FIU CHM 3410 - Homework 6 Solution on Physical Chemistry (1) (4 pages)

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Homework 6 Solution on Physical Chemistry (1)



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Homework 6 Solution on Physical Chemistry (1)

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Pages:
4
School:
Florida International University
Course:
Chm 3410 - Physical Chemistry I
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Solutions 6 sh is ar stu ed d vi y re aC s o ou urc rs e eH w er as o co m 1 The volume of the laboratory is V 5 0 5 0 3 0 75 0 m3 a water p 24 Torr n pV RT 24 Torr 75 0 103 L 62 364 L Torr K 1 mol 1 298 15 K 96 855 mol m nM 96 855 mol 18 02 g mol 1 1745 3 g 1 7453 kg b benzene p 98 Torr n pV RT 98 Torr 75 0 103 L 62 364 L Torr K 1 mol 1 298 15 K 395 49 mol m nM 395 49 mol 78 11 g mol 1 30890 8 g 30 89 kg c mercury p 1 7 10 3 Torr n pV RT 1 7 10 3 75 0 103 L 62 364 L Torr K 1 mol 1 298 15 K 0 00686 mol m nM 0 00686 mol 200 59 g mol 1 1 376 g 2 a p1 10 Torr T1 358 95 K and p2 40 Torr T2 392 45 K Use the Clausius Clapeyron equation to calculate vapH p2 vap H 1 1 ln p1 R T1 T2 vap H p3 vap H 1 1 ln p1 R T3 T2 R ln p3 p1 T3 1 1 T1 H vap R ln p2 p1 1 T1 1 T2 vapH 8 3145 J K 1 mol 1 ln 40 10 1 358 95 K 1 392 45 K 48 469 kJ mol 1 b p3 760 Torr T3 Th T3 1 1 358 95 K 8 3145 J K 1 mol 1 ln 760 10 48 469 103 J mol 1 489 47 K 216 32 C c vapS vapH Tb 48 469 103 J mol 1 489 47 K 99 0 J K 1 mol 1 3 fusH 2 292 kJ mol 1 T 234 3 K fusV 0 517 cm3 mol 1 p 101 325 kPa The additional pressure on the bottom of mercury column p p p gh 13 6 103 kg m3 9 81 m s 2 10 m 1334160 Pa https www coursehero com file 11467740 Homework 6 Solution on Physical Chemistry fus H T p p ln fusV T p p fusV T ln T fus H ln T T 1334160 Pa 0 517 10 6 m3 mol 1 2 292 103 J mol 1 3 0 10 4 T 234 3 K exp 3 0 10 4 234 37 K 4 We assume that a liter of seawater contains roughly 1000 g of water Then nwater 1000 g 18 02 g mol 1 55 5 mol nsolutes 2 0 50 mol 1 00 mol xwater nwater nwater nsolutes 55 5 56 5 0 982 pwater xwater pwater 0 982 2 338 kPa 2 296 kPa sh is ar stu ed d vi y re aC s o ou urc rs e eH w er as o co m 5 Check whether pB xB is equal at least approximately to a constant KB x 0 005 0 012 0 019 3 3 p x 6 4 10 6 408 10 6 410 103 Hence KB 6 4 103 kPa 6 We plot h c against c c 3 221 4 618 5 112 6 722 h 5 746 8 238 9 119 11 99 h c 1 7839181 7838891 7838421 783695 1 784 1 78395 1 7839 1 78385 1 7838 1 78375 1 7837 1 78365 y 7E 05x 1 7842 R2 0 9039 2 4 6 8 Th 0 The intercept is 1 7842 10 2 m4 kg 1 Intercept RT gM M RT g Intercept 1 1 M 8 31451 J K mol 293 15 K 1000 kg m 3 9 807 m s 2 1 7842 10 2 m4 kg 1 13 93 kg mol 1 13 93 kDa 7 Ebullioscopic constant T KxB RT 2 K vap H https www coursehero com file 11467740 Homework 6 Solution on Physical Chemistry T K bb nB n n mCCl4 xB B B bM CCl4 nB nCCl4 nCCl4 mCCl4 nCCl4 RT 2 M CCl4 Therefore K b KM CCl4 vap H Kb 8 3145 J K 1 mol 1 349 9 K 2 153 81 10 3 kg mol 1 30 0 103 J mol 1 5 22 K kg mol 1 Kf Cryoscopic constant fus H sh is ar stu ed d vi y re aC s o ou urc rs e eH w er as o co m RT 2 M CCl4 1 Kf 8 3145 J K mol 250 3 K 153 81 10 3 kg mol 1 2 47 103 J mol 1 32 4 K kg mol 1 8 1 2 According to Raoult s law pA xA pA Therefore we can calculate mole fraction of benzene xA pA pA 386 Torr 400 Torr 0 965 nA mA M A mA n A nB m A M A m B M B m A M A m B M B xAM A MB mB 1 x A m A xA 9 y N2 0 78 p 760 Torr According to Dalton s law the partial pressure of N2 is calculated as p N2 0 78 760 Torr 592 8 Torr Henry s law p N2 K N2 x N2 x N2 p N2 K N2 7 6 x N2 592 8 Torr 6 51 10 Torr 9 1 10 b N2 x N2 M H2O see problem 4 for the relation between mole fraction and molality b N2 9 1 10 6 18 02 10 3 kg mol 1 5 05 10 4 mol kg 1 0 51 mmol kg 1 y O2 0 21 p 760 Torr p O2 0 21 760 Torr 159 6 Torr p O2 K O2 x O2 x O2 p O2 K O2 7 x O2 159 6 Torr 3 30 10 Torr 4 8 10 6 b O2 x O2 M H2O b O2 4 8 10 6 18 02 10 3 kg mol 1 2 68 10 4 mol kg 1 0 27 mmol kg 1 Th MB 0 965 78 11 g mol 1 19 g 1 0 965 500 g 82 g mol 1 10 Kf water 1 86 K kg mol 1 https www coursehero com file 11467740 Homework 6 Solution on Physical Chemistry Th sh is ar stu ed d vi y re aC s o ou urc rs e eH w er as o co m In 250 cm3 of water approximately 250 g m water 250 g 1 M sucrose M C12H22O11 342 3 g mol n sucrose m sucrose M sucrose 7 5 g 342 3 g mol 1 0 0219 mol b n sucrose m water 0 0219 mol 0 25 kg 0 0876 mol kg 1 T Kfb 1 86 K kg mol 1 0 0876 mol kg 1 0 16 K The freezing temperature decreases by 0 16 K or 0 16 C https www coursehero com file 11467740 Homework 6 Solution on Physical Chemistry Powered by …


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