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322af06_plq1_2_key

322af06_plq1_2_key - CHEMISTRY 322aL...

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Unformatted text preview: CHEMISTRY 322aL dfinfiflr€%r”%&&& .aéfifififlfififififif pmcfi‘q {Mffir FIRSTLABQUIZ - 11—2// 0/6 ’ By Page éEf.\a/ NAME 1. (8) _ 2.(12) _ Lab time 3.(12) _ T.A. 4.(12) 5.(10) _ 6 (6) _ TOTAL (60) __ This test comprises this page and six numbered pages. If a question says to answer in fewer than a certain number of words, DO SO. Deduction for wordiness. -. xi 1 h. e “ iz -. ar 0 ic- ho s .nd a-" up at. ”say m'r ' o -0 coen/Ic .: “ I, ‘/ gr.6ing cuestion-, pl-.se .e- ouv T‘ f st. NH -1- ”(71: 1. (2) A 14/20 male (inner) standard taper joint (like those on your . distillation apparatus) is Imade of glass 1.0 mm thick. Calcu— 31/ late its Inside Diameter at its narrow end. 4”“ £9 /‘( “7:3“ "1)mm 8:74” 9‘; [ ac mgamfl‘sQ/(M (“1&1wa Weld/14 :UO MMW mm L . 2. (2 A circular groove in a flat iece of material has Outside Diameter = 12 mm, and both width and depth = 1.5 mm. Calcu— late the ID of a non-stretchable O—ring which will just fit. £0: Q1 —<',Ln M’flmm =7 “1% 3. (2) Complete the following sentence, mining the boiling point ”LN/{Eng Z (in general) of a liquid, A. The bp of A is [WWW +2... M .f wag; W3, 4/1.? 760 {am “Mm M «fQYWQ/{flrwaqpe Below is the distillation curve for a 100 g of a solution of L (lower boiling) and H. There is no azeotrope in this system. IIIIIIII IIIIIIIIIIIIIIIIII IIIIIIIIIIII H3. IIIIIIIIIIIIIIIII=IIIIIIIIIIIIIII-IKIIIIIIIIIIIIIII IIIIII I IIII I II IIIAIIFiEIEQIFfiflfiUII I'vf'ififlflflféafiiflfifll I IIIIIIIIIIIIIIIIIIIIIIIIIII IIIIIIIIIIIQESIIDELSIIII IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII-III“JZamfillEIIIII III-IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIAIIIIIIIIIIIIIII IIIIIIIIIIIIIIIIIIIIIIIIIHIIIIIII IIIIIIIIIIIIIIIIIIIIIIIIV‘IIIIIII I IIIIIIIIIIIIIIIIIIII-II"IIIIIIIIIIIIIIIIIIII IIIIIIII-IB’ZZZZ-ul I IIIIIIIIIIIIIIIII nuns-IIIIIIIIIIIIII IIIIIIIEIIIIIIIIII IIIIIIIIIIIIIIIIIIIIII :IIIIIIIIIIIIII=II latch Below, circle one of the items -oarated by / in each set. From the curve, H's bp isé / a most 84°, and the ' 55 original mixture was about 6O /6/ 7O / 75 / 85 % L. “Hm—M “2 NW” 5. (2) Acetic Acid (AA), CH3COOH, has strong H-bonding in the liquid. (1) It has abnormally low enthalpy and entropy of vaporiza— tion; and (2) its vapor density corresponds to MW z double its formula weight. Correlate (1) and (2) using <15 words. Wm .2 Mar/7 WW’ cg AW (Li/0’71 (A; oV/ézx (A3704 Uta/4,44%" 6. (3) A pure organic liquid containing C, H, and 0 shows strong C—H .l. absorption in the IR, but no other strong absorption above 1550 cm'l. Name or draw two oxygen—containing groups of atoms MI W by this re\sult. 0H 0 6% 7% m c=0 6L " ' j (0%,). (AW 2 «1“ 144/, 11/451 (3) An aqueous solution containing 20.0 g isopropyl alcohol (IPA, .78) 's "salted out". The upper layer weighs 16.0 g and awry d is 85% IPA, based on its measured density. Calculate what rac ion the original 20 g IPA is in the upper layer. :‘we/evaf “L k“ [6.06 KW O u 1, ‘ U46 . (2) Compound C, MW = 108, is steam distilled when Pexternal = 760 torr. If P°W(ater) = 600 torr at the mixture's bp, calculate the mass ratio in the distillate; MW = 18. , MM, )1 E N men {1: . .34 §\ \W k water ___9__ z 400M x/‘Wm mm? 9 69099999ch W 6ng 9. (2) W(ater) and odorous cpd D are mixed. Two liquid phases exist at equilibrium--one is 1.0 mole % D, the other is 5.0 mole % (géyTvy W. Assuming that in each phase, Raoult's Law applies to the Hajority component and Henry's Law to the minority one, give Ptotal for the W—rich phase, in terms of P°W and P° . moi; D 0 W999 0.927 70"..) +— 0.75% MW Omen? {WT/2A. 03?? A 03!.(17 17’") .M .03” -_._..a.—____.__ ”’3;- ’IV"«v' fl\7 ‘5 ' / L,r 10.(4)(a)(2) A Drying Agent forms only a 2 hydrate; the DA/DA-2W has rel humidity = 5.0%. For a 3—fold DA excess added to a 3 9 W per L Org Solv saturated soln, calculate the final W conc. oflw NW 6505(1) 3c}/L ‘0'“. MM 2 O {fa/L (b )(2) For the solution above treated with a DA of capacity 0.20 9 W per 9 DA, calculate how much DA is needed to dry 600 mL W— satd W/fléx OS For this calcn, assume all the W is removed. "Lila/1.