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Unformatted text preview: silva (jrs4378) Homework 3 Sutcliffe (51060) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. This is the remainder of the material for Exam 1. 001 10.0 points Find the normal boiling point of 4 . 3 m CaCl 2 (aq), assuming it undergoes complete dissociation. The boiling point constant of water is 0 . 51 K kg / mol. Your answer must be within 0.3% Correct answer: 106 . 579 C. Explanation: m = 4 . 3 m k b = 0 . 51 K kg / mol T b = ik b m For CaCl 2 , i = 3 and T b = 3 (0 . 51 K kg / mol) (4 . 3 m ) = 6 . 579 K = 6 . 579 C . The boiling point will be 100 C + 6 . 579 C = 106 . 579 C . 002 10.0 points Glycerol (a nonelectrolyte that can be used as antifreeze in a car radiator) has a density of 1.261 g/mL and a molecular weight of 92 g. If 100 mL are mixed with 1 kg of water, what is a good estimate of the temperature at which this solution will freeze? ( K f water = 1 . 861 C/ m .) 1. 0.0 C 2. 12 . 4 C 3. . 23 C 4. 4 . 8 C 5. 2 . 6 C correct Explanation: density glycerol = 1.261 g/mL m water = 1 kg MW glycerol = 92 g V glycerol = 100 mL K f H 2 O = 1 . 861 C /m m glycerol = 100 mL glycerol 1 kg water 1 . 261 g glycerol 1 mL glycerol 1 mol glycerol 92 g glycerol = 1 . 371 m T f = T f T f = T f K f m = 0 . 001 C (1 . 861 C /m ) (1 . 371 m ) = 2 . 55 C 003 10.0 points Polyacrylamide is a watersoluble polymer whose aqueous solution containing 25 g/L de velops an osmotic pressure of 0.54 torr at 25 C. Find the approximate molecular weight of the polymer sample. 1. 860,000 g/mol correct 2. 8,600,000 g/mol 3. 1133 g/mol 4. 350,000 g/mol 5. 35,000 g/mol Explanation: = 0.54 torr density = 25 g/L T = 25 C + 273 = 298 K Here we can use the equation = M RT for osmotic pressure, to our advantage. How ever, to use the standard value of R , well have to convert our pressure from torr to atm: 0.54 torr parenleftBig 1 atm 760 torr parenrightBig = 0.00071 atm Then use the equation for osmotic pressure: . 00071 atm = M parenleftbigg . 08206 L atm K mol parenrightbigg (298 K) = 2 . 9 10 5 mol / L silva (jrs4378) Homework 3 Sutcliffe (51060) 2 However, we know the solution is 25 g/L, so we can determine the molecular weight of the polymer with a simple division: 25 g / L 2 . 9 10 5 mol / L = 860,000 g/mol 004 10.0 points10....
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 Fall '10
 Sutcliffe

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