solution_pdf

# solution_pdf - Exam 1 holcombe(51055 This print-out should...

• Notes
• 10

This preview shows page 1 - 3 out of 10 pages.

– Exam 1 – holcombe – (51055) 1 This print-out should have 31 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The vapor pressure of benzene at 25 C is 94.6 torr and its enthalpy of vaporization is 30.8 kJ · mol 1 . Estimate the normal boiling point of benzene. Assume the enthalpy of vaporization is independent of temperature. 1. 640 K 2. Not enough information is given. 3. 358 K correct 4. 624 K 5. 470 K Explanation: T 1 = 25 C + 273 . 15 = 298 . 15 K P 1 = 94.6 torr P 2 = 1 . 0 atm = 760 torr At ANY liquid’s normal boiling point, the vapor pressure = 1.0 atm. Using the Clausius-Clapeyron equation, ln parenleftbigg P 2 P 1 parenrightbigg = Δ H vap R parenleftbigg 1 T 1 - 1 T 2 parenrightbigg R Δ H vap ln parenleftbigg P 2 P 1 parenrightbigg = 1 T 1 - 1 T 2 1 T 2 = 1 T 1 - R Δ H vap ln parenleftbigg P 2 P 1 parenrightbigg 1 T 2 = 1 298 . 15 K - 8 . 314 J / (mol · K) 30 . 8 × 10 3 J / mol × ln parenleftbigg 760 torr 94 . 6 torr parenrightbigg = 0 . 00279156 T 2 = 1 0 . 00279156 = 358 . 222 K 002 10.0 points Solution A contains 0.5 grams of solute A and solution B contains 0.5 grams of solute B (both A and B are nonelectrolytes). Other than the solutes, the solutions are identical (volume, temperature, etc .). Now you mea- sure the osmotic pressure of each solution and find that the osmotic pressure of solution B is twice that of solution A. What is the relation- ship between the molecular weights of solutes A and B? 1. The molecular weight of solute B is four times that of solute A. 2. The molecular weight of solute A is twice that of solute B. correct 3. The molecular weight of solute A is four times that of solute B. 4. The molecular weight of solute B is twice that of solute A. 5. The molecular weight ratio cannot be determined from this experiment. Explanation: Because the solution with B has twice the osmotic pressure, it must also have twice the number of dissolved particles (number of moles of B). The only way to have more moles for B (knowing the same mass of A and B was used) is that B has a smaller molecular weight than A. For the osmotic pressure to be ex- actly twice that of A, the molecular weight of B must be half that of A which is the same as A being twice that of B. 003 10.0 points Consider the following system at equilibrium. H 2 (g) + I 2 (g) 2 HI(g) + heat Which response includes all the following that will shift the equilibrium to the left, and no others? I) increasing the temperature II) decreasing the temperature III) increasing the pressure IV) decreasing the pressure V) removing some HI VI) adding some HI

Subscribe to view the full document.

– Exam 1 – holcombe – (51055) 2 VII) removing some I 2 VIII) adding some I 2 1. I, VI, and VII only correct 2. II, V, and VIII only 3. II only 4. II, IV, VII, and VIII only 5. I, III, V, and VII only Explanation: Increasing temperature adds heat and will shift equilibrium to the left. Changing pres- sure will not affect equilibrium because the number of gas molecules is the same on both sides of the equation. Removing a reactant or adding a product will shift equilibrium to the left.

{[ snackBarMessage ]}

Upload your study materials now and get free access to over 25 million documents.

Upload now for FREE access Or pay now for instant access
###### "Before using Course Hero my grade was at 78%. By the end of the semester my grade was at 90%. I could not have done it without all the class material I found."
— Christopher R., University of Rhode Island '15, Course Hero Intern

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern