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322a lab quiz 1 fall 02 key

322a lab quiz 1 fall 02 key - CHEMISTRY 322/5aL FALL 2002...

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Unformatted text preview: CHEMISTRY 322/5aL November 13, 2002 FALL 2002 FIRST LAB QUIZ BY RESE‘ l /'\/ Kfi ( NAME 1.(12) 2.(ll) Lab time } 3.(10) T.A.____________________________ . 4.(10) 5.(11) This test comprises this page 6.(10) and six numbered pages. TOTAL (64) If a question says to answer in fewer than a certain number of words, Do SO—-deduction for wordiness. gas will have quizzes at (make-up) lab and office hours. Labs Nov 14 — Nov 19 are make—up only; TAs will stay one—half hour if no one is doing a lab. Morning labs convene at 2LQQ. GC make-up will be only Mon Nov 18, 6:00 p.m. For grading questions on this test, SEE YOUR TA FIRST. —1— ”flit/L ‘ 1. (2) The wall of a 14/30 male (inner) standard taper joint is 1.0 1L mm thick. Calculate the joint's minimum inside diameter (ID). /n m.» M «mom (31:2) -—= ”39.6%th MM 6“ 1:10 w m (1’ 1%“ 2. (3) Note the saturation values (w/wsoln) for Water (MW = 18) and Ether (MW = 74): E in W 5. 0%; W in E 1. 2%. Using the sim- Dlfist dilute Soln approx (a) calculate the mole fraction solubility of each in the other, and (b) tell which liquid dissolves more of the other, molecule for molecule. @ EMA“)! XE: _co¥ak% :[m‘lo $0.013 M' ‘L’ * ____’_;’(;L The name on a bottle says "4-Methyl—2—pentano..."; the follow— ing 2 — 8 characters are bleached out. The IR spectrum of the Jt‘/4L carefully dried liquid contents shows strong absorption at 1720 cm and a strong, broad band from 3100 - 2600 cm‘l. ~‘\’LC- Assuming the liquid is a single mongfiungtignal compound, tell its likely identity; give reasoning (< 15 words) Yqu/tdl C I 6:0 c ‘ Lit/16¢“ /~2 .. w (‘qu act/jéf/ZM” 40/74 1220417M°mfl7 J~ /7Z.— yf‘rqu/7 #«éW 0HC7’W’7’ffiq, 4. (2) Ether has nbp = 35°(308 K), MW = 74; air has nbp~ —190°(83 K), effective MW = 30. Accordingly, the ratio of the density of ether vapor to t air is abou ( 'rcle one answer)-- 12 c), 71‘ 30.7 , 3.7 0.4 0.27 ' 1/3 that of water vapor to a1r 5. (2) An aqueous solution made from 30.0 mL isopropyl alcohol (IPA, d = 0.78) is "salted out". The resulting 35.0-mL Upper Layer has density: 0.87 g/mL and is therefore 66 % IPA (by mass). Calculate the fraction of the original IPA in the UL. M ovauvm/l}. jacuL 1‘ °7x%(,=27,%01f/V ML (221744 3.!" 0 mong7‘7/hL .7016? MW Pia/7' 0.: (AL :31)qu :a66‘gi=.20.l} fifl‘? Fm‘yzfl rngMLV z/—3 :Lgéaza -2._ A7 221/. 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. 0 (31‘? ”0.170" nu flannel. O IIIIIIII IIIIIIIIIIIIIIIIIIIIIIIIIIIIIII In IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII-IIIIIIIIIIIIIIIII IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII‘Ca-I IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII-"‘4IIII IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII-'-"._gun-IIIIIII IIIIIIIIIIIIIIIIIIIIIIIIU""_:_-u-uflIIIIIIIIIIIIIIII IIIIIIIIIIIIIIIIIF’:Qua-IIIIIIIIIIIIIIIIIIIIIIIIIII IIIIIIIIIIIIIIIV‘IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII IIIIIIIIIIIIII’1IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII -IIIIIIIIIIIII'lIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII IIIIIIIIIIIIII ,IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII IIIIIIIIIIIII AIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII IIIIIIIIIIIII IIIIIIIIII-IIIIIIIIIIIIIIIIIIIIIIIIIII IIIIIIIIIIIII ,IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII IIIIIIIIIIIIIII'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII