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2. (M) Substituting Cl for H makes the molecule heavier (and thus increases London forces) and polar, which results in the formation of dipole—dipole interactions. Both of these l
effects make it more difﬁcult to disrupt the forces of attraction between molecules, ll
increasing the boiling point. Substitution of Br for C1 increases the London forces, but makes the molecule less polar. Since London forces in this case are more important than
dipole—dipole interactions, the boiling point increases yet again. Finally, substituting OH for Br decreases London forces but both increases the dipoledipole interactions and creates opportunities for hydrogen bonding. Since hydrogen bonds are much stronger than
London forces, the boiling point increases even further. ' ‘1 12. (E) Both surface tension and viscosity deal with the work needed to overcome the attractions
between molecules. Increasing the temperature of a liquid sample causes the molecules to move
faster. Some of the work has been done by adding thermal energy (heat) and less work needs to
be done by the experimenter, consequently, both surface tension and viscosity decrease. The vapor “
pressure is a measure of the concentration of molecules that have broken free of the surface. As ' thermal energy is added to the liquid sample, more and more molecules have enough energy to
break free of the surface, and the vapor pressure increases. PV 1.00 atmxl.17L RT 0.08206 L atm moi—1 K” x (273.2 +816) K 1.00 k]  
AH,”p = m = 24.9 kl / mol acetomtrlle 22. = 0.0402 mol acetonitrile ( ) nacemm‘rrii'e 36. (M) The ﬁnal pressure must be 742 mm Hg and the partial pressure of CCl4(g) is constant at '
261 mm Hg, thus, the partial pressure of N2 must be 481 mm Hg. Using Dalton’s law of
partial pressures, PNZ/Pmm. = Vlethal, the total and ﬁnal pressures are known. Since the temperature does not change, the volume occupied by N2 is constant at 7.53 L. I Hence, V”, = VN x PM! = 7.53 LxE—g = 11.6 L.
2 PN2 481 mmHg 42. (M) P1 % 40.0 Torr T1 : 7.1 °C (266 K) and AHM, = 29.2 kJ met1 3
P2 = TOIT T2 = ‘ l
rum0) gm) __1 4) T.:3423Km~69 °c 40.0 83145, J___ 266K I; K mo]
(The literature boiling point for trichloromethane is 61 0C.) i it r 3; (E) Substances that can exist as a liquid at room temperature (about 20” C) have critical
temperature above 20° C, 293 K. Of the substances listed in Table 12.5,
C020; = 304.2 K), Hem; = 324.6 K), NH,(2; = 405.7 K), so,(2; : 431.0 K), and = 647.3 K) can exist in liquid form at 20 0C. In fact, CO2 exists as a liquid in
CO2 ﬁre extinguishers. z; 58. (M) The heat gained by the ice equals the negative of the heat lost by the water. Let us use this fact i.
in a step—bystep approach. W 0.0 °C and then the heat given off when the temperature of the water is lowered to 0.0 0 C. Mass of ice = 54 g;
(M) (M) (a) 335 pm; 2 radii or 1 diameter. Hence Po diameter = 335 pm
(b) 1 Po unit cell 2 (335 pm)3 = 3.76 x 107 pm3 (3.76 x 10‘23 cm3) per unit cell. (C) to heat the ice = 54 cm3 x 3 to cool the water = 4100.0 cm3 x (a) (b) density n = 1, d = 335 pm and 7t =1.785 ><10'10 or 178.5 pm sin 6 l
c ﬁrst compute the heat needed to raise the temperature of the ice to l 0.917 g X201 J g7] oC—l oc+25_0 0C):2.5X103J 1 cm f. 0.9983g X418 J g3 oC—i X“, «ac—32.0 DC) : 53.4x103 J lcm 1 .9 7 I .
cm3><0 1 E’ . 49.5 g = 50. g ice 1 1 cm“ We expect Si and C atoms to alternate in the
structure, as shown at the right. The C atoms are on
the corners (8 x l / 8 = l C atom) and on the faces
(6 x 1/2 : 3 C atoms), a total of four C atoms/unit
cell. The Si atoms are each totally within the cell, a
total of four Si atoms/unit cell. To have a graphite structure, we expect Sp2 hybridization
for each atom. The hybridization schemes for B and N
atoms are shown to the right. The half—filled 3,02 hybrid orbitals of the boron and nitrogen atoms overlap to form
the 0' bonding structure, and a hexagonal array of atoms. The 2 pl orbitals then overlap to form the Jr bonding orbitals. Thus, there will be as many a" electrons in a
sample of BN as there are in a sample of graphite,
assuming both samples have the same number of atoms. B [H6] 5;): N [He] spz 2 Km V 3.47x10'23g
3.76x10"“ cm3 = 9.23 g cm'3 Solve for sin 6, then determine 8. hi m (l)(l.785 X 10"”) _ —_ 2(335x10'7”) 20'266ﬂ or 6:114? 2d 76. (M) Ge unit cell has a length of 565 pm. volume 2 (length )3 : (565 pm)3 = 1.80 x 108 pm3 3 3
1 10'12 100 t _ '
Next we convert to cm3: 1.80 x 108 pm?” = 1.804 x 10 22 cm3 1 pm 1 m
Now we determine the number of Ge atoms per unit cell.
