Exam 1-solutions

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

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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) ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 9

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online