Exam 1 2006 key - CHM 331 Exam I February 25, 2006 Time: 50...

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Unformatted text preview: CHM 331 Exam I February 25, 2006 Time: 50 min Dr. Jumean Equation sheet provided All questions carry equal credit F: 96500 Coulomb R: 3.314 I K“ mol" 1. The conductivity (K) of a saturated solution of Ca(103)2 at 250°C is 1.25 x 104 S cm'1 and that of the water used is 1.4 x 10'5 S 0111'}. The molar conductivities at 250°C for the Ca+2 and 103’ ions are 118.9 and 27.1 S cm2 mol'l. Use DHLL to obtain the K5]: for Ca(103)2 at 25°C. t1 #8“ f a _...- f.£4)!)0 Kaéj/(LL , /.«7/$’_){Ii\r (I 2: [4{}fl\0 gm [(065%)] :2 //$’-€ 4— 6223'”: f4?! 2. The solubility of the salt MX3 (where M is a metal ion) in water is 000200 mol dm'3 at 25°C and 0.00250 mol drn'3 at 35°C. For this solution the mean ionic activity coefficient (at both temperatures), Vi, is related to the ionic strength by the expression log n = — 0.740 + 2.45 I (a) Evaluate K51, at each of the temperatures (25°C and at 35°C). ,3 . _ (f ti K539 : “S (SS) 2:? f: (“458(j/w __ I. 0 ,__ _' '3 I} fo-eot \L’SL‘ Gama“ 0 ol ., ’ .. + — M272 I It - - ( \ "a. 0 /*f 1'... Diff \q R igfjtp (LIL/MO '00 5g 0— F K” H “M: f #r—I—Jirf—O'1HW 3““er 0-0-03}?xe : v 9% 3?“; if“ ’3: gaufrzgfuumf j :- I U 3 r'La- 7/; f 001% Y ’ézwo '1: i r 16? " 2'3 iCo-W'I TO)L{(O,¢-'qg F (b) Calculate AH for the process 5 ZS MX3 (s) = M+3 + 3 X' m ., t i A 6L ’50 199:: :2 " gig/2T2, ‘ - (2, ,L I AL ’— b / 201% /@/h i611“) g;- / 0% flaw) fir 3. Use the single ion form of DHLL to estimate the potential of the cell at 250°C Ptl Mn(s) | MnClg (2.00 m) | 1 EC] (0.0100 m) 102 (1.00 atm) |Pt Given: Mn(s) = Mn+2 (aq) + 2e 13“ = 1.185 v 2H20(1) = 02(g) + 411* (aq) + 46 E" = 1229 v (Note that the anode and cathode solutions are separated by a salt bridge) 0 0 AM + I? L’ H ‘- [tr (7 4014*; 2,141,194 ZN“ “0?”? 2 flfh + L [0.011001% (2' 00 fl ’2.— o ( am” E???” awfifif‘ bf ..—-——-———‘ L\ I 61/. Ly. Quin/(4. mam U 7" , em“ 4. At 25°C, E‘J = 0.4085 V for the cell an 211304011), PbSO4 (s) l Pb When the cell contains 0.0005 11101 dm'3 zinc sulfate it has E= 0.61 14 V. a. From the above data, calculate the mean ionic activity coefficient of zinc sulfate when m = 0.0005 mole Kg}. '1 AVHHC A4»: 2740‘ —L‘;? 27/71 L 1 e k_ 50 4." W’LM (“9)le (§'_}’C'Ld A t, u H} 1-, fig r-(Jflififlzflur («(304 1 -L_. I! J fill-XI? 061/91” 0 Hoff,“ ra- 4‘Lé-occsv L h. Calculate the same mean ionic activity coefficient using DHLL. Comment on the difference. 5. (20 points) A solution of silver nitrate was contained in a Hittorf cell between two silver electrodes. A current of 0.020 ampere was passed through the cell for 2.00 hours. (Given: the transport number of silver ion (t+) in this case is 0.47 Atomic mass of Ag = 108) 3) Calculate the change in mass of each of the silver electrodes. a) Annfl Ayn Aj-Ea A5** fl mm W a a .44 r £- We!» bitch/ll; _"_ if" .F 001% ZIiGQYGU b) Calculate the change in mass of silver in the solution surrounding the anode‘ “ M f4, 1' jgfif t; . 4’ [m [sjprayfiepj <63 (jyt‘fixa, AI I’M—c .a f flffizr'v’ r slit ‘ __ 5 “fl! flyw-M*wwflfl 0- G '3 3 a” "Mfflr-j—Ef rarvw ,3 C!_¥003{l "a; A («e- -/£r'u. WT Aflflkpvfialj {- Girl-lifting 10% Z: 0.0.7. '5 b _ ...
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This note was uploaded on 12/17/2011 for the course CHEMICAL E 304 taught by Professor Darwish during the Spring '11 term at American University of Sharjah.

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Exam 1 2006 key - CHM 331 Exam I February 25, 2006 Time: 50...

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