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LQ1 Sp05 - CHEMISTRY 322aL/3 115d Prat{MIL-Er FIRST LAB...

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Unformatted text preview: -. CHEMISTRY 322aL/3 115d Prat {MIL-Er FIRST LAB QUIZ - ‘Fql/ O 5‘ ——. BY 2393 NAME < t 1.(12) 2.(12) Lab time 3.(12) T.A. 4.(10) 5. (9) This test comprises this page and five numbered pages. TOTAL (55) If a question says to answer in fewer than a certain number of words, DO SO. Deduction for wordiness. t‘~ wrw/ /)J i ’ UL ' 1 ' -/ (4/ . 1. (2) A 24/40 male (inner) standard taper joint (like that ’ -=‘ ’ I) , lower end of your reflux condenser) is made of glassfi W thick. Calculate its outsi e diameter (CD) at its narr w end. 4"" Joe»? 04,444 £20 MM 40 [ M K m, N C raw) my? M 4 m -Q—Lé ’+ MM MM 2. (2) An 8—mm ID O-ring just fits in a groove of OD = 11 mm. Calcu— late the O-ring's thickness. \ K [l‘ K 3. (2) A bulk grade liquid sample shows moderate absorption at 3400 cm'l. After treatment with anhydrous CaClz, the 3400 cm'1 absorption is very weak. Explain in < 12 words. \ \ V arc/1 w a {was 5“” ‘0 ~ or . (AM 4. (3) An aq soln made up from 15.0 mL isopropyl alcohol (IPA, densi- ty = 0.78 g/mL) is "salted out“. The upper layer weighs 12.0 g; its density corresponds to 75 wt % IPA. Calculate what fraction of the original 15.0 mL IPA is in the upper layer. MW I‘m/1‘64! f/#~= {s’ML Y0.75 7/ml. : /I.7 . Ila “L La 75%@£] ‘ 01L. I‘M =3 10.7 “‘ flfi/ 2 477% (0.77) n, 5. (3) Define normal DQilng DQLQL (nbp) by completing the following AL sentence in <15 words: The nbp of any liquid is .. “Mr" {Lg Up a flL frmlaaw/‘M “f“ q (M‘CA [(74 {(7 «(2653 V90” 7:441 7é0 é cm (W W 5 dfwmpkre‘) p.71”1'2' 6. (4) Heavy water (D20), formic acid (FA, HC(O)OH), and n—BuBr all have nbp's ~ 101°; see drawing to the right. In the blanks below, write the letter of the VP vs T curve corresponding to each: D20 é FA A n-BuBr é_ At pressures belo 0 torr, e.g., 740 torr, which liquid has @the lowest bp? B C (Circle one letter.) Compound C, MW = 162, is steam distilled when Pexternal = 720 torr. If P°W(ater) = 640 torr at the mixture's bp, calculate the mass ratio, in the distillate; Mwwater = 18. mlewm 1%.fieffi / c. ‘6‘“) 1/ “”4““6/ g/Mlec/ 1w (agxééjrv) 73C (2) W(ater) and odorous cpd D are mixed. Two liquid phases exist 8. ,4 fit A at equilibrium-—one is 2.0 mole % D, the other is 6.0 mole % KL. ' 5W. Assuming that in each phase, Raoult's Law applies to the M dL majority component and Henry's Law to the minority one, give 0 L4/ Ptotal for the D-rich phase in terms of P°W and P°D. .4 ‘ O 0 ngdcflfl Ptot, D-rich phase = 0.?g FM) + (9-?‘11/00 Home . 9. (2) A D ing Agent forms only a trihydrate; the system DA/DA-3W has relative humidity = 15%. One adds a slight DA excess to a half-saturated soln which contains 2.0 mole % Water. Calcu— }M late the final W cone, in m‘ole %. 0,3c ,.‘———‘ 10. (2) A Water—saturated Organic Solvent is 0.50 wt % W. For a DA with capacity 0.30 9 W per g DA, calculate how much DA is W needed to dry 3 kg m OS. Assume all the W is removed. {Wade} 47900.} W 30003—561“ ’° 3"?” mm] ,at I00 (74.1 A7 ¢¢/$ -3- 11. (3) An unrinsed Single Laundry Load (SL) retains 5.0 % soln (FSRSL = 0.05) on spin drying, with detergent residue pg; mafia = C0,SL' One rinse—and—spin thus leaves C1,SL = 0.05 C0 SL' For a Quadruple Load, calculate CQL in terms of C0,SL for Lhree ras. Take C0,QL = 4 C0,SL and FSRQL = 4 FSRSL. Tell what your calculation means regarding the final detergent concentrations for (SL, 1 ras) compared to (QL, 3 ras). You 6m WM {714A fWA: 3 23 Comm -. (om-(FSKQL) ; ‘tQ‘SLZ—H‘FIQL] ‘ 4C).