Chem 162-2011 Chapter 12-Some Prop Solns pract probl

Chem 162-2011 Chapter 12-Some Prop Solns pract probl -...

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PRACTICE PROBLEMS CHEM 162-2010 EXAM I CHAPTER 12 - PROPERTIES OF SOLUTIONS SOLUTION CONCENTRATIONS CONCEPTS (Molarity, mole fraction, etc.) SOLUTION CONCENTRATIONS CALCULATIONS (Molarity, mole fraction, etc.) ENERGETICS OF SOLUTIONS AND SOLUBILITY CONCEPTS ENERGETICS OF SOLUTIONS AND SOLUBILITY CALCULATIONS VAPOR PRESSURE OF SOLUTIONS CONCEPTS (Henry’s Law, Raoult’s Law) VAPOR PRESSURE OF SOLUTIONS CALCULATIONS COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE CHANGES) CONCEPTS COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE CHANGES) CALCULATIONS MISCELLANEOUS E. Tavss, PhD 1
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SOLUTION CONCENTRATIONS CONCEPTS 2
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SOLUTION CONCENTRATIONS CALCULATIONS 36 Chem 162-2007 Final exam + answers Chapter 12 – Properties of Solutions Solution Concentrations Calculations g solute /g solution → M An automobile antifreeze contains equal volumes of ethylene glycol (d = 1.114 g/mL); FW = 62.07) and water (d = 1.00 g/mL) at 25 °C. The solution has a density of 1.06 g/mL. What is the molarity of ethylene glycol in the solution? [Hint: the volumes are NOT additive!] A . 9.00 B. 5.68 C. 4.50 D. 0.323 E. 8.97 V EG = V H2O D EG = 1.114g/mL D H2O = 1.00g/mL Molarity = moles ethylene glycol/1000 mL solution Let’s say that there is 1 mL each of EG and H 2 O present. Therefore, there is 1.114 g EG and 1.00 g H 2 O present. 1.114gEG/1.00gH 2 O 1.114gEG/2.114gsolution Convert gEG/gsolution into moleEG/Lsoln. First do numerator, then denom. 1.114gEG/62.07gmol -1 = 0.0179475molEG 2.114gsolution x (1.000L/1060g) = 1.99434 x 10 -3 L 0.0179475molEG/(1.99434 x 10 -3 L) = 8.9992 molEG/L = 9.00M Note: Having 9.00 and 8.97 as options is anal retentive. With a very small rounding- off error, one can easily get an answer closer to 8.97. 3
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6. CHEM 162-2007 EXAM I + ANSWERS CHAPTER 12 - PROPERTIES OF SOLUTIONS SOLUTION CONCENTRATIONS CALCULATIONS (MOLE FRACTION) What mass of sucrose, C 12 H 22 O 11 , must be dissolved in 1 L of water (d = 0.998 g/ml) to obtain a solution with 0.025 mole fraction C 12 H 22 O 11 ? A) 165 B ) 486 C) 362 D) 246 E) 452 mole fraction = mole solute /mole solution 0.025mol sucrose/mol solution → gsucrose/1Lsolvent Numerator: 0.025 mol sucrose x 342.34gmol -1 = 8.5585g sucrose Denominator: mol solute + mol solvent = mol solution mol solvent = mol solution – mol solute mol solvent = 1mol solution – 0.025mol sucrose = 0.975mol solvent 0.975mol solvent x 18.02g/mol = 17.5695 g solvent 17.5695gH2O x (1 mL/0.998g) = 17.60 mL solvent = 0.01760Lsolvent numerator/denominator: 8.5585gsucrose/0.01760Lsolvent = 486.3gsucrose/Lsolvent B 2. CHEM 162-2007 EXAM I + ANSWERS CHAPTER 12 - PROPERTIES OF SOLUTIONS SOLUTION CONCENTRATIONS CALCULATIONS (Molarity) The density of a solution that is 20.0% HClO 4 by mass is 1.138 g/mL. Calculate the molarity of the HClO 4 . A) 1.75
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Chem 162-2011 Chapter 12-Some Prop Solns pract probl -...

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