Fall 2009 Lab Exam #1 key

Fall 2009 Lab Exam #1 key - CHEMISTRY 322/5aL FALL 2009...

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Unformatted text preview: CHEMISTRY 322/5aL October 28, 2009 FALL 2009 Poff/N G— FIRST LAB QUIZ BY £398 NAME 1.(12) 2.(10) Lab time 3.(11) T.A. mdt‘m : 4.(10)_____ Ava, :— ¥9~C70%) 5'(8"““ This test comprises this page 6. (9) and six numbered pages. TOTAL (60) If a question says to answer in fewer than a certain number of words, DO SO——deduction for wordiness. 133 will have quizzes at (make-up) lab and office hours. Labs Oct 29 - Nov 3 are make-up only; TAs will stay one-half hour if no one is doing a lab. Morning labs convene at Qian. IR make-up will be only Mon Nov 2, 6:00 p.m. For grading questions on this test, SEE YOUR TA FIRST. 3. 5. (2) (4) fl 12/; A 29/45 female (outer) standard taper joingfhas a 1.25 mm thick wall. Calculate its narrow end outside diameter. 19 Murrow and) 500399-4‘5—MM 27 M” :zfirmofl = lint 9.0”) but} <9 =@ section 2.0 nulwide and 1.0 mm deep; see drawing. is to have half its thickness outside the groove. IDxOD in mm, should be (circle one answer)-- 3%.- - 8x9 8x10 9x11 9x12 9x13 39 (Not to s ale) 4—Methyl-2—pentanol is made into 4-methyl-2-pentanone. An o-ring Its size, (a)(2) Circle near what IR frequency (cm'l) one should look to see it much unreacted starting material is in the product. 3800 3000 2800 2200 2000 1700 1500 (b)(2) Tell whether the product sample must be well dried for this IR (2) test to be useful. Explain in < 8 words. YM; {mag 14) a/Wvé: of 35‘00 ago. One rinses a flask with water. Taking MW it a 30, bp ~ -200°, tell whether any residual liquid film wili evaporate faster with the flask neck up or neck down. Explain in <10 words. Mecl (4/0. An aq soln containing 30.0 g isopropyl alcohol (IPA, d a 0.78 is "salted out". The 40-mL upper layer's density or 0.84 g/mL means it is 79 (wt) % IPA. Calculate (to 2 sig rigs) what percentage of the original IPA is in the upper layer. Answer THIS question only. (‘rr% app/4a on. =- around” . .7 5"” = ‘- *09*w3/"L "0 7% 16$ {MM g 307r’#“ ‘SSXO/o (353$?!) 0.12m é a/O ‘2 “W60 60mm :3 6. (2) Using < 15 words, define the term hailing mint (in general) 0 ~ g “dtfor a mixture of two liquids, A and B, in tems of PA and PB. Assume their vapor mixture is an ideal gas. (a 5/0“) m thwem/ZLM W 4 d‘whfck fl/f-‘f’pg 2‘ MfoQV‘VlJW. 7. (4) (a) (2) Note the distillation curve below for 100 g of a solution .of L (lower boiling) and H. L and H do not form an azeotrope. 7° ::=:::==::::::-'.3:222:33;::=:::::::::::::::::: “‘4 Y’W" III-III... I... I' 4IIIIIIIIIIIIII-IIIIIIIIIIIIIIIIIII III-IIIIIIIIII '- I-IIIII-III-n (an IIIIIIIIII'.I :IIIIIIIIIIIIIII GD usually-rul-“Issues-annals.- [ From the curve, the bp of. I. at least / ~55°, and , the mixture was about 70 / / 78 / 83 / 9 percent H. (b) (2) 0n the diagram above, draw a distillation curve for the same mixture distilled mm 3.152111. ( 3J0“) (gun/Q “g 7/0 30 "2'57 {W 6a.! flu/VIM fqéulf a. (4.) Water) and cpd D are mixed. At equilig, 91 is 2.0 mole % w, e2 is 0.50 mole a: D. Raoult's Law applies to the major, and Henry's Law to the minor component in each «I. P"w - 80 13°13. (a) (2) Calculate the Henry's Law constant for D in water, HLCD_1n_w, in terms or P°D. (b) (2) Explain why the conc or n as the W W-rich phase evaporates in the open. (4/0 =0..g "=/0 :0.00 ° @ 2 “5' 7 pa 0"" (Hm/{FD “LO/“P3 “Noam @ Dmc W “A 4%, ‘QUqflON/fa 6k. 0!; VP V} have raft/flue fiat: meoééév ‘ 3‘ i - // 9. (4) For de E, KvB/VA, i.e., z = 4. Calzuzate the fraction of B unextracted [we = 1/(1 + (z/n))n] for (a) division of the extracting solvent B into two portions, and (b) for division of B into a very large number of equal portions. OK to take e = 2.7, e2 = 7.4, e3 = 20. (a) (2) I (b) (2) I; 7‘ RN; 3 r,“ (1%)" «m L 11;)" H r W A 2 = e + 10. (2) Recall the system water/i—PrOH/NaCl. in i-PrOH, tell how one might extract using i-PrOH. Use < 20 words. flags! gflrO/vl 7‘0 62?, “In and. “fa” ref? wifl N4 5/ (arm 00/144). 