This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 Δ T b = ik b m For CaCl 2 , i = 3 and Δ 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 − Δ 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: π = 0.54 torr density = 25 g/L T = 25 ◦ C + 273 = 298 K Here we can use the equation π = M RT for osmotic pressure, to our advantage. How ever, to use the “standard” value of R , we’ll 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....
View
Full Document
 Fall '10
 Sutcliffe
 Chemistry, Equilibrium, Reaction, Chemical reaction, Kc

Click to edit the document details