{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

322a lab quiz 1 fall 04 key

322a lab quiz 1 fall 04 key - [0r CHEMISTRY 322...

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

View Full Document Right Arrow Icon
Background image of page 1

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

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3

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

View Full Document Right Arrow Icon
Background image of page 4
Background image of page 5

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

View Full Document Right Arrow Icon
Background image of page 6
Background image of page 7

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

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

Unformatted text preview: [0r CHEMISTRY 322 aL , #7 (C‘h‘CL fuf 40% FIRST LAB QUIZ -—-———_ NAME Lab time This test comprises this page and seven numbered pages. If a question says to answer in fewer than a certain number of words, DO SO——deduction for wordiness. e hours. Labs Nov 11 — ake-up only; TAs ' stay one—half a lab. Morning 1 GC make-up wil BY 2&9: 1.(11) 2 (8) 3.(1o).______ 4.(11) 5. (9) 6. (6) 7.(1o) TOTAL (65) Nov 15, 6:00 p.m. -1- ' Mi- l/ 1. (2) A 24/40 male (inner) standard taper joint has a 1.5 mm thick wall. Calculate its narrow end inside diameter (ID). 1% _D= +-%0)H=l9” .1. \/ ‘3 (3 “‘ “‘ 1:0 =[,zo—- lC/,5‘)3Am={/7 W412 2. (2) Calculate the thickness of an O—ring of ID = 6 mm, OD = 9 mm. (c , 3. e solubility, w/wsoln' for Water (MW = 18) in an Org' o v Llfl Lfm (MW = -- '1.0%. Using the i _. a- ' - ,_ soln approx ‘ calculate the ‘o.* oerce n the sd - , Ceynof A 7"3‘ ( fl ’ 60‘ l. ., f- " . 05 [14qu) (3) A bottle's partly torn off label says "4—Methy pentan...". The carefully dried liquid contents' IR spectrum shows strong .1 absorption from 3100 - 2600 cm'1 and at 1720 cm'l. Assuming the liquid is a single mongiungtignal compound, tell its 0 “C 0" likely identity; give reasoning (< 15 words). -ol‘oV‘ f,( :7 M -—Coo,///- C/X a3» 4-M7F M4 [if/”WON dcz‘6{. 5. (2) An aqueous solution made from 35.0 mL isopropyl alcohol (IPA, d — 0.78) is "salted out". The resulting 35.3—mL Upper Layer weighs 30.0 9. Using the table below, calculate ONLY the mass of IPA in the UL. ZERO EEI for doing any OTHER problem. d i . M Alcohol,wt% ‘ _§Q%§2LL 6 &&M4 7/ Ml. 2 _;%Z£¥£. ;:(0.X5fi 0 .5 313333 33:3 3L3mb 09“» 36A 3333 33:3 :? [A 7%»7%(% EM =6) 08E” 613 0%W m3 07¢, 23:32 32.23 3N4 “L w 3 02" am- 33333 33:33 _ 11,2 42w ’4“ V ‘M 0.7863 100.00 6. (2) /0J7 2:. SE: Note the diagram, suppos- edly the normal pressure liq-vapor phase diagram for cyclohexane(CH)/fur— fural. In < 15 words, tell why it must be incorrect by considering what vapor it says is in equilibrium with boiling solns of XCH>0.25 0(‘47 raw '5? {41% = ()3, N0 \f “1° 0 I‘ o 0.5 V\\/\ a r /(- b VflAM 015—" Mole fraction of cyclohexane ( X C H) wgflm 50/[email protected]< MM QCW lrm(mf~1um/)) ~ 7. (6) Recall that [SVP per degree, dP/dT, = P(£3$vap/RT). Acetone's nbp (bp760 torr) = 56.5°C (329.7 K); take [X 21.5 cal/(mole-K). A tilde, no means per mole. :9 Svap = (a)(3) Calculate dP/dT (3 sig figs) at