6‘44 («Y goflp' fflfiiwn‘clv [Qt—k AM MT‘ —. %%1?fiii}- 1:: I If cpd F is very soluble F from an aqueous soln 11. (2) G has mp a 120°; H has mp a 170°. The most probable melting point for a 90:10 G:H mixture is (circle one answer) -- 45° 75° 125° 145° 155° 165° 175° 12. (3) Consider the following claim: "We have made superior purity de J than that made by the standard procedure. Our J has mp = 130°—132°. Mixture mp with J of mp a 128°-130°, prepared by the industry standard method of Hickory, Dickery, and Dock, was 129°-131°". Assuming the data are correct, evaluate the probability that the claim of higher purity is correct, giving your reasoning; use < 20 words. (3 ——9 C/cu‘m V3 (644401124 val/‘4’. finuaazao new T740 <9 ( MIA vf 0M flowefo v‘x A‘s/tat“; .z W' 3““,de 33$".w3‘d/1QWV «4— ‘h a; 1 “w 13.(5)(a)(3) A solid sample comprises 8.0 g M n 10.0 g N, whose v solubilities in a solvent S, bp a 65°, are independent. The gsltflD solubility of each in not 3 is 10.0 g per 100 g S. In gold S, %N M's solubility is 1.0 g per 100 g S; N's is 6.0 g per 100 g S. I 1‘ A 94 The entire sample is dissolved in 200 9 hot S and then.cooled. Calculate how much of each substance precipitates; put a num- L~)T‘ 'qfr in each blank. N0 credit without calculation/explanation. ' éto 9” 3th M .1067 2 C lfi/[00a 5) __Q___gN W? was ‘67/75 CdUW' we I; MW} Kala, > {77‘ 450%) @g 1113A) =7 4s A) we (b)(2) The entire sample woul y dissolve in <100 9 hot S. Explain in <15 words why this is NQI the right amount of solvent to use to get pure M. would. {om A) 0M, 50!) Mxfi 14. (2) A student recrystallizes material from alcohol, sucks it dry, and spreads it out to dry. Tell how s/he can tell when it is in fact dry. Use <12 words. J [4, (>1 68 WW i fizzy/070.23. :32ch Q ‘ ____ I a ’- 15. (3) The Wicked Witch of the West (WWI) in the "The Wizard of Oz", doused with water. suffered a fatal attack of melting point () ‘ depression ("I'm melting, I'm melting'l). Given that WWW weighed 81 lbs and that 9 lbs water depressed her mp to 20°: dlvvoaléhlo Describe this result in solubility terms inx. State a tem- N/éw perature. the nature of the soln using an: of the following terms: concentrated, dilute, saturated, unsaturated; and give 1;” : the solution's composition as weight M0 20% A6 20°, 4 safumf’é/ («an refinpf wow)“; qomr/o cam). 2 = g 16. (4) Recall the n—BuI prep: acetone n-BUBr + NaI —-—-——————-)-. n-BuI + NaBr+ {? MW 2 137 FW = 150 MW = 184 46 mm d : 1.27 d = 1-51 ‘l .‘a/ Suppose one wishes to be sure of getting 4.6 g n-BuI and is confident of obtaining 260 % yield. Calculate the minimum [ “_ volume (in mL) of n—BuBr (limiting) with which one must start to assure this result. ‘176 - Ltd) 3‘ £6“; :_ i ~0.0ZJ’ W 2 (0.0W7w/e) 0.63 4 3 “L 710 “L 4'03er = 0-0 ‘H7 W! ‘ (NJ/"WI 1.;737WL 3‘0 ml. 1v 6am 17.(4) Of the four butyl bromides, t—Bu—Br reacts the fastest with AgNO3 in EtOH but the slowest with NaI in acetone. Explain this difference in reaction rate in mechanistic terms, referring both to the solvent and the inorganic solute. Also, draw the structure of the neutral organic product for the faster reaction and explain its formation. Use < 25 words total, but just saying 8N2 and/or 3N1 is not enough. E‘éoH a) / galvada/rr‘ni-mfi'm w m “Wm f.“ MAE}. W 4cm 0': ' Mi”? 18. (5) Em A}; 115ml, gas phase (1) g-Bu-Br -—--> n—su+ + Br' + 173 (2) NaI -—--> Na+ + I‘ + 164 (3) n-su+ + 1‘ -—--> g-su~I — 171 (4) Na+ + Br' —--—> NaBr — 174.5 Sum: n-Bu-Br + NaI -—-—> n—Bu-I + NaBr - 3.5 Keq - 200 (5) NaCl —--—> Na+ + Cl‘ + 183 (5) n-Bu-Cl -—--> n-Bu+ + Cl‘ + 135 Sum: n-Bu—Cl + NaI -—--> n-Bu-I + NaCl - 5.0 Keq ~ 2000 they do not take into account 43', the entropy change, or solvation--for these reactions. Tell why one can still get a reasonable AGmrall. Use < 30 words; refer to the reactants and products,- no credit for saying just "everything cancels". (b) (3) Calculateaa for g-Bu-Cl + NaBr ----> n—Bu-Br + NaCl, and say 1% what it means. AG < -1.3 kcal means Iceq > 9.0. ~ 5 W e (a “GP AI?) z: +( is: +(71‘.5‘—(7K*/Z3 z W “""‘ “at :2 +1 - 3.57 =K‘IJAJ'? 770?; firifdewué‘ 19. (3) Regarding the drying of n-BuI as done: CaC12 (the DA) can absorb almost 1 9 W droplet: per 9 DA, but this leaves about 30* of the W amount of water, only 2 tug/g n-Bul, dissolved in the product. CaC12 can cut this by a factor of 80 by using 10 mg DA per mg W. Explain why the n-BuI was dried in 3:19 steps. Use < 20 words; consider what would happen if. one attempts to 222mb all the water in one ste . 7‘0 More, (0 wk mu. (7‘ <4) cu‘H. So Mclv a) AL flash) «bforM- 75002“ W M . ‘ {wt 04 M cw/loal/‘brtkm later “CV/k! 61% ...
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Fall 2009 Lab Exam #1 key - CHEMISTRY 322/5aL FALL 2009...

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