HW02-solutions - kamalska(mk23835 HW02 Quigley(104001 This...

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kamalska (mk23835) – HW02 – Quigley – (104001) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points What is the maximum possible value of the van’t Hoff factor for a solution of the strong electrolyte FeCl 3 that is so dilute that no ion association occur? 1. i = 2 2. i must be an odd number. 3. i = 0 4. i = 6 5. i = 4 correct Explanation: FeCl 3 Fe 3+ + 3 Cl Each mole of FeCl 3 can produce 4 moles of ions. 002 (part 1 of 3) 10.0 points For the next three questions, consider the following pressure diagram for a binary liquid containing A and B. 0 1 x A 90 torr 60 torr If 1760 g of A (110 g / mol) and 1995 g of B (105 g / mol) are mixed, what is the vapor pressure of B? 1. 27 torr 2. 48 torr correct 3. 76 torr 4. 90 torr 5. 60 torr Explanation: We have 1760 g 110 g / mol = 16 mol A and 1995 g 105 g / mol = 19 mol B, so the mole fraction of A is x A = mol A total mol = 16 mol 16 mol + 19 mol = 0 . 46 0 1 0 . 46 60 torr 48 torr 003 (part 2 of 3) 10.0 points What is the corresponding vapor pressure of A? 1. 90 torr 2. 27 torr correct 3. 60 torr 4. 48 torr 5. 76 torr Explanation: 0 1 0 . 46 60 torr 27 torr 004 (part 3 of 3) 10.0 points
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kamalska (mk23835) – HW02 – Quigley – (104001) 2 What is the corresponding total vapor pres- sure? 1. 90 torr 2. 60 torr 3. 48 torr 4. 76 torr correct 5. 27 torr Explanation: 0 1 0 . 46 60 torr 76 torr 005 10.0 points Chemical processes, like dissolving solids, are favored by (decreased, increased) potential energy and (decreased, increased, unchanged) disorder. 1. increased; increased 2. decreased; decreased 3. increased; decreased 4. decreased; increased correct 5. increased; unchanged Explanation: Chemical energy is potential energy. Exothermic reactions release energy as heat to the surroundings. In an exothermic re- action, this means the process has decreased its potential energy by giving some of it to its surroundings. This is one of the ways by which a chemical process can be favored. The other way is by an increase in disorder (en- tropy). These two considerations, when taken collectively, will indicate when a process will be favored or not. 006 10.0 points A well-behaved (non-volatile, non-ionic) so- lute is dissolved in water to yield a solution that freezes at - 0 . 930 C. At what temper- ature would this solution boil under 1.0000 atm pressure? Note that K f = 1 . 86 C/ m , K b = 0 . 512 C /m . 1. 99.744 C 2. 101.394 C 3. 100.930 C 4. 100.256 C correct 5. 100.000 C 6. 99.070 C Explanation: T f = - 0 . 9306 C K f = 1 . 866 C /m K b = 0 . 512 C /m First, we have to determine the molality of this solution. Knowing the freezing point will lead us to the molality. Using the equation Δ T f = K f m and knowing that the freezing point of water is 0 C, we know that the freezing point has been depressed 0.930 C. This means the molality of this solution is 0 . 930 C = (1 . 86 C /m )( m ) m = 0 . 500 m Then we can use the equation Δ T b = K b m to determine the boiling point elevation and the eventual boiling point of this aqueous so- lution.
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