nbp; take R = 1.98 cal/mole—K. JL I’ghggw that the units come out as torr/K. fix /0 :éléigqfi : 21mg ¢~gf1=fii , 760 km {fig wait ”a” - 7r” ~ M (299:. (b)(3) A solution of a non-volatile solute (NVS) in acetone has a conc in mole fraction terms, XNVS' = 0.050. Assuming dP/dT is constant and that acetone obeys Raoult's Law, calculate the nbp of the solution. (First calculate xacn' then the VPsoln at pure acn's nbp, then apply the result of (a) to the soln.) @ Xacn “-3 LOW ‘ quf = (.000 ~0.0S’D = 0.9570? / : 7é07L'VY): Oq‘r: 7ZL7LM/V‘ {alhf‘r ‘ / < . W14 AV = 7w @ ”MAX 7‘1 “W“ 4” V 31%,“, -mww w m :7 M = ,4 19 : mm :9 A4,, {car wry. a; 3227+ {.f‘ -3- [)j Z:[O 8. (3) de A steam distills at a mixture hp at which P°W = 8/9 the external pressure. (1) Calculate the ratio, molesW/molesA, in the distillate. (2) If the distillate has 17 gA per 18 gw, ’t’ f i.e., per mole W, calculate the molecular wt of A. /’ 89 ~ ) *(0w/mfi—o ./%= g:/(“J°°‘fg ? buw{€([f ’ _, fl4‘¢5¢ (HM/W '5? (77¢: y[M{e%fi 3: (7kgt/3/6 9. (7) Notelthe data below. "m-n" in the left column means "a mix- ture of CaClZ'm H20 and CaClz'n H20 in equilibrium". CaCl2 hydrate mixture Relative humidity (rh), % 1.25 4.2 14 4- 21 6—saturated aq soln I (a)(2) The solubility of W in an Organic Solution is 4.0 g/L. One has 100 mL of W—satd OS, which also has 3.0 9 W droplets. 0n adding 3 - 4 g CaClz, it forms a clumpy solid plus some gooey aq soln. Calculate the final mass of W in the 100£Ei:¥? ( M — 00MP~ (00 ML w—:«// 05 km 4‘08" ~ . flu = 0.110 0&0; (Ma/)w, (9 40;; x Ow my (b)(2) The OS is separated from the solid and aq soln. A new ow capacity, only 0.015 gW/gDA, but Very Efficient Drying Agent is added. Calculate how much VEDA is needed to dry the CaC12< treated soln above, assuming it removes all the water. 304W= 0"” WWW 0.0lS-aufyor 0 l 2 o vle-4 (c)(3) Leslie the Lab Loser (L3) decided to save time by using enough VEDA to remove all the W completely in one step. Calculate needed just to remove the W droplets from the original 11 why L3 lost nearly all his/her OS. gVEDA mixture and te -4- 10. (3) An unrinsed Single Laundry Load (SL) retains 8.0% soln on spin fiEC‘ drying with detergent residue per mass = CO. A Multiple Load - of size n-SL retains soln fraction and detergent residue per taCn For two rinse-and—spins, calculate n for 1?;. ’0‘- mass in proportion. P \ - . (n is not necessarily a w le number.) I‘M 62::2’1;:;‘ :;:¥‘;£(é =? y\ (:5 l:f" <:£::;/€}Lj:] :%?€7 MLQ 1.. 3 51:51:14/104= n 6.. n (0.02)] =L0-03)”m 064 L!— 9(‘V’:§$ ( H 2 2': W 11. (2) Methyl t—butyl ether (MTBE), bp — 55°, is often a convenient solvent for extracting organics from aq soln. To isolate an organic from MTBE easily by distillation, how should it giffie; substantially from MTBE? Answer in <10 words. (fey/1444M Xcfiém setéfftufh/é) W t ; pure K melts at 95°; a certain sample x melts 84° - 88°. A mixture of equal parts J and X melts 72° - 74°; a similar mixture of K and X melts 86° - 89°. Tell what these data say about the likely major constituent of X. Give your reasoning concisely; use no more than 15 words. ‘ \ woe Mm; Ka / J (cr‘ ((erK 3M7 ‘5 MP. 12. (3) Pure J melts at 91° 13. (3) A