017 10 points The vapor pressure of pure CH 2 Cl 2 molecu lar weight 85 gmol is

# 017 10 points the vapor pressure of pure ch 2 cl 2

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017 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 = 89 torr 2. vapor pressure = 105 torr 3. vapor pressure = 144 torr 4. vapor pressure = 93 torr correct 5. vapor pressure = 72 torr 6. vapor pressure = 124 torr Explanation: For CH 2 Cl 2 , P 0 = 133 torr MW = 85 g / mol
spano velasco (ds28282) – hw 02 – Fakhreddine – (52395) 6 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 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? 018 1.0 points Consider two liquids A and B. The vapor pressure of pure A (molecular weight = 50 g/mol) is 225 torr at 25 C and the vapor pressure of pure B (molecular weight = 75 g/mol) is 90 torr at the same temperature. What is the total vapor pressure at 25 C of a solution that is 25% A and 75% B by weight? 1. 108 torr 2. 225 torr 3. 76 torr 4. 191 torr 5. 115 torr 6. 335 torr 7. 124 torr 8. 135 torr correct 9. 203 torr Explanation: For A, P 0 = 255 torr MW = 50 g / mol For B, P 0 = 90 torr MW = 75 g / mol The mole fractions are 1 3 for A and 2 3 for B. parenleftBig 1 3 parenrightBig (225) + parenleftBig 2 3 parenrightBig (90) = 135 torr 019 1.0 points At a given temperature, pure A has a vapor pressure of 140 torr and pure B has a vapor pressure of 60 torr. At this temperature, we
spano velasco (ds28282) – hw 02 – Fakhreddine – (52395) 7 make a solution in which the mole fractions of A and B are equal. We then allow this solution to reach equilibrium with its vapor.

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• Fall '07
• Holcombe
• Chemistry, Mole, Vapor pressure, Molecule, Torr, spano velasco