Exam 1-solutions - Version 211 – Exam 1 – vanden bout...

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Unformatted text preview: Version 211 – 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 0.0 points Which of the following statements about col- ligative properties of solutions is FALSE? 1. Osmosis is a colligative property. 2. The higher the concentration of solute in the solution, the lower the vapor pressure of the solvent. 3. Colligative properties arise from the con- centration of the solute but not the inter- molecular forces of the solute 4. Colligative properties are identical for all solvents correct 5. Colligative properties assume ideal solu- tions 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 002 0.0 points The vapor pressure of benzene (C 6 H 6 ) is 120 torr at 27.0 ◦ C, and its normal boiling point is 80.1 ◦ C. What is the molar heat of vaporiza- tion of benzene? 1. 2 . 49 × 10 2 J/mol 2. 1 . 31 × 10 3 J/mol 3. 4 . 95 × 10 4 J/mol 4. 4 . 56 × 10 3 J/mol 5. 3 . 07 × 10 4 J/mol correct Explanation: P 1 = 120 torr P 2 = 760 torr T 1 = 27.0 ◦ C + 273 = 300 K T 2 = 80.1 ◦ C + 273 = 353.1 K Here we use the Clausius-Clapeyron equa- tion: ln P 2 P 1 = Δ H vap R 1 T 1- 1 T 2 Plugging these values into the Clausius- Clapeyron equation, we have Δ H = R ln P 2 P 1 1 T 1- 1 T 2 = 8 . 314 J mol · K ln 760 torr 120 torr 1 300 K- 1 353 . 1 K = 3 . 07 × 10 4 J / mol 003 0.0 points A mixture consisting of 0.250 M N 2 (g) and 0.500 M H 2 (g) reaches equilibrium according to the equation N 2 (g) + 3 H 2 (g) → 2 NH 3 (g) . At equilibrium, the concentration of ammonia is 0.150 M. Calculate the concentration of H 2 (g) at equilibrium. 1. 0.425 M 2. 0.275 M correct 3. 0.150 M 4. 0.0750 M 5. 0.350 M Explanation: 004 0.0 points ”Like dissolves like” refers to the fact that two compounds are likely to spontaneously form a mixture when the two compounds Version 211 – Exam 1 – vanden bout – (51640) 2 1. have similar entropies 2. have intermolecular forces of the same strength 3. have intermolecular forces of the same type correct 4. have similar molecular weights Explanation: This refers to the compounds having the same type of intermolecular forces. polar liquids dissolving (mixing) with polar com- pounds. Compounds with the same strength of in- teraction will have similar vapor pressures (and boiling points) but they will not nec- essarily be miscible. 005 0.0 points Consider the reaction 3 Fe(s) + 4 H 2 O(g) → 4 H 2 (g) + Fe 3 O 4 (s) ....
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Exam 1-solutions - Version 211 – Exam 1 – vanden bout...

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