322af07_lq1_key

322af07_lq1_key - CHEMISTRY 322/5aL November 14, 2007 FALL...

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Unformatted text preview: CHEMISTRY 322/5aL November 14, 2007 FALL 2007 FIRST LAB QUIZ BY Bags NAME 1.(12) 2.(ll) Lab time 3 (9) T.A. 4. (9) 5. (9) This test comprises this page 6. (6) and seven numbered pages. 7 (9) If a question says to answer in fewer than a certain number of words, D0 SO—-deduction for wordiness. TOTAL (65) 153 will have quizzes at (make-up) lab and office hours. Labs Nov 15 — Nov 20 are make—up only; TAs will stay one—half hour if no one is doing a lab. Morning labs convene at 2439. GC make—up will be only Mon Nov 19, 6:00 p.m. For grading questions on this test, SEE YOUR TA FIRST. -1- flk713/9x A 14/20 female (outer) standard taper joint (like that on your 25 mL RB flask) has 1.0 mm thick walls. Calculate its inside 1. (2) diameter (ID) at its narrow end. ‘ La/ 7 "‘7 Hum a. £6 «fwd T «a 30 M .l. Mun/6.44. [0 am =5 51wa ed «a?! ‘- \ro = (Hug? = (I2 m ’:'7naww.u./QD=I~(- mm. ' 2. (2) An O—ring just fits in a circular groove of outside diameter 14 mm and width 1.5 mm. Its ID in mm is (circle one answer)-- [1‘1" 104‘qu 7.5 9 @ 12.5 15.5 17 3. (4) 4-Methyl-2—pentanol is made into 4—methyl—2-pentanone. (a)(2) Circle near what frequency (cm'l) in the IR should one look to see if there is considerable unreacted starting material. 3800 @ 3000 280 (2200 2000 1700 1500 "‘ h! 4, L4de (b)(2) Tell whether the product ample must be well dried for this IR test to be useful. Explain in <12 words. Yzm mf- hat/(y {Cm-e (A) 4Z0 45(0ré; 3%00 0:: 4. (2) One rinses a flask with pentane, n-C5H12, bp = 35°. Taking MWair = 30 and bp z —200°, tell whether any residual liquid will evaporate faster if the flask is held neck is up or neck down. Explain in <10 words. fL (aw MUMMAW wMLJZe/E‘f’fl-afim o ' V4 M M zzduset ‘ ..... “I M f #211 No4 4 5. (2) An aq soln containing 25.0 g isopropyl alcohol (IPA, d = 0.78) is "salted out". The 35—mL upper layer's density of 0.86 g/mL means it is 70 (wt) % IPA. Calculate (to 2 sig figs) what percentage of the original IPA is in the upper layer. Answer THIS question only. mi=// 6. (4) 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. Io‘ .uIn":=======nn:::========:=:===:========= Diff; [hfiou “4.4 {fa/pd III-III..-III-III..-III-III..-III-IIIIIIIIIIIIIIIII III-IIIIIIIII-III...III-IIIIlI-IIIIIIIIIIII-II‘Ch-I III-III..-III-III..-III-IIII-IIIIIIIIIIIIII-’ggl-II IIIl-I-IIIIIIIIIII-I III-IIl-III-I---‘T_----IIlI-II IIII-l-IIIIIIIIIIIII Il---’:_——---IIIIIIIIIIIIIIII III-II-III-II-III-’: ail-III-III-IIIIIIIIIIIIIIIII --.------.---.-"-l. III-III.-III-IIIIIIII-IIII-III III-III..-III-HIIIIIIIIIIIII-IIII-IIIIIIIIIIIIIIIII III-Il-IIIIIIIIIII III-III-III III-Il-IIIIIIIIIII III-III...- II-ll III-IIIIIIIIIIIIIIIIIIII III-I I‘ll-IIIIIIIIIIIIIII-III III-I III-IIIIIIIIIIIIIIIIIIII III-IIIIIIIIIIII III-II IIIIIJIIIIIIIIII III-II III-IIIIIIIIIIIIIIIIIIII IIIKIHIIIIIIIIIIIIIII-II ' III-IIIIII-II I'.IIH-IIIIIIIIIIIIIIIII I III-IIIIIIIIHIIIIIIIIII-III..-III-IIIIIIIIII-III-I III-II_IIIIII-III-IIIIIIIIIIIIIII-IIIIIIIIIIIIIIIII 14:72t.‘=57’7SqL)lJ-_———::—-J:: 123 lul‘f7flazk“’ 1” ‘0 co loo Below, circle one of the items separated by / in each set. From the curve, the bp of L is at least / 55°, and the mixture was about 70 /®/ 78 / 83 / 90 percent H. 7. (3) Recall that AVP per degree, dP/dT, = Mag/RT) . For OCT, VP per degree = 3.7 torr/K atfl95°C (368 K), where VP of pure OCT = 100 torr. Calculate Asva in cal/(mole—K), "eu"; take R = 1.98 cal/(mole—K) , and show ghat the units come out. 1//,\ £2!