“, (OJ-)3 = 460,”. (0.00%) : .' 0.03am“. amm M 42:... A190} McAuL/o—M W W Sin-LW. 12. (2) Ordinary alcohol, ethanol, is cheap and readily available, volatile, and relatively non-toxic. Tell in <10 words why it is not used routinely for extracting a eous solut'ons./ MCfa‘é/e wrfl 4) & “Jfl’nlj W [4YJL./ 13. (2) G has mp = 100°, and H has mp = 140°. The most probable mp for a mixture of 95% G and 5% H is (circle one answer)—— 80° 95° 105° 120° 135° 145° 14. (2) A solid sample J has mp = 69 - 72°. A purification process . gives good recovery of material of mp = 71 - 73°. Did the process likely purify J? Explain your answer in < 10 words. 1% YMJ mlz‘w‘d r {Mt/Kw W cm M 51ch 15. (3) Sucrose (ordinary table sugar) has mp = 175°. Addition of 1 9 water to 3 9 sucrose depresses its mp to 65°. Restate these data in solubility terms; refer to a temperature, the nature of the saIfiETSE_TaiTEEE, conc, saturated), and to solution composition in terms of wt fraction sucrose. 4695.4M m 7—: [0 16. (4) A solid sample comprises 9.0 g M and 3.0 g N, whose solubili- ties in a solvent S are independent. The solubility of each in hQL S is 6.0 g per 100 g S. In cold S, M's solubility is 1.0 g per 100 g S; N is four times as soluble. (a) Calculate the amount of S needed (a) to just dissolve the whole sample hot, AND (b) to just keep all the N dissolved in )after boiling down filtrates to recover W“ ' =(/r0 3] WWW % (7 M, (0% a. 29%;! “WE 17. (2) A solid sample on treatment with hot recrystallizing solvent gives a colorless solution with suspended gray powder. What should one do next? Answer :?§g_fiigally' 1n < 6 words. #07! 7"”7 f/w‘ c252»; 18. (8, 4 on Recall the preparation of n—BuI: this page) acetone n—BuBr + NaI -———————e’— n—BuI + NaBr MW = FW = 150 = 184 = 1.61 «0 mod «fwim (a)(4) Suppose one starts with 6.85 g n—Bqu/;nd 9. 0 g NaI. Calcu- late the theoretical yield of n—Bu— I in mL. ”L 257M444 £56451“ 2 08‘. : Otamm/b ” fl??? c0) WA) W yrs/J % A—Igwr = m mud/JO WT ullKN, MLWflu‘; : 0-mm{~wx%: 18. (contd) - 5 — /1j 6: 7” (M3 lkgall (b)(4) Reaction (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(Br): n-Bu-Br + NaI -—--> n-Bu-I + NaBr - 3.5 Keqezoo (5) NaCl -——-> Na+ + c1' + 183 (6) n-Bu-Cl ----> n—Bu+ + Cl‘ + 185 Sum(Cl): n—Bu—Cl + NaI ——-—> n-Bu-I + NaCl — 5.0 Keq~2000 ’l—- ——49 Tell how the solvent acetone facilitates the reaction as done. State specifically whether its effect is mainly on the equi; librium or on the rate. Then using < 20 words, explain how the solvent affects whichever aspect you chose. Khan/{i 03 cm v96 [Ag-Wuq/ {44/ lc‘fl’fe ‘mg/aewmfi , M6 “fit/m Lcfl/(mm a! View? ”4.5,ou7 “W“: ~ @ -5 43., 7/ ML11/‘WJ I fiWfl. “l . 19. (5) J 3 Both n- and t-Bu—Br react rapidly with NaI in ethanol, each giving NaBr in near 100% yield. Explain in mechanistic terms: (a) why t-Bu—Br does not react in acetone but does react in EtOH, and (b) what the main substitution product t-BuBr gives with ethanolic NaI-—it is not t—BuI—-and why. C47 3.57% 643A 24m flak/0M5 .mc/fm 5U; W1» 7‘77“” 0““ “,4” Now) fl/anc‘fd ‘6) fl 5774/ erq-Igr cm (WFQLC ( «A f’flwe ((#wa WW4 A’lfifigfflwa ’ mug/22‘ 406/74 ‘47) hf/c . 6'6“, 90:} em! tr£q~0~éf7 ~ ée (7: ngb“) ...
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