Homework 2-solutions

# Homework 2-solutions - Version 218 – Homework 2 –...

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Unformatted text preview: Version 218 – Homework 2 – Walker – (53985) 1 This print-out should have 40 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 1.0 points What can best explain the anomalous prop- erties of water? 1. resonance 2. molecular weight 3. isotopes 4. hydrogen bonding forces correct 5. dipole-dipole forces Explanation: 002 1.0 points How many moles of HCl are present in 40.0 mL of a 0.035 M solution? 1. 0.25 mol 2. 0.0060 mol 3. 0.0012 mol 4. 0.0014 mol correct 5. 0.012 mol Explanation: V = 40 . 0 mL M = 0 . 035 M ? mol HCl = 40 . 0 mL soln × 1 L 1000 mL × . 035 mol HCl 1 L soln = 0 . 0014 mol HCl 003 1.0 points How many grams of CaBr 2 are needed to prepare 4.65 L of a 7.65 M solution? 1. 5.26 g 2. 0.00303 g 3. 7110 g correct 4. 122 g 5. 4270 g 6. 0.178 g Explanation: M = 7 . 65 M V = 4 . 65 L ? g CaBr 2 = 4 . 65 L × 7 . 65 mol CaBr 2 L soln × 200 g CaBr 2 1 mol = 7110 g 004 1.0 points A student was asked to prepare exactly 250 mL of a 0.500 M aqueous potassium hydroxide solution. What mass of potassium hydroxide (molar mass = 56.10 g/mol) must the student dissolve in the 250 mL of solution? 1. 28.1 g 2. 7.01 g correct 3. 56.1 g 4. None of these 5. 3.0 g 6. 14.0 g Explanation: V = 250 mL M KOH = 0 . 500 M Calculate the number of moles of KOH needed: n KOH = 0 . 250 L soln × . 500 mol KOH 1 L soln = 0 . 125 mol m KOH = 0 . 125 mol KOH × 56 . 1 g 1 mol KOH = 7 . 0125 g KOH Version 218 – Homework 2 – Walker – (53985) 2 005 1.0 points What is the molarity of a 2.073 L solution that is made from 11.10 g of NaCl? Correct answer: 0 . 0916249 M. Explanation: V solution = 2.073 L m NaCl = 11.10 g M = ? M = mol solute L soln ? mol NaCl = 11.10 g NaCl × parenleftbigg 1 mol NaCl 58.44 g NaCl parenrightbigg = 0 . 18994 mol NaCl ? M = mol NaCl L soln = . 189938 mol 2.073 L = 0 . 0916249 M 006 1.0 points What is the molarity of a solution composed of 3.938 g of HCl in 0.1515 L of solution? Correct answer: 0 . 712929 M. Explanation: m HCl = 3.938 g V soln = 0.1515 L [HCl] = ? 3.938 g HCl × 1 mol HCl 36.46 g HCl = 0 . 108009 mol M = . 108009 mol HCl 0.1515 L soln = 0 . 712929 M HCl 007 1.0 points A 0.2 M aqueous calcium hydroxide solution contains 1. 0.2 M calcium ion and 0.2 M hydroxide ion. 2. 0.1 M calcium ion and 0.2 M hydroxide ion. 3. 0.2 M calcium ion and 0.4 M hydroxide ion. correct 4. 0.2 M calcium ion and 0.1 M hydroxide ion. 5. 0.4 M calcium ion and 0.2 M hydroxide ion. Explanation: 008 1.0 points How much acetone (CH 3 COCH 3 ) would be needed to obtain 7.85 L of acetone? The density of acetone is 0.851 g/mL. 1. 6680 mol 2. 0.535 mol 3. Not enough information 4. 115 mol correct 5. 135 mol 6. 0.115 mol Explanation: V = 7 . 85 L density = 0 . 851 g / mL ? mol C 3 H 6 O = 7 . 85 L × 1000 mL 1 L × . 851 g C 3 H 6 O mL × 1 mol C 3 H 6 O 58 g = 115 mol 009 1.0 points Draw the Lewis electron dot structure for SF 2 ....
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## This note was uploaded on 05/01/2008 for the course CH 305 taught by Professor Sutcliffe during the Spring '08 term at University of Texas at Austin.

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Homework 2-solutions - Version 218 – Homework 2 –...

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