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Homework 03

# Homework 03 - Version 094 Homewrok 3 Fakhreddine(53605 This...

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Version 094 – Homewrok 3 – Fakhreddine – (53605) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 1.0 points Several interesting observations from the world around you are listed below. Which of these is NOT explained by a colligative property? 1. Antifreeze is added to a car radiator to keep the car from overheating. 2. A lobster will die when placed in fresh water. 3. The freezing point of water is lowered when salt is added. 4. At high altitude it takes longer to cook spaghetti. correct 5. Water boils at a higher temperature when salt is added. Explanation: Colligative properties of a solution depend on the number of solute particles in solution, not the type. Boiling point variations due to pressure changes have nothing to do with solutions and colligative properties (boiling point variations due to particles in solution, etc.). 002 1.0 points The vapor pressure of pure CH 2 Cl 2 (molecu- lar weight = 85 g/mol) is 133 torr at 0 C and the vapor pressure of pure CH 2 Br 2 (molecu- lar weight 174 g/mol) is 11 torr at the same temperature. What is the total vapor pres- sure of a solution prepared from equal masses of these two substances? 1. vapor pressure = 144 torr 2. vapor pressure = 105 torr 3. vapor pressure = 93 torr correct 4. vapor pressure = 89 torr 5. vapor pressure = 124 torr 6. vapor pressure = 72 torr Explanation: For CH 2 Cl 2 , P 0 = 133 torr MW = 85 g / mol For CH 2 Br 2 , P 0 = 11 torr MW = 174 g / mol This is a combination of Raoult’s Law and Dalton’s Law of Partial Pressures. The an- swer does not depend on what the masses are, as long as they are equal. You can choose any mass you like, but to speed up calculations, it is convenient to choose the mass the same as one of the molecular weights given, so that the number of moles for one of the components is exactly ONE. So, for argument’s sake, choose 85 g to be the mass of each of the components. That way you have: (85 g CH 2 Cl 2 ) parenleftbigg 1 mol CH 2 Cl 2 85 g CH 2 Cl 2 parenrightbigg = 1 . 0 mol CH 2 Cl 2 Now calculate the moles of the other compo- nent. (85 g CH 2 Br 2 ) parenleftbigg 1 mol CH 2 Br 2 174 g CH 2 Br 2 parenrightbigg = 0 . 49 mol CH 2 Br 2 Once you have the two values for moles you can calculate the mole fraction of each com- ponent. n total = 1 . 0 + 0 . 49 = 1 . 49 mol X CH 2 Cl 2 = 1 . 0 mol 1 . 49 mol = 0 . 67 X CH 2 Br 2 = 0 . 49 mol 1 . 49 mol = 0 . 33 Then use those values in Raoult’s Law to get the vapor pressure for each component. Raoult’s Law states that: P A = X A P 0 A P CH 2 Cl 2 = (0 . 67)(133 torr) = 89 torr P CH 2 Br 2 = (0 . 33)(11 torr) = 3 . 6 torr

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Version 094 – Homewrok 3 – Fakhreddine – (53605) 2 Add the two together to get the total vapor pressure (Dalton’s Law). P total = P A + P B + · · · = 89 torr + 3 . 6 torr = 93 torr You might want to check to see that you get the same answer no matter what value you assume as the equal masses of the two components. As an additional challenge, can you solve this problem WITHOUT assuming a definite number of grams, by setting the mass of each component equal to an algebraic variable?
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Homework 03 - Version 094 Homewrok 3 Fakhreddine(53605 This...

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