{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

322a lab quiz 1 fall 05 key

322a lab quiz 1 fall 05 key - CHEMISTRY 322/5aL November 9...

Info icon This preview shows pages 1–8. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

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

View Full Document Right Arrow Icon
Image of page 4
Image of page 5

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

View Full Document Right Arrow Icon
Image of page 6
Image of page 7

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

View Full Document Right Arrow Icon
Image of page 8
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CHEMISTRY 322/5aL November 9, 2005 FALL 2005 FIRST LAB QUIZ A V BY Big—e NAME < L 1.(13) 2.(11) Lab time 3. (8) T.A. 4. (8) 5.(10) This test comprises this page 6. (8) and seven numbered pages. 7. (8) If a question says to answer in fewer than a certain number of words, DO SO——deduction for wordiness. TOTAL (66) 155 will have quizzes at (make—up) lab and office hours. Labs Nov 10 — Nov 15 are make—up only; TAs will stay one-half hour if no one is doing a lab. Morning labs convene at 2;QQ. GC make-up will be only Mon Nov 14, 6:00 p.m. For grading questions on this test, SEE YOUR TA FIRST. (like that on yourao alculate its inside 1. (2) A 24/40 female (outer 100 mL RB flask) has diameter (ID) at its narrow eno. flemv‘ W“ m //0 —> wao‘s ‘W0 “W ”i“ (MWM ML mrrM—zm/ =9 jnl/L {14“ch hm 1301‘“ ‘H MM "- 7/0 MMZ WWW/cg 2. (2) O- ring size is given by inside and outside diame ers (IDmm an OD). Calculate the OD for ID = 8 mm and th_gk_ess= Aémf face, )c-secfivc): C(‘roqlw, 00 1 2 w {- ‘L : [1 MM. MM 00. g M ¢§OMM [F4 3. (4) 4—Methyl-2—pentanone is made from 4-methyl-2—pentanol. One wants see if the product has 510% unreacted starting material. (a)(2) Near what frequency (cm'l) in the IR should one look to see if there is considerable unreacted alcohol (circle one number)? 3800 3000 2800 2200 2000 1700 1500 (b)(2) Explain in <12 words why the product sample must be thoroughly ~«~05! (fr (NW ML 41“ 4. (2) One flushes a flask with ordinary ether, bp = 35°, MW = 74. Taking MWair = 30 and bp z —200°, tell whether any residual liquid will evaporate faster if the flask neck is up, or is down fiyn <10 words. gm val/oar j/ /(mm4) W flaw 41V(la; 5. Summarize the difference(s) among DMSO, propanoic acid, and methanol as pure substances, by circling the state(s) in which each shows significant intermolecular H—bonding at 1 atm: (Circle at least one item in each row.) Ljf M DMSO (CH3)ZS=O) liquid vapor Propanoic Acid (CH3CH2COOH) w A vapoa neither /égn~( Methanol (CH3OH) liquid vapor neither H‘m “Twa- n 13U 6. (4) Note the distillation curve below for 100 g of a7solution of L (lower boiling) and H. L and H do not form an azeotrope. elow, Circle one of the items separated by / in each set. From the curve, the m@/ at most ~89°, and the mixture was abou 20 ‘39 30" 75 / 90 percent L. /' 7.(5)(a)(2) A solvent has a constant .AVP/AT = 16.0 torr/K from its nbp (hp at 760 torr) to nbp + 4°. A solution of a an-Vola- ct. qcojwptile Solute boils at nbp + 2.50°. Assuming S obeys Raoult's Law, calculate mole frac S, XS, and from this, xNVS' PM 5‘5 V10 01‘ n6; +2.W‘ = (740 + 2,3770%wa = $700 fuw g X51 % = (9‘98‘0 0’". X/UVJ: 0.050 (b)(3) The solution consists of 12.0 g NVS in 1.40 moles S. Calcu— late (1) molests; use XNVS = (molesNVS)/(molesNVS + moless); then (2) the MW of the NVS. M-JMgVJ mom 2 - 0.0m WM 'L nix-«m5 HM 9/ ”V5 + 0.050010) : whom; a W “N VI: (1 ,_ ~ I" Mqu! “' “Jay” ’ 9. (2) An aq soln containing 25.0 g isopropyl alcohol (IPA, is "salted out". The 35-mL upper layer's density of means it is 70 (wt) % IPA. Calculate (to 2 sig figs) what percentage of the original IPA is in the upper layer. ‘ '3 Siifll. Cllfi p0 :3J‘M¢L‘ ‘x (9.36 ayztl. :: (W - 0 M 5’04“ “L: 3mg K70% = 11.07 / C ‘9“ on} fflflAdL:/¥%71 . _ 3 - (£113 10. (4) W(ater) and odorous de D are mixed. At equi i rium, one phase is 4.0 mole % W, the other 0.80 mole % D. Assume Raoult's Law applies to the major component and Henry's Law to the minor component in each phase. (a)(2) Calculate the Henry's Law constant for D in water, HLCD—in-WI in terms of P°D. (b)(2) Then explain why the con— centration of D drops as the W—rich phase evaporates in an open vessel even if P°W = 50 POD' Refer to D's conc in the vapor compared to its conc in the liquid; use <15 words. @‘2m (9 0.7é[0% : HLCD .4”) (0.002) w % (100) 0 (271.1/ HLCDJ‘CW = 19.0 100 O) (a) 40...,“er P20. v I? “me .. L M 0 m WC Wit" .2, 77 :45va Fir-9 .4?de MFR/>7 M {d/K 014,04. 