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Unformatted text preview: CHM 331 Exam III Equation sheet provided This exam has FOUR questions. Total points: 100 21 December, 2008
Time: 50 min
Dr. Jumean 1. (25 points) The following results refer to the adsorption of acetic acid from 100 cm3 of
aqueous solution on 2.0 g charcoal. (Molar mass of acetic acid = 60.0 g/mol.) Initial molarity of solution 0.520 ' 0.260 0.056
Molarity of solution at equilibrium 0.484 0.231 0.042 a) (18 points) Calculate, from a suitable graph, the Langmuir isotherm constants. Express
the amount adsorbed as grams acetic acid per gram charcoal, and concentration as molarity. .L
.  M mum mm L .
ﬁanbul) Newt) ° JawI Mag/3 [173 x, LA'] I Glance4
'ﬁ
OS‘Ie 04,579 0.05.1. 0.13035 aoorS 204 ygé
0'16“ {27.3} 0.02? 0.0013 043")"; 9'33 67%
0’056 675“; 0'0/7 0.05!!! 9575“; 33‘3 l‘127
K (9 . , I . ,. l 1
“54.02  ,w. Jim»: +4.
”(1.7% «bahJ 14ij ’4 K K943 k b) (7 points) How many molecules of acetic acid are required to completely cover 1.0 g of
charcoal? 0 5 10 15 20 2. (22 points)
a) (8 points) The adsorption of nitrogen on charcoal obeys the Langmuir isotherm. Show that l at constant surface coverage (i.e. 6 constant): I KP = constant (p; KP
M‘Cr” 2Q, KWQH: MOO«Q g ./_ l4? {9 c9 “to Moat/3'74
"“ ﬂaw b) (14 points) The following temperature pressure data (corrected to STP) were obtained for
the adsorption of nitrogen on charcoal at constant 9. Calculate the enthalpy of adsorption under these conditions. T/K 225 I 273
Lat: P/atm 11.5 35.4
W ' , l
A}: k“ &\ )CL. )— .— bu /‘L' (“1
’ can”? ———— c / 7.. ’7!
p“? K KrDU‘ —"“"_
3%’ﬁ“ 3”“ W“ g 53
‘4 2. 3. (25 points) The surface tensions , in N m'], at 18°C for aqueous solutions of nbutnoic acid of molar concentration c are given by
y = 0.0715 m 0.0298 log (1 + 19.6 c) a) (12 points) Calculate the value of the surface excess concentration I” for c = 00100 M. z 303
if ”ooz‘i‘h’ﬁ(i 3’2;sz 3—0 2J1.
0!» 24¢; (”It“) 1+“ch
4
P: “f— ??I 5 ° 01"“ (*U’UI :a’Js’x/Io
KT ac {Jinn/1?! “”L 5“ b) (8 points) Calculate the concentration at which the area occupied by a single molecule
of butanoic acid on the surface is 35.0 nmz. , t ,_ Je ,, .
r, I I/ﬂ ’ 6.1.13 3119131 SfWHO ifm‘”) ,5? (
’ 0.1M >(I0 r“ “‘ ’— 9.461 two”? 2:. w—"
33MXZ7/ (pr/966) ‘r
a; u€HVIB M c) ( 5 points) If e could increase indeﬁnetly, what would be the limiting value of f as 0
becomes inﬁnite? W TIT?“ Mr Lg’lc l76C77/ P : 3
Wrié §rjs(v'0 4. (28 points)
a) (21 points) A cylindrical tube is 2 mm long and has a diameter of 2 um. A liquid of
viscosity 4.0 x 10'3 kg m'1 s'1 ﬂows through the tube at a pressure drop (PlP2) of 20 Torr. (i) Calculate the volume of the liquid passing per second in the tube. ELI: ’ h‘ﬂqmp; jT (Loxﬁifixzo' K1313
/ ’ ,.—————"'"' _______._._—~——' 1,.
d6 griL 3’ x 2’Xlo’3xulbg T
_ 6:" ‘L ,l
j; 13‘! \(l m 5 (ii) Calculate the average linear speed of flow of the liquid in the tub. _
{74 L: n— ( l X ‘9 ‘L. 31)”)“0’ (iii) Calculate the maximum linear speed that any cylindrical layer of the liquid can
travel in the tube. ’1.
MH’M _.—~~ """"""" ’ 34_#—_ﬂ '3
W11 axsrxxa “um b) (7 points) The relative viscosity of a polymer solution containing 1.00 g of polymer in
100 cm3 is 2.800. A solution half as concentrated has a relative viscosity of 1.800. Calculate the intrinsic viscosity [1}].
(Hint: assume that a graphical treatment would give a straight line.) FA 50%, 475,0: ’2: (VII/hf, ____..——ﬁ
:2 i— ('ZW’O/F'gv
c9 Irrb , ‘5
745V), ,J—— (.g7oa—">__l
0 03—9“
v
s
(gal; C 17"? M
0 \,G O
OMVIrW ﬂ AKMAWH oo§o° ‘g a
6 0.14.0
CW“ M 7.... Z axial/“m’ i“ ...
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