“ W“ “L 09’3qu 05 AMA) 061 r ELQ. \ 04 a For cpd F, KvB/VA, (i.e., z) = 6. Calculate the fraction of F 3 I" unextracted [FUE = l/(l + (z/n))n] for (a) use of B all at once, and (b) use of B so FUE is minimized. Take e3 = 20. (a) (2) fl :1 (b) (2) AAOO ,; ' flqe— (1“ I x We (+6 “"V’ (u-é)” = ‘/7 (at+3..) =e*‘ «C6 70%..” W0 (: 0001:) (6 \clfaooozth) (c)(2) For 2 > 6, the superiority of (b) to (a) (circle one choice)-— gets even better. stays the same. decreases. 12. (2) J has mp = 100°, and L has mp = 130°. The most probable mp for a mixture of 85% J and 15% L is (circle one answer)—— 98° 105° 115° 125° 135° . CUM/61034.. I)! flf W “'40 Oflt/V -‘t- IV a . ., 13. (3) A certain sample X, of mp = 75—80°, is suspected to be either largely cpd M or largely cpd N (but could contain neither). A mixture melting point with pure M of mp = 82~83° is 68-69°; a mmp with pure N of mp = 90-91° is 79-82°. Tell what these ‘ data say about the likely composition of X; give reasoning. [Lfl (0W WM 4/004 a?” “\(7{ WWW? flaéafl A771w7Wa/Lufal‘4u m “A 14. (6) A solid sample comprises 10 g P and 5.0 g Q, whose solubili— ties in a solvent S are independent. The solubility of each in not S is 5.0 g per 100 g S. In cold S, P's solubility is 1.0 g per 100 g S; Q is four times as soluble. (a)(2) Calculate the minimum amount of S needed to keep all the Q in solution after cooling. Show calculation. Jada 6.9% 42 WC 40/1 J 14. iaa/Wou 14.14%“: “Vi (b)(2) Remembering that it is wéLsirable% to get the entire sample dissolved, calculate how much hot S one should use initially. JM% (db/J) M “4AM MM /:d%tufl =L;L005 be c)(2) If the initial dissolution in hot S gives a colorless solution .1, with some brown goo sticking to the flask walls, tell what one ,( ’oL should do nex Use <10 wor f“ I: MC/L gated/fl {d/Vwafi O ’/,,/157’T§7—Slucose (dextrose, grape sugar) has mp ~148°. Addition of 1 g 7 water to 6 9 glucose depresses its mp to 95°. Restate these er/O data in solubility terms; refer to a temperature, the nature of the solution (dilute, conc, saturated), and to solution composition in terms of them __as_s_r Latig ofjlucose to water. 4f 750 4 (J6 “/14 (% %ZW+U (7’04 W a: (4)) m (ytww w W“ 18V? (4) A qggAflé mew.) ———l——____—‘___ -5- 16. (10) Recall the preparation of n—BuI: fl; 6 "[0 acetone n-BuBr + NaI —————>. n—BuI + NaBr MW = 137 ® MW ‘sz " ‘1 184 1.61 l‘vrez (a) (5) Suppose one starts with 5.48 g n—BuBr and-. Calcu— late what volume (in mL) of n—BuI corresponds to a 75% %. Ame“ A 6 fir 2 5‘4 X : 0.040 “‘1‘" ._, 9‘! fig “-1346? M c (b) (5) Both t-Bu—Cl and n—Bu-Br react rapidly with NaI in ethanol, giving ~100% yields of NaCl and NaBr respectively. Explain the following in mechanistic terms, using <35 words total: (1) The rxn rate of L—BuCl is independent of [I'] , while that of n-BuBr does depend on [I'] , and (2) the yield of n—Bu—I is unchanged by even a high cone of L—BuCl in the same rxn soln. “— -5- 16. (10) Recall the preparation of p-BuI: /f g 1[0 acetone NaI ——>- r_1-BuI + NaBr MW = 137 ® MW .‘EW‘ " Q g-BuBr + 184 1.61 (Wye/6 (a) (5) Suppose one starts with 5.48 g n-BuBr and-. Calcu- late what volume (in mL) of n—BuI corresponds to a 75% g A '80! 16v : $.45 : 0.040 ““1“ “Z" — fir} Mm M f, (b) (5) Both t-Bu—Cl and n-Bu—Br react rapidly with NaI in ethanol, giving ~100% yields of NaCl and NaBr respectively. Explain the following in mechanistic terms, using <35 words total: (1) The rxn rate of t—BuCl is independent of [I‘], while that of n—BuBr does depend on {1‘}, and (2) the Liglg of n—Bu-I is unchanged by even a high cone of t-BuCl in the same rxn soln. 17. (6) Note the data below: Reaction as (kgal) (1) n—Bu-Br ----> _I_1-Bu+ + Br’ + 178 (2) NaI —-—-> Na+ + 1' + 164 (3) n—Bu+ + I‘ ~---> n-Bu-I — 171 (4) Na+ + Br‘ —--—> NaBr — 174.5 (5) NaCl ————> Na+ + or + 183 (6) n—Bu—Cl ----> n—Bu+ + or + 185 (a)(3) Calculate (kaoverall for the reaction: n—Bu—Cl + NaI —-——> n-Bu—I + NaCl, showing your calculation. +W,/~’——”::>V\“’Z figflq'({f' N4; #7 "6‘1": +U“C/ f‘ (b)(3) Given that Keq for the reaction of (a) above is highly favora— ble, tell why alkyl chlorides are worthless in a practical sense for making alkyl iodides by the process used in this lab. You need not do calculations but do refer to how 5 changes along the reaction path. ...
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