IIIIIIIIIIII IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII IIIIIIIIIII' IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII IIIIIIIIIII'AIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII IIIIIIIIII’.IIIIIII-IIII-IIIIIIIIIIIIIIIIIIIIIIIIIII ' IIIIIIII”4IIIIIIIIIIIII-IIIIIIIIIIIIIIIIIIIIIIIIIII III-”:L‘IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII :aIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII IO :0 “Jihfitéw ‘70 $0 ca .101? Below, circle one of the items separated by / in each set. “1356, From the curve, the- / maximum p of H is about 94°, and mixture was about 10 / 15 / 20 /~ 2 / 30 percent L. 7. (5) Regall thatA VP per degree, dP/dT, = P(Afivap/RT2) and — AHvap/T- Acetone's bp760 torr = 56.5°C (329.7 K); s _. va 2§vag = 21.0 cal/(mole—K) . A tilde,~ , means per mole. (a) (3) Calculate dP/dT (3 sig figs) at acn's nbp, .i.e., at 760 torr; take R = 2.00 cal/mole-K‘. Show hey the units come out. g Sukfi‘hh? \ duo/d7“: {9 AS“, 2 76.0 éowx 11.0 , l 2“ /I< 0» MW) (b) (2) A solution of a non—volatile solute, (NVS) in acetone has nbp = 57.8°. Assuming dP/dT is constant from 56° - 58° and that C acetone obeys Raoult's Law, calculate XNVS'the mole fraction - “477‘ of the NVS (You must first calculate xacn') . [(2 {1.2 * a (.3054/40‘» ilk/o ¢ Am, from , 6%: of) AF/AT' c4101 «4mm qu’c“ 4:?‘5‘7. [0: from (4) 760M” lilUfiflwr t 75¢}.réovr. From Rama/1": ”3 LM¢P=I°°X$XM=¢~2016§ W” XMK :2 0. 0‘f (‘1‘ male 66) 8. (2) Consider an ALcohol, a Ketone, and a Carboxylic ACid each having nbp ~ 140°. Circle the correct descending order for their AVP/deg near-their nbp: AC>AL>K AC>K>AL Ample”). K>AC>AL K>AL>AC m 29/0 9. (3) de A steam distills at a mixture hp at which PW‘= 6/7 the , external pressure. (1) Calculate the ratio, molesW/molesA, in ”- the distillate. (2) If the distillate has 1.2 gA per 1.0 gW, calculate the molecular wt of A. (Recall MWW = 18 0) 10. (4) Note the data below. "m- n" in the left column means "a mix- ture of CaClz'm H20 and CaCl2 n H20 in equilibrium". All the hydrates are solids up to at least 50°. CaC12 hydrate mixture Relative humidity (rh), % 0-1 1.25 1-2 2- 4 (P m: D c.‘ H $4 «1 0.. in IQ 8 |-‘ :3 N H: E? 3 fix 3» QAS=d 5? -‘r .F e~< K 5.7%. & ~t (a)(2) A W— saturated Organic Soln has 0.50 g W per L of soln. 0n adding anhyd CaClz, it hydrates, and part dissolves to a separate aq phase. Calculate the final W cone in the OS. 7.22:7; Jay/7:» I:(:4\b)(2)3 A difiierent Drying Agent can remove 0.20 9W per gDA. Calculate how much DA is needed to dry 3 L of the Original so n of Lpart (a). Assume all the water is removed. (far 6%“ 44‘W4M 4 “:0‘ am 11. (3) An unrinsed Single Laundry Load (SL) retains 10% soln on spin drying with detergent residue per mass = Co; a DOuble Load (DL) retains 20% soln and detergent residue per mass = 2 C0. One now does a single rinse—and—spin on SL and two ras's on DL. Calculate Cfinal for SL and for DL, in terms of C0. (f?) ”"9‘ (:JF“:‘“Z J L, :2 C9. /(9 (:0 Q .7. (mm:— -—= 2©('o.zo)”‘= 0.086. —4— ”a 12./0 12. (2) Ethyl acetate (EA), of bp = 80°, is cheap, relatively non- toxic, and efficiently extracts many organics from water. If one wishes to isglate a given org, what physical property should the egg have in relation to EA? Answer with <12 words. ff (MM mf Ui/ "66(— 90° 13. (3) Pure J melts at 95°; pure K melts at 90°; a certain sample X melts 82° — 86°. A mixture of equal parts J and X melts 87° - 90°; a similar mixture of K and x melts 70° — 71°. Tell what these data say about the likely major constituent of X. Give [1546 your reasoning gongisely; use no more than 20 words. ‘ 3r #:‘Neév I fact 41/“ I; T 7‘? X WM“ ”’“fll ”7° Cw‘Wv/ fm WNW/A" (TQLI&ZVkfl¢vZ§‘ CZ 6L4,a 3/47 .jifl [Al/if. t:' J/OciaL my) I( .le 4 mp X3 14. (3) A solid sample comprises 12.0 g M and 7.0 g N, whose solubili~ ties in a solvent S, bp = 65°, are independent. The solubili- f¢é7"~*{f ty of each in not S is 10.0 g per 100 g S. In cold S, M's solubility is 2.0 g per 100 g S; N is twice as soluble. 