hw2solution_pdf

hw2solution_pdf - So(dds785 – Homework 2 – Sutcliffe...

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Unformatted text preview: So (dds785) – Homework 2 – Sutcliffe – (52440) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Read the questions carefully. Look for whether youve been give or asked for moLaR- ity or moLaLity! The last half of the questions require you to consider the vant’Hoff factor i, for solutes that might dissociate into several ions, so read carefully. Assume ideal behavior of solutions in questions 13 through 18. This covers the remaining material on Exam 1. We will cover chapter 6 on Exam 2. 001 10.0 points 26 . 6 g sodium nitrate is dissolved in water to make 550 g of solution. What is the percent sodium nitrate in the solution? 1. 100% 2. 4 . 83636% correct 3. . 0483636% 4. 5 . 08216% Explanation: m NaNO 3 = 26 . 6 g m soln = 550 g percent = mass solute mass solution × 100 % = 26 . 6 g NaNO 3 550 g solution × 100 % = 4 . 83636% 002 10.0 points What is the molarity of a solution composed of 7.074 g of HCl in 0.5874 L of solution? Correct answer: 0 . 330304 M. Explanation: m HCl = 7.074 g V soln = 0.5874 L [HCl] = ? 7.074 g HCl × 1 mol HCl 36.46 g HCl = 0 . 194021 mol M = . 194021 mol HCl 0.5874 L soln = 0 . 330304 M HCl 003 (part 1 of 2) 10.0 points A student investigating the properties of so- lutions containing carbonate ions prepared a solution containing 5 . 647 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 . 168 mmol Na 2 CO 3 ? Correct answer: 33 . 6345 mL. Explanation: m Na 2 CO 3 = 5 . 647 g V = 250 mL = 0 . 25 L n Na 2 CO 3 = 7 . 168 mmol = 0 . 007168 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 = 5 . 647 g (105 . 99 g / mol) (0 . 25 L) = 0 . 213114 M Na 2 O 3 V = (0 . 007168 mol Na 2 CO 3 ) × parenleftbigg 1 L Na 2 CO 3 . 213114 mol Na 2 CO 3 parenrightbigg = 0 . 0336345 L = 33 . 6345 mL 004 (part 2 of 2) 10.0 points What volume of solution should be dispensed from the buret to provide 7 . 705 mmol CO 2 − 3 ? Correct answer: 36 . 1543 mL. Explanation: n CO 2- 3 = 7 . 705 mmol V = 0 . 007705 mol CO 2 − 3 parenleftBigg 1 mol Na 2 CO 3 1 mol CO 2 − 3 parenrightBigg × parenleftbigg 1 L Na 2 CO 3 . 213114 mol Na 2 CO 3 parenrightbigg = 0 . 0361543 L = 36 . 1543 mL So (dds785) – Homework 2 – Sutcliffe – (52440) 2 005 10.0 points 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. 18 m 4. 31 m 5. 26 m Explanation: 006 10.0 points Toluene (C 6 H 5 CH 3 ) is a liquid compound similar to benzene (C 6 H 6 ). Find the molal- ity of toluene in a solution that contains 35.6 grams of toluene and 125 grams of benzene....
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hw2solution_pdf - So(dds785 – Homework 2 – Sutcliffe...

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