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Key to exam III

# Key to exam III - CHM 331 Exam III Equation sheet provided...

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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) ° Jaw-I Mag/3 [173 x, LA']- I Glance-4 'ﬁ O-S‘Ie 04,579 0.05.1. 0.13035 a-oorS 2-04 ygé 0'16“ {2-7.3} 0.02? 0.0013 043")"; 9'33 67% 0’05-6 67-5“; 0'0/7 0.05!!! 9-575“; 33‘3 l‘127 K (9 . , I . ,. l 1 “54.02 -- ,w. Jim»: +4. ”(1.7% «bah-J 14-ij ’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 n-butnoic 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 Sf-WHO ifm‘”) ,5? ( ’- 0.1M >(I0 r“ “‘ ’— 9.461 two”? 2:. w—" 3-3MXZ7/ (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 l7-6C77/ 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 (Pl-P2) 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. 3-1)”)“0’ (iii) Calculate the maximum linear speed that any cylindrical layer of the liquid can travel in the tube. ’1. M-H’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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Key to exam III - CHM 331 Exam III Equation sheet provided...

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