solid sample comprises 11.0 g M and 7.0 g N, whose solubili— — 65°, are independent. The solubili- 0.0 g per 100 g S. In cold S, M's The entire sample is dissolved in 200 9 hot S and then cooled. /(i> Calculate how much each of M and N precipitate; put a num— ber in each blank. NO credit without calculation/explanation. l4 — 5 - [€7’ ii—ta‘7 (3) In the "The Wizard of Oz", Wicked Witch of the West (WWW) is very Water—soluble. Assume at 22°, a satd soln has 50 kg WWW in 5 kg W. Restate this fact in terms of mp depression by giving a wt % W added to WWW to lower her mp to a certain value. Use <15 words; zero if refer to soln or solubility. A) 7’» mm [é/LW “/7 fl lzd~ \ \ Lrowtflfi/ IA‘W3Q*H<- (6) The dibromoalkane shown below reacts with either NaI in ace— tone or with AgN03 in EtOH, giving one equivalent of an organic product. Write the structure of the principal organic product formed in each case. If and onl if reaction occurs at a stereocenter, state whether the center retains, W . . racemizes, or inverts; otherw1se write.“NA". 15. , and £33 NaI in acn ----- > 1 NaBr+ + org prod (R)‘Br' C'CHz 'CH2 -CH2 'Br I AgNO3 in EtOH —-—-> 1.e&du1 + org prod CH2CH3 {‘J( Product Stereochem Max)? M (a) Sing; (,6) 3r“ (C " C (“A CL/LCé/lr A) M as a Product Stereochem - 6 - [e7' 2: ‘z;¢; 16. (16 pts, 6 on this page, 10 on the next) Recall the n-Bul prep: 6 7,6 M 74.1%.. + acetone n-BuBr + NaI ——————————er— n—BuI aBr+ (er (a)(4) Suppose one is confident of obtaining a 40% yield and wants to get at least 30 mmol n—BuI. Calculate the minimum volume (in mL) of n-BuBr (limiting) which one must use to assure this result. You may leave your answer factored. 30M / - mm! Mm Wéw “” M“: 0-4 mw(W‘A€«L : 75‘de A‘gbtgr- (MIM(W%'I"‘ MW = 137 FW = 150 Q = 1.27 (b)(2) 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. he Na+ so I' is reactive. the Na+ so NaBr can precipitate. the Br' so NaBr can precipitate. 16. (contd) - 7 - flf icfo (c)(10) Reaction fifgkcall (1) n—Bu—Br —-—-> r_1—Bu+ + Br' + 178 (2) NaI -—--> Na+ + I‘ + 164 (3) Q-Bu+ + I' ----> n-Bu-I - 171 (4) Na+ + Br‘ -—--> NaBr - 174.5 Sum(Br): n—Bu—Br + NaI -—--> n—Bu—I + NaBr — 3.5 Keq=200 (5) NaCl —---> Na+ + c1~ + 183 (6) n—Bu-Cl -—-—> n-Bu+ + Cl‘ + 185 Sum(C1): n—Bu—Cl + NaI ---—> g—Bu-I + NaCl - 5.0 Keq~2000 Sum (1)+(3) I'+ n-Bu-Br -—-—> Q—Bu—I + Br" + 7 Sum (6)+(3) I'+ n—Bu-Cl ————> n—Bu—I + C1" + 14 (1)(3) Note the overall equilibrium constants given above for start— ing with R-Br compared to R-Cl. For each, tell approx what fraction of the reactants turn into products at equilibrium. 2, ~ “If?" ‘4' @J a, 202) :9 7 4281! (2313/ @az’i kwwfi ~77flr%waw(gfil} (2)(4) Given that the rate determining step (rds) is I' + n-Bu-X ——-—> n—Bu-I + x;, (i) state in <8 words what the last two numbers in the [3G column, "+ 7" and "+ 14", are, and (ii) tell in <12 words what their importance is regarding the behavior of the reaction. (3)(3) As a practical matter, which are more important here, the K 's or the last t o,A? e ies? Explain in < 20 words. eq (2ffid442V‘fi:y - f%1uifL _faa‘\’ AQ‘I/UMC/ «MN MW ,' e6 NS“ \——9 7 iidd M1 47/ Jiotf/rélffbtwl . ...
View Full Document

{[ snackBarMessage ]}