“ A339,: 191' 000/”? Sizuwa ‘ ('7? .fié-qz ' 3.7 8. (4) W(ater) and cpd D are mixed. At equilib, cpl is 3.0 mole % W, <I>2 is 0.50 mole % D. Raoult's Law applies to the major, and Henry's Law to the minor component in each 4’. P°w = 80 P°D. (a) (2) Calculate the Henry's Law constant for D in water, HLCD_in_W, in terms of P°D. (b) (2) Explain why the cone of 12 mp3 as the W-r1ch phase evaporates in an open vessel. . :21, A 0 W) 3377 ,00" = HLCflfiu (0.00573) 1: HLC - [W lop (A) I” WM, 4), whfi/n‘; g o, /1+/’,3) 77 Whig/ff) 4 (A),¢‘~'r . M {=7 9. (3) Note the data below. "m—n" in the left column means "a mix- ture of CaClz'm H20 and CaClz'n H20 in equilibrium". All the hydrates are solids up to at least 50°. CaCl2 hydrate mixture Relative humidity (rh), % 0—1 1.25 1—2 4.2 2-4 14 4-6 21 Exactly 100 9 Organic Soln has 2.0 g dissolved W(ater) and 3.0 g W droplets. Adding CaC12 gives a solid hydrate and an ag soln. Calculate the final W conc dissolved in the OS. ——-. 10. (2) A certain very efficient Drying ‘ per 9 DA. Calculate how much DA one needs to dry 1000 g 08 W containing 0.30 wt % W; assume complete W removal. % 0.30% ¢ loco? : 3.0;6J ” 3.0 U - MWK. - 11. (4) (a) After washing, Single Laundry Load (SL) retains 10% of the tub soln on spinning with Soap Conc Co; a triple load (TL) has Fraction Soln Retained and SC in proportion. Calculate Cfinal for one rinse-and—spin for SL, and four ras for TL. (b) Com- pare the result with SL, one ras: much better, a little better, nearly the same, worse. Use formula Cfinal = n-C0[n-(FSRSL)]#raS; n = the mass ratio of a multiple load to that of a single load. <: h 5‘, f (9" (:6 4' ~‘f (£564,rt == 3(0C3(0"B : 3"“) <9 3 fl :: 0.01‘F3Co. Maud 5217f“, C; < V‘I- C054. “(ML M7 1 mmhfi‘i‘r‘wr w 3 wukl~3m %‘ 36M! NB '4' flji:7 12. (2) G has mp = 110°, and H has mp = 160°. The most probable mp for a 90:10 G:H mixture is (circle one answer) -— 45° 75° 115° 125° 135° 155° 165° 13. (2) Sometimes the melting point difference between 93% J and 99% J cannot be easily seen. But there is usually an observable difference in behavior on heating. State this difference in <10 words. (Wide) MULch ram ¢ 6‘3 ‘Yo CAIN Ax aanflfL/CLVJM. 14.(5)(a)(3) A solid sample comprises 8.0 g M and 9.0 g N, whose solubilities in a solvent S, bp = 65°, are independent. The solubility of each in hot S is 10.0 g per 100 g S. In cold S, M's solubility is 1.0 g per 100 g S; N's is 5.0 g per 100 g S. The entire sample is dissolved in 200 9 hot S and then cooled. Calculate how much of each substance precipitates; put a num— ber in each blank. NO credit without calculation/explanation. 4 3,003 MM 6‘ «NM [to/c/ 97M _ yam. ,agg-ugwr M? 0 M w “a s m-// 1.14 We Mwa w? WfdlnM‘tlf 7] IV “mm/.1 .. “if! fang; Q; Ufl/é plain in <15 words why this is flQT the right amount of solvent to use to get pure M. (303 f an MBA cuc‘f/ A] (w cof/J M137 15. (3) The Wicked Witch of the West (WWW) in the "The Wizard of Oz" suffered a fatal attack of melting point depression ("I'm melting, I'm melting"!) Given that WWW weighs 90 lbs and that 10 lbs of water (about 5 qts) depresses her mp to 24°—- Describe this result in solubility terms. State a tempera— ture, the nature of the soln using one of the following terms: concentrated, dilute, saturated, unsaturated; and give the [or- soln components' weight ratio, i.e., wtwww/wtwater. ATM“; I46 74'", « sdarlvf‘q/C‘tfj MFM% (u/‘t of WWW (In 7!