This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Version 100 – Exam 3 – quy – (50970) 1 This printout should have 30 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. Be SURE to completely bubble in your scantron form. Your score may be penalized for incompletely filling in the name, UTEID, or version number spaces of the form. 001 10.0 points A sample of N 2 gas occupies a volume of 746 mL at STP. What volume would N 2 gas occupy at 155 ◦ C at a pressure of 368 torr? 1. 2415 mL correct 2. 566 mL 3. 588 mL 4. 312 mL 5. 3295 mL 6. 323 mL 7. 983 mL 8. 1792 mL Explanation: V 1 = 746 mL T 1 = 273 . 15 ◦ K P 1 = 1atm = 760torr T 2 = 155 ◦ C = 428 K P 2 = 368torr Combined Gas Law: P 1 V 1 T 1 = P 2 V 2 T 2 V 2 = P 1 V 1 T 2 T 1 P 2 = (760torr)(746mL)(428 . 15K) (273 . 15 K)(368torr) = 2414 . 9mL 002 10.0 points A sample of water vapor is allowed to reach equilibrium in a closed container at 127 ◦ C. The equilibrium pressure is measured as 8 . 2 atm. What is the molar concentration parenleftbigg mol L parenrightbigg of the water vapor? 1. . 80 mol · L − 1 2. not enough information is given 3. . 25 mol · L − 1 correct 4. 1 . 16 mol · L − 1 Explanation: T = 127 ◦ C = 400 K mol L = n V = P RT = 8 . 2 atm . 0821 L · atm · mol − 1 · K − 1 · 400 K = 0 . 25 mol · L − 1 003 10.0 points A mixture of gases at 25 ◦ C and 1 atm. pres sure contains 10.0 L of oxygen gas, 20.0 L of nitrogen gas, 30.0 L of cabon dioxide, and 15.0 L of hydrogen gas. The percent (by volume) of oxygen gas is 1. 7.5% 2. 40.0% 3. 13.3% correct 4. 10.0% 5. 75.0% Explanation: Percent = Part Whole × 100% Part = 10 L Whole = 10 L + 20 L + 30 L + 15 L = 75 L Percent = 10 L 75 L × 100% = 13 . 3333% Version 100 – Exam 3 – quy – (50970) 2 004 10.0 points What is the molar mass of a compound that takes 1 . 9 times as long to effuse through a porous plug as it did for the same amount of XeF 2 at the same temperature and pressure? 1. 433.408 2. 308.549 3. 1280.33 4. 643.763 5. 611.173 6. 819.412 7. 380.925 8. 677.2 9. 1100.87 10. 857.081 Correct answer: 611 . 173 g / mol. Explanation: Let Eff 2 be the effusion rate of the com pound and Eff 1 be the rate of the XeF 2 : Eff 1 = 1 . 9 Eff 2 MW 1 = 169 . 3 g / mol By Graham’s law of effusion, Eff ∝ 1 √ MW , so Eff 1 Eff 2 = radicalbigg MW 2 MW 1 MW 2 = MW 1 parenleftbigg Eff 1 Eff 2 parenrightbigg 2 = 169 . 3 g / mol parenleftbigg 1 . 9 Eff 2 Eff 2 parenrightbigg 2 = 611 . 173 g / mol . 005 10.0 points What volume will 20.0 L of He at 50.00 ◦ C and 1201 torr occupy at STP? 1. 20.0 L 2. 12.8 L 3. 31.1 L 4. 26.7 L correct 5. 18.6 L Explanation: P 1 = 1201 torr P 2 = 760 torr V 1 = 20 L T 2 = 273.15 K T 1 = 50 ◦ + 273.15 = 323.15 K Using the Combined Gas Law, P 1 V 1 T 1 = P 2 V 2 T 2 and recalling that STP implies standard tem perature (273.15 K) and pressure (1 atm or 760 torr), we have V 2 = P 1 V 1 T 2 T 1 P 2 = (1201 torr) (20 . 0 L) (273 . 15 K) (323 . 15 K) (760 torr) = 26 . 7151 L...
View
Full
Document
This note was uploaded on 12/03/2011 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas.
 Fall '07
 Fakhreddine/Lyon

Click to edit the document details