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Exam1BKey - 1(18 A solution is prepared by dissolving 10.5...

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1) (18) A solution is prepared by dissolving 10.5 g of NaCl in 125 g of water. What is the freezing point of this solution? Δ T f = k f m c 10.5 g NaCl x NaCl g 58.44 NaCl mol 1 = 0.1797 mol NaCl m c = O H kg 0.125 ions mol 0.1797) x (2 2 = 2.87 m Δ T f = (1.86 o C/m) ( 2.87 m) = 5.34 o C freezing point = - 5.34 o C 2) (22) 580 mg of a protein (assume non-ionic) were dissolved in enough water to make 5.00 mL of solution. The osmotic pressure of the solution was measured to be 1.85 atm at 25 o C. What is the molar mass of the protein? Π = M c RT 1.85 atm = M c (0.0821 L-atm/K-mol) (298 K) M c = 0.0756 M 5.00 mL x mL mmol 0.0756 0.378 mmol M m = mmol 0.378 mg 580 = 1500 g/mol molar mass = 1500 g/mol Pledge: I have neither given nor received help during this exam. Signature: ______________________________ Name (print): ______________________________
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3) Isooctane (C 8 H 18 , M m = 114.2 g/mol) is one of the principle components of gasoline. a) (16) Write a balanced equation for the combustion of isooctane, where the only products are carbon dioxide and water. 2 C 8 H 18 + 25 O 2 barb2right 16 CO 2 + 18 H 2 O b) (22) Based on your equation in part a, how many liters of carbon dioxide, collected at 1 atm pressure and 25 o C, can be formed from the combustion of 1 gallon (2.62 kg)
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