/ wUU‘wa/Jt. "flab/or {A [(0’wa -§eg-butyl bromide, (S)-CH3CH2CH(Br)CH3, the SM, reacts with NaI in EtOH giving two non-isomeric organic products. All the Br forms NaBr. The rate of Product A's formation = kA[SM][NaI]; that of Product B = kB[SM]. Draw each product's structure, stating the stereochemistry (R, S, or racemic——need not draw the s-chem). Explain how each product forms, correlating its structure, including stereo— chemistry, and the rate law with the relevant mechanism(s). Use < 30 words total. Product A Stereochemistry ‘ u (#164,166ng “‘ “K/ 3.) 1. "‘ 1— W4 Product B Stereochemistry Mcemrc Cd cakCél-C 3 I i i=6 17. (15 pts, 6 on this page, 9 on next) Recall the ESQZI prep: acetone n—BuBr + NaI ——————————e-— Q-BuI + NaBrl MW = 137 FW = 150 MW = 184 Q = 1.27 d = 1-61 (a)(3) Suppose one wishes to be sure of getting 9.2 g n-BuI and is confident of obtaining 270 % yield. Calculate the minimum volume (in mL) of n—BuBr (limiting) with which one must start to assure this result. f 242%», MW! agent 2 1713;):le =f0mm' 5M W Hue? 35mm“! n4 mnméfflar may?) 70 "MI ,9qu 0: @747 mo u w! our 137 “L New, = 0‘0 719‘ ml 50 Wei (4? <7 L = M 7.7OML2 = MW( n-8uir (b)(3) As done, the reaction rate did not depend on the amount of acetone, but the rate is near zero without it. Tell under what conditions the rxn rate would depend on the amount of acetone but be independent of the amount of NaI. Explain using < 25 words. ach ERR, «my: odd/yfi‘cjxywuffl” “W a mm M Mu: only Many U41: J: m5“ M m OLCM am! if “an M and,“ d~fl4f [Ma/(N’tfl/“f. 17. (cont) (c) (9) Beam AG l_<;a_)_k l (l) g-Bu-Br —-—-> Q-Bu+ + Br“ + 178 (2) NaI -——-> Na+ + 1' + 164 (3) n-Bu+ + I" --—-> g—Bu—I - 171 (4) Na+ + Br' -———> NaBr - 174 Sum: n-Bu—Br + NaI ———-> g-Bu—I + NaBr — 3. (5) NaCl --——> Na+ + (31' + 183 (6) n-Bu—Cl ———-> Q-Bu+ + (21‘ + 185 Sum: n-Bu—Cl + NaI ———-> n—Bu—I + NaCl — 5.0 Keq z 2000 This is a calcn of part of A5 for n—BuCl —> n-BuI, based on the solubilities of NaI and NaCl in acetone. (l)(3) The sum of rxns (2) and (5) is-- NaI(sld) + c1"'——-*",‘ NaCl(sld) + 1‘ Take NaCl's solubility in boiling acetone as 1.0 x 10-6 M. Calculate its Ksp, = [Na+][Cl‘] (assuming it is a 1:1 electro— lyte). Then calculate [Cl‘] in soln for [Na+] = 1.0 M (the approximate effective [Na+] in acetone saturated in NaI). Kip ‘- [(fl‘10‘63t‘ /0‘n' (=[Ndf3ZCr]) ff [Mat]: lml {Lu [Ct-3: {043% (2)(2) Taking concs of pure solids as 1, write an expression for Ke then calculate its value; again take [I'] = 1.0 M. __ [r- l (ca? - 3 = -(L [0 [Cr] (0 N (3)(4) Calculate (;G = —RT(ln Keq), or = —2.3RT(loglo Keq), taking RT = 0.66 kcal/mole. Compare your result with Ag for the reac- tion of part(1) based on the bend energies in the table. Tell whether the assumption that acetone as a rxn solvent has little effect on AG is justified. Explain in < 10 words. L% Ag: _’L3£r({oa (0%): “z” -13 (0.69%) {L .. A8; = “@4433 WWW = Lib/é’Z/Me -.- —(q,lc~(/u4alc {0W (“f/({ h (d; ql ...
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This note was uploaded on 04/25/2008 for the course CHEM 322AL taught by Professor Jung during the Fall '07 term at USC.

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322af07_lq1_key - CHEMISTRY 322/5aL November 14, 2007 FALL...

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