solution3_pdf - toupal (rgt374) Homework 3 Sutcliffe...

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Unformatted text preview: toupal (rgt374) Homework 3 Sutcliffe (52410) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. Read the last third of the questions care- fully; you may have to take into account the vant Hoff factor, i. 001 10.0 points 58 . 7 g sodium nitrate is dissolved in water to make 616 g of solution. What is the percent sodium nitrate in the solution? 1. 9 . 52922% correct 2. 100% 3. . 0952922% 4. 10 . 5329% Explanation: m NaNO 3 = 58 . 7 g m soln = 616 g percent = mass solute mass solution 100 % = 58 . 7 g NaNO 3 616 g solution 100 % = 9 . 52922% 002 10.0 points How much CH 2 O is needed to prepare 445 mL of a 2.65 M solution of CH 2 O? 1. 17.6 g 2. 35.4 g correct 3. 179 g 4. Not enough information is given. 5. 0.0393 g 6. 45.6 g 7. 79.6 g Explanation: M = 2 . 65 M V = 445 mL = 0 . 445 L ? g CH 2 O = (0 . 445 L) 2 . 65 mol CH 2 O L soln 30 g CH 2 O 1 mol = 35 . 3775 g 003 (part 1 of 2) 10.0 points A student investigating the properties of so- lutions containing carbonate ions prepared a solution containing 6 . 749 g Na 2 CO 3 in a flask of volume 250 mL. Some of the solution was transferred to a buret. What volume of so- lution should be dispensed from the buret to provide 7 . 558 mmol Na 2 CO 3 ? Correct answer: 29 . 6737 mL. Explanation: m Na 2 CO 3 = 6 . 749 g V = 250 mL = 0 . 25 L n Na 2 CO 3 = 7 . 558 mmol = 0 . 007558 mol FW Na 2 CO 3 = 2 (22 . 9958 g / mol) + 12 . 011 g / mol + 3 (15 . 9994 g / mol) = 105 . 99 g / mol M Na 2 CO 3 = 6 . 749 g (105 . 99 g / mol) (0 . 25 L) = 0 . 254703 M Na 2 O 3 V = (0 . 007558 mol Na 2 CO 3 ) parenleftbigg 1 L Na 2 CO 3 . 254703 mol Na 2 CO 3 parenrightbigg = 0 . 0296737 L = 29 . 6737 mL 004 (part 2 of 2) 10.0 points What volume of solution should be dispensed from the buret to provide 3 . 604 mmol CO 2 3 ? Correct answer: 14 . 1498 mL. Explanation: n CO 2- 3 = 3 . 604 mmol toupal (rgt374) Homework 3 Sutcliffe (52410) 2 V = 0 . 003604 mol CO 2 3 parenleftBigg 1 mol Na 2 CO 3 1 mol CO 2 3 parenrightBigg parenleftbigg 1 L Na 2 CO 3 . 254703 mol Na 2 CO 3 parenrightbigg = 0 . 0141498 L = 14 . 1498 mL 005 10.0 points Do not forget the moLaLity is moles of solute per kg of SOLVENT!! Calculate the molality of perchloric acid in 9.2 M HClO 4 (aq). The density of this solution is 1.54 g/mL. 1. 15 m correct 2. 21 m 3. 26 m 4. 31 m 5. 18 m Explanation: 006 (part 1 of 2) 10.0 points Calculate the molality of an aqueous 4 . 08% by mass HCl solution. Correct answer: 1 . 1667 m . Explanation: MW H = 1 . 0079 g / mol MW Cl = 35 . 45 g / mol %m HCl = 4 . 08% 1.0 kg of solution will contain 40 . 8 g HCl and 959 . 2 g H 2 O....
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solution3_pdf - toupal (rgt374) Homework 3 Sutcliffe...

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