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COPPIN STATE COLLEGE DEPARTMENT OF NATURAL SCIENCES Name ______Last First Chemistry 103-101 Examination V: Chapters 6 States of Matter: Gases, Liquids, and Solids Dr. Alfred N. Amah

This examination consists of two parts: Part A (50 points) consists of 10 multiple choice questions; carefully read each question and all the possible responses. Choose the correct answer by cycling the alphabet before your choice. Part B (50 points) contains short structured questions. Please turn in all examination papers. GOOD LUCK!

Part A (50 points). Multiple Choice Questions.

1. What is the equivalent of one standard atmosphere?

a. 760 mm Hg. b. 101,325 Pa. c. 760 torr d. All of the above

2. A pressure of 740 mm Hg is equal to?

a.1 atm b. 0.974 atm c. 1.03 atm d. 0.740 atm

3 A pressure of 2 atm is roughly equivalent to how many pascals? [1 atm = 101,325 Pa] a. 1.0 x 105 b. 1.0 x 103 c. 2.0 x 105 d. 2.0 x 107 2

4. The air humans exhale consists chiefly of nitrogen, oxygen, carbon dioxide, and water vapor. Under a total atmospheric pressure of 745 mm Hg, PN2 = 558 mmHg; PO2 = 114 mm Hg; PH2O = 46 mm Hg; What is the PCO2 ?

a. 42 mm Hg b. 27 mm Hg c. 16 mm Hg. d. 718 mm Hg.

5. A sample of nitrogen gas at 20.0 oC and in a volume of 875 mL has a pressure of 730 mm Hg. If at 20.0 oC, the volume of this gas is changed to 955 mL, what is the new pressure?

a. 750 mm Hg b. 658 mm Hg c. 797 mm Hg d. 669 mm Hg

6. A sample of gas is at 740 mmHg and 25 oC and occupies a volume of 600 mL. The gas is heated at 740 mm Hg to 30 oC. The new volume is.? [ Hint the pressure is constant] a. 600 mL b.610 mL c. 720 mL d. 590 mL

7. Which of the following conditions are known as STP (standard conditions of temperature and pressure)?

a. 0 K and 1 atm b.273 K and 760 mm Hg c. 273 oC and 760 mm Hg d. 0 oF and 760 mm Hg 3

8. The vapor pressure of four liquids at room temperature are: A = 100 mm Hg; B = 225 mm Hg; C = 300 mm Hg; and D = 455 mm Hg. Which one of these liquids has the lowest boiling point?

a. A b. B c. C d. D

9. Where could water boil at 100 oC in an open pan?

a. On the shore of the Atlantic Ocean. b. On Mount Washington (5000 ft) c. On Mount Whitney (15,000 ft) d. On top of the Empire State Building (1000 ft)

10. If you know the pressure, temperature, volume, and the identity of a gas (i.e. the number of moles of the gas), you can calculate the mass of the gas if you apply what law? a. Charles' Law b. Boyle's Law c. the universal gas law d. none of the above 4

PART B (50 POINTS). Short Structured Questions. For each of the following problems show your calculation and setup. In both your setup and answer show units and follow rules of significant figures.

1. (Credit 15). A sample of an anesthetic gas with a volume of 925 mL at 20 oC and 750 mm Hg pressure is warmed to 37.0 oC at a pressure of 745 mm Hg. Calculate the final volume of the gas. [ Please show your work for proper credit].

2. (Credit 10) Calculate the volume that is occupied by 64.0 g of oxygen gas, ( O2) at STP. [ O2 = 32 g/ mol. [STP means 0.00 oC, 1.00 atm} 5

3.(25 points). Osmosis is a colligative property; it is dependent on the concentration of solute particles in solution. For example a solution of 1 M NaCl produces two moles of + - particles per liter of solution (one mole of Na and one mole Cl ). A solution of 1 M CaCl2 produces 3 moles of particles (one mole of Ca+ and two moles of Cl-). By convention, the molarity of particles in solution is called osmolarity, abbreviated as Osm, in osmotic pressure calculations.

-3 (a) Calculate the osmolarity of 5.0 x 10 M Na3PO4 [ Hint: Na3PO4  3Na 3- + PO4 ] Please show your work.(10 points)

(b) For the solution described in question 1(a) above, calculate the osmotic pressure (?). Assume a temperature of 25 oC. (Hint: Osmotic pressure  = MRT ) Please show your work.(15 points). 6

Some Useful Information

P1V1 = P2V2. P1V1/T1 = P2V2/T2

Pv = nRT (universal gas law) R = 0.08206 L. atm/ K. mol

 = MRT (osmotic pressure equation)

V1/T1 = V2/T2 M = moles of solute/volume of solution in liters

P1/T1 = P2/T2

PH + pOH = 14.0

PH = -Log [H+] pOH = -Log [OH-]

[H+] x [OH-] = 1.0 x 10-14 @ 25 0 C