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# HW2_pdf - leazenby(jl35875 Homework 2 Sutclie(52440 This...

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leazenby (jl35875) – 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 69 . 9 g sodium nitrate is dissolved in water to make 717 g of solution. What is the percent sodium nitrate in the solution? 1. 10 . 802% 2. 100% 3. 0 . 0974895% 4. 9 . 74895% correct Explanation: m NaNO 3 = 69 . 9 g m soln = 717 g percent = mass solute mass solution × 100 % = 69 . 9 g NaNO 3 717 g solution × 100 % = 9 . 74895% 002 10.0 points What is the molarity of a solution composed of 4.506 g of HCl in 0.5080 L of solution? Correct answer: 0 . 243282 M. Explanation: m HCl = 4.506 g V soln = 0.5080 L [HCl] = ? 4.506 g HCl × 1 mol HCl 36.46 g HCl = 0 . 123587 mol M = 0 . 123587 mol HCl 0.5080 L soln = 0 . 243282 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 6 . 967 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 9 . 929 mmol Na 2 CO 3 ? Correct answer: 37 . 7628 mL. Explanation: m Na 2 CO 3 = 6 . 967 g V = 250 mL = 0 . 25 L n Na 2 CO 3 = 9 . 929 mmol = 0 . 009929 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 . 967 g (105 . 99 g / mol) (0 . 25 L) = 0 . 26293 M Na 2 O 3 V = (0 . 009929 mol Na 2 CO 3 ) × parenleftbigg 1 L Na 2 CO 3 0 . 26293 mol Na 2 CO 3 parenrightbigg = 0 . 0377628 L = 37 . 7628 mL 004 (part 2 of 2) 10.0 points What volume of solution should be dispensed from the buret to provide 6 . 234 mmol CO 2 3 ? Correct answer: 23 . 7097 mL. Explanation: n CO 2 - 3 = 6 . 234 mmol V = 0 . 006234 mol CO 2 3 parenleftBigg 1 mol Na 2 CO 3 1 mol CO 2 3 parenrightBigg × parenleftbigg 1 L Na 2 CO 3 0 . 26293 mol Na 2 CO 3 parenrightbigg = 0 . 0237097 L = 23 . 7097 mL

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leazenby (jl35875) – 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. 31 m 2. 18 m 3. 15 m correct 4. 21 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. Correct answer: 3 . 0909 m . Explanation: m toluene = 35.6 g m benzene = 125 g m toulene = n toulene kg benzene = 35 . 6 g 92 . 14 g · mol 1 toulene 0 . 125 kg benzene = 3 . 0909 m 007 10.0 points Formalin is a solution of 40.0% formaldehyde (H 2 CO), 10.0% methyl alcohol (CH 3 OH), and 50.0% water by mass. Calculate the mole fraction of methyl alcohol in formalin. Correct answer: 0 . 0706436.
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HW2_pdf - leazenby(jl35875 Homework 2 Sutclie(52440 This...

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