mass you like but to speed up calculations it is convenient to choose the mass

Mass you like but to speed up calculations it is

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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? 006 10.0points You place two beakers into an evacuated chamber. One beaker has 100 mL of a 0.1 M solution of NaCl in water and the other has 100 mL of a 1.0 M sugar solution. When the system comes to equilibrium, 1. all of the water will be in the sugar solu- tion. 2. the beaker with the sugar solution will have a larger volume. correct 3. the two beakers will have identical vol- umes. 4. the beaker with the salt solution will have a larger volume. 5. all of the water will be in the salt solu- tion. Explanation:
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gamboa (gg8745) – Homework #2 – holcombe – (51160) 3 Both liquids will evaporate and condense from each beaker. The chamber will come to equilibrium when the two beakers have the same concentration (same vapor pres- sure). The salt solution will become more concentrated (lower volume) and the sugar so- lution will become less concentrated (higher volume). 007 10.0points Which of the solutions below will have the greater boiling point and what will it be? K b = 0 . 512 C / m for water. I) 135 g of glucose (C 6 H 12 O 6 ) dissolved in 0.5 kg water II) 35 g of NaCl dissolved in 0.5 kg of water 1. Solution II with a boiling point of 100 . 6 C 2. Solution I with a boiling point of 118 . 4 C 3. Solution I with a boiling point of 100 . 8 C 4. Solution II with a boiling point of 101 . 2 C correct Explanation: First calculate the molality of each solution, which is mol solute per kg solvent. To do this, you will need to convert from mass to mols, using molar mass. Remember also that i = 2 for NaCl and i = 1 for glucose. Using Δ T = imK b you will find that Δ T = 0 . 768 C for the glucose solution and 1.227 for the salt solution, which means the boiling point of the salt solution will be 101.2 C.
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