11. (2) 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 M2 4—6 21 6-satd aq soln 30 The solubility of W(ater) in a certain 0(rganic) S(olvent) is 6.00 g/L. After drying, the solid's average composition is CaC12'3.9W. Calculate the final W concentration in the OS. ’71“ NW '—"- 6.00J/L. )C 0.1% '—‘ 12. (2) A certain Drying Agent has a capacity of 0.20 9 W per 9 DA. $00Kv-1’of‘ Calculate how much DA one needs to dry 800 mL ya;gr;sa;ura;gg , OS of question 11 above; assume complete W removal. {09, (a ”r000; goo “L MFG/9C 0! Am {@151 anL L. #00le .. IMO co. +xoyw Ammwmfi£ «Ac/“ohm W 211% 4% I'Ji‘fi‘ 13.(4)(a)(2) After washing, Single Laundry Load (SL) retains 15% of the tub soln on spinning with Soap Conc Co; a double load (DL) has Fraction Soln Retained and SC in proportion. Calculate Cfinal for one rinse- and— spin for SL, and three ras for DL. Use formula Cfinal = n- Co[n (FSR SL)]#ras; n = the mass ratio of a multiple load to that of a single load. (b)(2) As FSR for SL increases, at some value the larger load/multi— ple spin Cfinal gets worse. For the general case of SL/l ras compared to DL/3 ras, calculate at what FSRSL the Cfinal' s for SL and DL are equal. Oflt F53: 14. (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) -- 40° 83° 6 110° 125° 135° 155° 165° 15. (2) Sometimes the melting points of 93% J and of 99% J appear to be the same. Even so, there is usually an easily observable difference in their melting behaviors overall. State this difference in <10 words. (that!) WM“? '32}? %— 17%? w«'// 44 Myer. _5— N1=10 16. (5)(a)(3) A solid sample comprises 7.5 g M and 9. 5 g N, whose M‘s WrngN‘tfl' 2000’ [flfwc/W 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. 5 g per 100 g S, N's is 5.0 g per 100 g S. The entire sample is dissolved in 200 g hot S and then cooled. Q» Calculate how much of each substance precipitates; put a num— ‘ her in each blank. NO credit without calculation/explanation. m LooaIwfl/w’o lp/‘W an «(4 9%" Mafia/1f. (b)(2) The entire sample would easily dissol e in 150 9 hot S 6 17. 18. Explain in <15 words why this is NQI the right amount of :10?qu feclveit f: uEZet bpuj‘ 5/ <4g4) [of-6- (2) A worker recrystallizes a material from a volatile solvent, suction filters it off, and spreads it out on paper to dry. Tell how one follows its drying and how one can tell when the material is Ucompletely dry. Answer in < 15 words. A ( /c m WA“ 1.145%” “3/ 37 (3) The Wicked Witch of the West (WWW) in the "The Wizard of 02" 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 WWW centration in wt%, i. e. , WtWWW/Wtsoln' Alf Ej’o 6L wan/4L So/I} M 1’g 19. (4) (R)—s§g-butyl bromide, (R)—CH3CH2CH(Br)CH3, reacts (slowly) with either NaI in acetone or with 2% AgNO3 in EtOH giving a near 100% yield of NaBr or AgBr, plus other product(s). Below, for each reaction, draw the principal organic product's structure, stating the stereochemistry (R, S, or racemic); you need not draw the stereochemistry. /zip (a) NaI in acetone: Product Stereochemistry /@ C/lyCl/LCI-H‘Chl; (0'! mlMVflrfi”) .t (p (b) AgNO3 in EtOH: Product Stereochemistry n QMI‘C @ a ' “ ((#3CI{.,,—CH-C H; C; baa/mdw’o) MMHF \’ V" ' p O~Et ._ a mat MW 20. (12 pts, 4 on this page, 8 on next) Recall the n—BuI prep: acetone n-BuBr + NaI - n-BuI + NaBr+ MW = 137 FW = 150 MW = 184 d = 1.27 d = 1.61 (a)(4) Suppose one wishes to be sure of getting 13.8 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. You may leave your answer factored. (tam/1‘- ...OM Maura/“EL 2 aovrwu ' 7 " ”6 i7? '0 20. (b) (5) Reaction éfilhcall (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+ + Cl‘ + 183 (6) n—Bu-Cl —---> _11-Bu+ + Cl' + 185 Sum: n—Bu—Cl + NaI ————> n—Bu—I + NaCl — 5.0 (1)(3) For the reaction n—Bu-X + NaI —————— > n-Bu-I + Nax, X = C1 or gr, show calculations of the free energies g: QQLL; gatign, 13G?§, for both cases, assuming the rate determining step is n—Bu—X + Na+ ----- > Q-Bu+ + Nax+. W "' ¢ x = He: «23 $36 + X=Br f’I7? “[14.~r-'T> 454-; :+3'S. (2)(2) Based on the fact(s) about the n-Bu-X/NaI—in-acetone reaction rate for X = Cl compared to that for X Br, e olain in <25 words whether the abov: uppose the n—Bu-I prep from n-Bu—Br and NaI in acetone is B done using a large excess of NaI but varying amounts of ace- tone, all too small to dissolve all the NaI. The reaction rate is then proportional to molesacetone but independent of molesNaI. Explain the rate dependence data and how they mean that acetone is catalytic by definition; use < 25 words total. ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern