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Unformatted text preview: CHEMISTRY 322aL ' W pWC‘fl‘CL W ‘éV' FIRSfLABzQUIZ  m KEY Lab time BY Eﬂgﬁ T.A. This test comprises this page and five numbered pages. If a question says to answer in fewer
than a certain number of words, DO SOdeduction for wordiness. anon £913 for calculations. TAs wi ave quizzes at office and makeup labs, week of Apr 2. evening lab convenes at 7:00, Tuesday morning convenes at ading questions, please see your ‘1‘ 13/3 /L‘H'elhvu_ 1. (2) A 19/35 male (inner) standard taper joint has a 1.0 mm thick ~'/
wall. Calculate its outside diameter (OD) at its narrow end. '
~ 0k)
H 00¢waLM/ m 17“”‘1‘ “a” W
ml 00 ()3 215'3r—“M W) 3?
SQ? "Inn—m 3! 1" rMMi
2. (2) Metric oring size is specified as "IDxOD", both in mm.
Circle the size of the Oring which will just fit in a circu
lar groove of ID = 9.0 mm and width = 1.5 mm.
7.5x9.0 9.0x10.5 6.0x9.0 6.0x7.5
3. (2) An organic liquid's IR spectrum shows only CH stretching
above 1550 cm‘l. IF it contains oxygen, name or describe what
functional group is indicated. Explain in <15 words.
Efﬁe/t C—OrC. 447 —O¢4/
T“,—
0, (5.0 “MM atéfo" d’ wrm on".
I CC=O ~110° 10.4 7000‘?WOJ
4. (2) An aq soln containing 25.0 g isopropyl alcohol (IPA, density =
0.78 g/mL) is "salted out". The 20.0 mL Upper Layer weighs
17.2 g. Using the table below, calculate ONLY what fraction
of the UL is IPA. ZERO if give information not on point.
mmno t5 (7.2. _ a?/
3:22: 3:3 00‘“:ch +ML  z—vﬁ ~ 0.860 4!.
cows :34 zaaxluq
mums 4&0
as 
3:3? 2? LLL w 70.322 L74 1
sumo 7&3
osun 78J8
memo 803!
can) 9t”
ojun "10
oJun lﬂlw
5. i“ (5) Heavy water (D20), formic
acid (FA, HC(O)OH), and nBuBr
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: FA A nBuBr 6 At pressures above 760 torr, e.g., 780 torr, which liquid has
the lowest bp? A B <:::> (Circle one letter.) WrannllllllllllzazI '_—" ____.____‘
.4..: (IIIﬂl'll‘=F‘l ' ___
aum __l._.__ D20 '“ n7£=ﬂ (2) Define nouns]. boiling point (nbp) by completing the following
sentence in <15 words: The nbp of any liquid is .. O n m ﬁemww 4f NattA (ED!
“0” VP ﬁxed 760 fun (0! CW 7. (3) Compound C. MW = 162, density = 1.50 g/mL, is steam distilled. Pexternal = 720 torr; PoW(ater) = 640 torr at bpmixture‘ Calculate the volume ratio in the distillate; MW 18. water = C
me/mLc= 6w AMMMX N was xswgl‘go’kc ‘‘ /‘§§MLL:)
guﬂdﬁw I627. 374cc. A0 W (“J
Mix M
W(ater)Kand OCT gives two equilibrium phases. 91 (aq)
is 1 mole % OCT, @2 (org) is 97 mole % OCT. P°w  6 P°OCT. ‘1 :;f After phase separation, which has the larger partial POCT: W /
3’5, (1) immediately after separation? o1 <92
(2) after ~20% of each has evaporated? ¢1 neither.
4M»
9. (3) A Drying Agent forms only a trihydrate;  * DA/DA3W‘vrel
l has relative humidity = 10%. One adds I00% DA exces;
f. onethird saturated soln, which contains .0 mo e Water.
f4f'““"’ Calculate the final W conc, in mole %.  1_ ﬂ 0.1%, a! a!” a, 6‘0 w/e % W. q0‘3‘i’o U WC J. 0.!(6.0M('¢)[0.60%
ué¢ 10. (3) The solubility of water in an Organic Solvent is 0.20 wt 8.
A DA has capacity 0.20 9 W per g DA. Calculate how much DA is
needed to dry 3 k9 ﬂ;satd OS. Assume all the W is removed. I/f’ﬂ 3d), 3%? waf—,( A! Woodsman, 11. (2) Acetone is cheap and readily available, relatively non—toxic,
and has nbp = 56° Tell in <10 words why it is not used rou—
tinely for extracting aqueous solutions. 3 lite? 12. (3) An unrinsed Single Laundry Load (SL) retains 10.0 % soln
(FSRSL = 0.10) on spin drying, with detergent residue per mass
= COISL. One rinseand—spin thus leaves C1,SL = 0.10 Co SL.