1:701L‘ The entire sample is dissolved in 200 g hot S and then cooled. W “1:3 Calculate how much each of M and N precipitate; pm; a. nun: L, be; in each blank. NO credit without calculation/explanation. 4% M w} w{/ Kala 1‘ 20/90;5*7’“35 ———r———9M ;%O&;:><(1,0~‘OJ; gogflflé m4, 4 A; .: “(was ‘va N60,: LQN Sahq W 2,009 5! {Cami/ma NM! Kane? gsU/W‘ 15. (2) A solid sample heated with recrystallizing solvent mostly dissolves to a colorless soln but leaves some suspended insol- uble brown flakes. Using < 10 words, tell what step(s) one may correctly omit from the complete procedure. (”0 MTL W) T‘Van wffl‘ AMI (ha/wad. —5- i c.— ryi // 16. (3) According to the movie "The WiZard of 02", Wicked Witch of the West (WWW) is very water-soluble. uSuppose at 22°, 8 pounds of ”1 I 41/ water just dissolves a 92 pound WWW. Restate this fact in \ \, ML 46*“! terms of mp depression by stating the mp of certain mixture in r- wt %. Use <15 words; zer ' to soln or ' i 6.) C3 . (M 0 #mm‘wwf Xwflfiwuk 7.1" wwww¥s #22:; a; _- - AflAWMfi 13*? 17. (4) seg-Bu-Br, CH3CH2CH(Br)CH3, reacts (slowly) with either NaI in . acetone or with 2% AgNO3 in 95% EtOH. Draw a structure of the ' main organic product, then state whether the product's stereo- L Pf chemistry is mostly inverted, retained, or racemized. (a) NaI in Product W Stereochem et ne mt .. . (MM/«#3 . Muevfi/ aA4W‘ 1: (b) AgNO3 in Product Stereochem 95% EtOH (#3 Cl/l-‘(yv-(lllj ‘ MQWIW 0 65/} (47‘; 18. (14 pts, 4 on this page, 10 on the next) Recall the n—BuI prep: acetone. n-BuBr + NaI n-BuI + NaBrw MW = 137 FW = 150 MW = 184 Q = 1.27 g = 1.61 (a)(4) Suppose one is confident of obtaining a 60% yield and wants to get at least 33 mmol n—BuI. Calculate the minimum volume (in mL) of n—BuBr (limiting) which one must use to assure this result. NW , 33MM0/ W : ‘RSrW/fl-flqiy 0.6 Mw/ W M ( («ml 4’4» k 2/ fame ’ KL flvflufik '2 ((MM/ X gill—a: ww( fl W215'73ML (b) (7) heal-ion AG 11:53.11 (1) n—Bu-Br -——-> n—Bu+ + Br‘ + 178 (2) NaI -—-—> Na+ + I' + 164 (3) n—Bu+ + I‘ —--—> n Bu I - 171 (4) Na+ + Br' -——-> NaBr — 174 5 Sum: n-Bu—Br + NaI -———> n—Bu—I + NaBr - 3.5 (5) NaCl ----> Na+ + (21‘ + 183 (6) { n—Bu-Cl --——> n—Bu+ + C17 + 185 Sum: n—Bu—Cl + NaI —-—s> n-Bu-I + NaCl - 5.0 (1)(4) Show calculations of the fires energies oi ag;i_atign, 415‘?;, or starting with (i) n- Bu- -Br, and (ii) n- Bu- Cl, assuming the rate determining step (rds) is I'+ n—Bu—X -———> .n—Bu-I + X‘. y-QGA n— —Bu- Br n Bu Cl ‘6 new!“ qua/860+ Ire m fl-é’u'CI -+ u *‘Cng- W 4-16 ~4- n—Aqr ~n :1 336+ £94 14w; H1! 543:4 fir +£6.14- [Lt-flap *IVG ‘ Agent-Cl face arid—7* H cce (2)(3) Calculate the rate ratio, kn —BuBr/kn- BuCl' from ern = :¢ k0 [exp(—AG /RT)]- , assume the ko's are equal. Take AG 's in cals (1 Ecal = 1000 cals), and RT = 660 cals; show how you set up the equation you solve. ’4 Ac*-AC 03,6 6-63 :«/£7’ 6 <4 1" ' 71—22%]; (+000 ’ 7000 :: '<Z dkfio 2 (c)(3) Circle one phrase: Acetone helps the reaction by solvating—— the Br in n—BuBr so it leaves easily. the I in n—BuI so it bonds easily. the I’ so it is reactive. the Na+ so I‘ is reacti 9. _ the Na+ so NaBr can precipitate. the Br‘ so NaBr can precipitate. ...
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