# test 1 - Version 087 Exam 1 Mccord(90540 This print-out...

This preview shows pages 1–3. Sign up to view the full content.

Version 087 – Exam 1 – Mccord – (90540) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. McCord CH302 1pm Henry’s Law Constants (M/atm) Gas k H hydrogen 8.5 × 10 4 nitrogen 7.0 × 10 4 argon 1.5 × 10 3 helium 3.7 × 10 4 carbon dioxide 2.3 × 10 2 001 10.0 points The “bends” result from the rapid release of nitrogen gas in the bloodstream when a scuba diver returns to the surface after breathing compressed air for an extended period. Re- placing the nitrogen in air with another gas can help avoid the bends. Would argon gas and/or hydrogen gas be suitable replacements for nitrogen gas in “compressed air” used for scuba diving? 1. Hydrogen would be more suitable. 2. Argon would be more suitable. 3. Either would be suitable. 4. Neither would be a suitable replacement. correct Explanation: Checking the Henry’s Law constants reveals that both hydrogen and argon are more sol- uble (larger value of k H ) than nitrogen which means that the bends would be even worse in those cases (more dissolved gas coming out of the blood). A better substitute would be helium which is about 1/2 as soluble. 002 10.0 points When the external pressure is decreased, the boiling point of a liquid 1. increases. 2. does not change. 3. may increase or decrease, depending on the liquid. 4. decreases. correct 5. disappears. Explanation: If external pressure is decreased, it will re- quire less energy and be easier for molecules to escape liquid phase, therefore the liquid will boil at a lower temperature. 003 10.0 points We dissolve 32 g of Ca(NO 3 ) 2 in 964 g of water. What is the boiling point elevation? Note that K b for water is 0.512 C/ m . Assume complete dissociation of the salt and ideal behavior of the solution. 1. 0.315764 2. 0.31071 3. 0.497337 4. 0.157148 5. 0.682669 6. 0.459833 7. 0.505954 8. 0.620361 9. 0.378413 10. 0.604437 Correct answer: 0 . 31071 C. Explanation: m Ca(NO 3 ) 2 = 32 g m water = 964 g K b water = 0 . 512 C /m CaNO 3 Ca 2+ + 2 NO 3 m = (30 g Ca(NO 3 ) 2 ) parenleftBig 1 mol Ca(NO 3 ) 2 164 . 1 g Ca(NO 3 ) 2 parenrightBig = 0 . 195003 mol Ca(NO 3 ) 2 Δ T b = K b m i

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Version 087 – Exam 1 – Mccord – (90540) 2 = parenleftBig 0 . 512 C m parenrightBigparenleftBig 0 . 195003 0 . 964 parenrightBig (3) = 0 . 31071 C 004 10.0 points A 4 . 13 g of sample of a molecular substance dissolved in 104 g of camphor freezes at 176 . 5 C. What is the molar mass of the substance? The freezing point of camphor is 179 . 8 C and the freezing point constant of camphor is 39 . 7 K · kg / mol. 1. 510.151 2. 237.001 3. 112.784 4. 594.858 5. 329.719 6. 218.615 7. 163.305 8. 135.748 9. 340.248 10. 477.742 Correct answer: 477 . 742 g / mol. Explanation: m X = 4 . 13 g t f , X = 176 . 5 C m camphor = 104 g k f = 39 . 7 K · kg / mol t f , camphor = 179 . 8 C Δ T f = 179 . 8 C - 176 . 5 C = 3 . 3 C = 3 . 3 K Δ T f = k f m 3 . 3 K = (39 . 7 K · kg / mol) m 3 . 3 K = (39 . 7 K · kg / mol) parenleftbigg 4 . 13 g MW X parenrightbigg 0 . 104 kg MW X = (4 . 13 g)(39 . 7 K · kg / mol) (0 . 104 kg)(3 . 3 K) = 477 . 742 g / mol 005 10.0 points Would you expect NaCl (table salt) to be soluble in a nonpolar solvent like hexane?
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### 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