09 J/g ◦ C C s , water = 4 . 18 J/g ◦ C C s , steam = 2 . 03 J/g ◦ C Δ H fus = 334J / g Δ H vap = 2260J / g 001 10.0 points Which of the following would decrease the solubility of carbon monoxide gas, CO(g), in the solvent ethanol? I) decreasing the temperature of the sample II) increasing the amount of solvent III) changing to a less polar solvent 1. I and III 2. III correct 3. II and III 4. I and II 5. I only 6. II only 7. I, II and III Explanation: The solubility of a gas is directly propor- tional to the gas’ pressure, inversely propor- tional to the temperature of the solvent. Con- sidering the principle of like dissolve like, a polar species such as CO(g) will be more sol- uble in polar solvents. The volume of the solvent does not impact the solubility, which is an intensive property; this is evident when one considers the unity of solubility such as grams per liter, etc. 002 10.0 points What is the temperature change when 8 g of MgBr 2 (with- 185 . 6 kJ / mol enthalpy of so- lution) is dissolved in 150 g of water? Assume that the specific heat capacity of the solution is 4 . 18 J K · g . 1. 99.843 2. 18.811 3. 11.7368 4. 95.502 5. 17.9428 6. 82.9613 7. 48.2333 8. 12.8622 9. 78.138 10. 104.666 Correct answer: 12 . 8622 ◦ C. Explanation: m MgBr 2 = 8 g m H 2 O = 150 g C = 4 . 18 J K · g MW MgBr 2 = 184 . 113 g / mol Δ H =- 185 . 6 kJ / mol =- 1 . 856 × 10 5 J / mol The number of moles is n = 8 g 184 . 113 g / mol = 0 . 0434516 mol , so the heat released is obtained from n Δ H = mC Δ T Δ T = n Δ H mC =- (0 . 0434516 mol) (- 1 . 856 × 10 5 J / mol) (4 . 18 J K · g ) (150 g) = 12 . 8622 K = 12 . 8622 ◦ C . 003 10.0 points Version 268 – Exam 1 – mccord – (51620) 2 Carbon Dioxide 3 1 4 2 A B 200 250 300 350 400 10 10 1 10 2 10 3 10 4 Pressure,bar Temperature, K A sample of carbon dioxide is stored at 10 4 bar and 250 K. This sample is then decompressed to 10 bar at constant temper- ature. Then, at constant pressure it is heated to 400 K. Next, it is compressed at con- stant temperature to 200 bar. According to the phase diagram, how many phase transi- tions has the carbon dioxide gone through, and what is its final state? 1. 2, gas 2. 2, supercritical fluid correct 3. 2, liquid 4. 3, supercritical fluid 5. 3, gas Explanation: Navigate through the diagram through each process. In the first decompression, the solid is decompressed and becomes a liquid and then a gas (2 phase transitions). Upon heating, it becomes a supercritical fluid (no phase transitions)....
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This note was uploaded on 04/22/2011 for the course CHEM 302 taught by Professor Mccord during the Spring '10 term at University of Texas.

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Exam 1 - Version 268 – Exam 1 – mccord –(51620 1 This...

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