For a Triple Load, calculate CTL in terms of COISL for three
ras. Take C0 TL = 3 COISL and FSRTL = 3 FSRSL. Tell what
your calculation means regarding the final detergent concen
trations for EL; 1 rag compared to IL‘ 1 ras. (TL‘JM = 3(0\5L(3'NIJL)J= 3 (0.3}3Qﬂft 7/ = 00" COJL. (Mg/,JLIIM mama, (.034 Mlqd M MW 4/ Lawn/I7.
(M. (mew/“41‘. 1 ftﬁc} 13. (3) Pure J has mp = 86°; pure K's mp = 92°; a certain sample x
melts 78° — 83°. Mixture mp J and X = 73°  76°; mmp K and x
melts 84°  87°. Giving your reasoning in <15 words, state
the likely major component of x. 14. (3) Sucrose (ordinary table sugar) has mp = 175°. Adding 1 g
water to 5 g sucrose depresses its mp to 100°. Restate these I data in solubility terms; refer to a temperature, the nature
“2/4"”\ of the solution (dilute, conc, saturated) and its wt fraction sucrose. ZERO for any reference to mp depression. '4' Ali=—(0 15. (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 hp; S is 5.0 g per 100 g S. In cold S, M’s solubility is
2.0 g per 100 g S; N is twice as soluble. Calculate the amounts of S needed (a) to just dissolve the whole sample hot, AND (D) to just keep all the N dissolved in
gold S, after boiling down filtrates to recover more M. (a) : /£0 2th W”) M W [email protected] 16. (2) A solid sample heated wit solvent gives a colorless solution
with suspended green flakes. In < 8 words tell what step
should be omitted from the recrystallization procedure. ﬂuﬁmf “(ﬂ ﬂed/aria»? céqrwtg 17. (10, 4 on Recall the preparation of nBuI:
this page)
acetone
nBuBr + NaI ——————’— n—BuI + NaBr MW=137 FW=150 MW=184 d = Ame (a)(4) Suppose one starts with 6.85 g n—BuBr and 9.0 g NaI. Calcu
late how many mL nBuI corresponds to a 70% yield. 6.3!? Q—iudr = I?) yam/e. ISVI¢ x 00:14
MMM. 1.4.: w; g__/: = 705x *4 Lag/‘1‘ 7 ‘fOO ML 17. (contd) (b) (6) Ream.1931 A5 ﬁscal). (1) nBu—Br —> n—su+ + sr‘ + 178
(2) NaI ~> Na+ + I‘ + 164
(3) n—Bu+ + I‘ > nBu—I  171
(4) Na+ + Br‘ > NaBr  174.5 Sum(Br): nBuBr + NaI —> n—Bu—I + NaBr  3.5 Keq~200
(5) NaCl —> Na+ + C1“ + 183
(6) nBuCl > n—Bu+ + Cl‘ + 185 Sum(C1): nBuCl + NaI > n—Bu—I + NaCl  5.0 Keq~2000 (i)(2) Given that the ratedetermining step is (x = C1 or Br) I + n—Bux  > n—Bu—I + X— calculate 4'5: for x = C1 and Br, assuming they each equal the diﬁﬁerenee between the nBux and nBuI Bond Dissociation
Energies. (Reverse the signs of the data above as needed.) C.: /$5 ' (73
[ l‘1/ A;O" “ 7/
17% WML (ii) (2) Relative rates are exponential in the in nat’s. For this rxn, each 1 kcal/mole deereaee in [N3 , increases the
g 57; rate by a factor of approximately 4.5. Using your results . from (i), calculate the rate ratio for x=Br t X=C1. a t
Mi A‘#« AG 5* f;1vv ': '1‘! l \ﬂ74f;
galmi Maud/(\Me “a: CH7 :‘C 37000 C (iii)(2) Even disregarding rates, tell in < 15 words why the differ
ence in the Keq's for X=Br and X=C1 is not important. 6m K ‘I “was 777%
/ a? crux/«Arm f \qu/oﬂ‘add: 18. (4) Both n—BuBr and tBuBr react readily with excess NaI in EtOH
giving ~100% yields of NaBr. Write the principal neutral
organic product for each reaction: 2‘ n—Bu—Br  > (4,3,4—3:
w/L ...
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 Fall '07
 Jung
 Organic chemistry, Solvent, Keq, VP vs T, Pexternal

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