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CH 302 HW 1

# CH 302 HW 1 - wei(cyw224 H01 Chapter 16.10-11 Mccord(52450...

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wei (cyw224) – H01: Chapter 16.10-11 – Mccord – (52450) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points How much NaNO 3 is needed to prepare 225 mL of a 1.55 M solution of NaNO 3 ? 1. 0.132 g 2. 0.244 g 3. 29.6 g correct 4. 4.10 g 5. 12.3 g Explanation: V = 225 mL M = 1 . 55 M ? g NaNO 3 = 225 mL × 1 L soln 1000 mL × 1 . 55 mol NaNO 3 1 L soln × 85 g NaNO 3 1 mol NaNO 3 = 29 . 6 g NaNO 3 002 10.0 points What is the final concentration of NaOH when 335 mL of 0.75 M NaOH are mixed with 165 mL of 0.15 M NaOH? 1. 0.82 M 2. 1.67 M 3. 0.45 M 4. 0.55 M correct 5. 0.18 M Explanation: V 1 = 335 mL M 1 = 0.75 M V 2 = 165 mL M 2 = 0.15 M Molarity is moles solute per liter of solution. Two solutions are being mixed together in this problem. We find the moles of NaOH in each of the individual solutions: ? mol NaOH = 335 mL soln × 1 L soln 1000 mL soln × 0 . 75 mol NaOH 1 L soln = 0 . 2512 mol NaOH ? mol NaOH = 165 mL soln × 1 L soln 1000 mL soln × 0 . 15 mol NaOH 1 L soln = 0 . 0248 mol NaOH The total moles of NaOH in the new solu- tion will be the sum of the moles in the two individual solutions: ? mol NaOH = 0 . 2512 mol + 0 . 0248 mol = 0 . 276 mol NaOH The total volume of the new solution will be the combined volume of the individual solu- tions: ? mL new soln = 335 mL + 165 mL = 500 mL soln To calculate the molarity of NaOH in the new solution, we divide the total moles of NaOH in the new solution by the total volume of the new solution: ? M NaOH = 0 . 276 mol NaOH 500 mL soln × 1000 mL soln 1 L soln = 0 . 552 M NaOH 003 10.0 points What is the molarity of a solution containing 3.50 grams of NaCl in 500 mL of solution? 1. 7 . 69 × 10 - 2 M 2. 5 . 98 × 10 - 2 M 3. 1 . 20 × 10 - 1 M correct

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wei (cyw224) – H01: Chapter 16.10-11 – Mccord – (52450) 2 4. 6 . 00 × 10 - 1 M 5. 1 . 54 × 10 - 2 M Explanation: m NaCl = 3 . 5 g V soln = 500 mL Molarity is mol/L. Thus we need to deter- mine the number of moles of NaCl that are present: ? mol NaCl = 3 . 50 g 1 mol NaCl 58 . 44 g NaCl = 0 . 0599 mol NaCl Then, to determine the molarity: ? M = mol solute L soln = 0 . 0599 mol NaCl 0 . 500 L soln = 0 . 12 M NaCl 004 10.0 points How many mL of a 0.75 N KOH solution should be added to a 500 mL flask to make 500 mL of a 0.300 M KOH solution? 1. 200 mL correct 2. 500 mL 3. 250 mL 4. 1250 mL 5. 50 mL 6. 100 mL Explanation: N 1 = 0.75 N N 2 = 0.3 N V 2 = 500 mL This is a dilution problem. 0 . 75 N KOH = 0 . 75 M KOH M 1 = 0.75 M M 2 = 0.300 M M 1 V 1 = M 2 V 2 V 1 = M 2 V 2 M 1 = (0 . 3 M) (500 mL) 0 . 75 M = 200 mL 005 10.0 points What is the mass of oxygen gas in a 15.2 L container at 26.0 C and 2.45 atm? Correct answer: 48 . 5451 g. Explanation: T = 26 . 0 C + 273 = 299 K P = 2 . 45 atm V = 15 . 2 L m = ? n = P V R T = (2 . 45 atm)(15 . 2 L) ( 0 . 0821 L · atm mol · K ) (299 K) = 1 . 51703 mol O 2 m = (1 . 51703 mol) 32 g mol = 48 . 5451 g O 2 006 10.0 points A mixture of oxygen and helium is 92.3% by mass oxygen. It is collected at atmospheric pressure (745 torr). What is the partial pressure of oxygen in this mixture? 1. 448 Torr correct 2. 299 Torr 3. 333 Torr 4. 688 Torr 5. 412 Torr Explanation: Assume you have 100 g of this mixture; calculate the number of moles: n O 2 = (92 . 3 g O 2 ) 1 mol O 2 31 . 9988 g O 2 = 2 . 91261 mol O 2 .
wei (cyw224) – H01: Chapter 16.10-11 – Mccord – (52450) 3 n He = (7 . 7 g He) 1 mol He 4 . 0026 g He = 1 . 92375 mol He .

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