Method: Find number of atoms of Ge per cm3, then using the unit cell volume
(calculation above), determine number of Ge atoms in the unit cell. atomsGe [5.36gGeI lmolGe 16.022x1023lz4445 ><10natomsGe cm3 2 cm3 72.61 g Ge lmol Ge cm3 atoms Ge 022 atoms Ge ] I 8.02 atoms Ge _ =(1.804><10‘22cm3) 4.445><1 —3—
umtcel] cm There are 8 atoms of Ge per unit cell. Ge must adopt a face centered cubic structure
with the 4 tetrahedral holes filled (diamond structure). unit cell 82. (M) According to Figure 1248, there are four formula units of KCl in a unit cell and the length
of the edge of that unit cell is twice the internuclear distance between K+ and‘Cl‘ ions. The
unit cell is a cube; its volume is the cube of its length. length = 2 x 314.54 pm = 629.08 pm 1m 100 cm
X 1012 pm 1m 3
volume = (629.08 pm)3 = 2.4895 x108 pm3[ J = 2.4895 ><10‘22 cm3 unit cell mass = volume x density = 2.4895 x 10‘22 cm3 x 1.9893 g'/ cm3 = 4.9524 x 10—22
= mass of 4 KCl formula units
_ 4 KCl formula un1t 74.5513 g KC] : 6.0214X1023 formula unit/mol NA ————*3§m)<
4.9524x10 gKCI 1molK_C1 86. (D) The cycle of reactions is shown. Recall that Hess’s law (a state function), states that the
enthalpy change. is the same, whether a chemical change is produced by one reaction or several. { K“) + “2 F2(g) formation KHS) ' sublimation ldissociaﬁon K(s) t 17(3)
ionizationl lelectron afﬁnity I Ktrg) + F'(g)n.+,__m/ Formation reaction: K(s)+ % F2 (g)—) KF(s) AH ” =—567.3 kJ /mol  if latice energy f
Sublimation: a K(g) Mm = 89.24 kJ / mol
_ Ionization: —~> K+( g)+ e" 11 = 418.9 kJ/mol
Dissociation: F2 (g) H) F(g) ' D.E.= (159/2) kJ/mol F
Electron'afﬁnity: F(g) + e' —> FA (g) E.A.= —328 k] lmol
AHID = AHM +11 + E.A.+ lattice energy (LE) . ' —567.3 kJ/mol = 89.24 kJ/mol+418.9 kJ/mol+[159/2) kJ/mol—328 kJ/mo]+ LE.
L.E.= —827 kJ / mol 108. (M) (a) The coordination number for 82' is 8, while for Lilr it is 4 (tetrahedral holes). (1)) 8 X Li+ within the unit cell(tetrahedral holes) and 4 X 82" (8 corners X 1/8) + (6 faces X % ) =
1 + 3 = 4 formula units per unit cell. ‘ Isa 28. (M) In the original solution, the mass of CH3OH associated with each 1.00 kg or 1000 g of  with water, both molecules contain reasonably large nonpolar portions, which will not interact (1x10E2 my (lOOcm)3 , ><———=1><10'3Gcm21
(1pm)3 (1171}3 1><10'30 cm3
3 (c) Note: 1 me = V: E3 = (5.88 x 102 pm)3 x = 2.03; x 10‘22 (31113
1pm
In a unit cell there are 4 L128 formula units. 1 (L128 —> 45.948 g mol'1 or 4 mol L128 —> 183.792 g) 1:.
Consider one mole of unit cells. This contains 45.948 g of Lizs and has a volume of (2.03 x
10'22 cm3) x (6.022 x1023) = 122.4 cm3 mass 1183.792gzl'50gcm3 ii:
122.4cm~ ' volume (E) (c) Formic acid and (f) propylene glycol are soluble in water. They both can form
hydrogen bonds with water, and they both have small nonpolar portions. (b) Benzoic acid and
(d) lButanol are only slightly soluble in water. Although they both can form hydrogen bonds strongly with water. (a) Iodoform and (e) chlorobenzene are insoluble in water. Although both
molecules have polar groups, their inﬂuence is too small to enable the molecules to disrupt
the hydrogen bonds in water and form a homogeneous liquid mixture. 20. (E) $311555
H3P04 molarity= u m: g iL 4 = 12M
100.0 g soln x x 1.57 g 1000 mi water is: 1.38 mol CH3OH x 32.04 g CH3OH
1 kg H20 1 mol CH3OH
Then we compute the mass of methanol in 1.0000 kg = 1000.0 g of the original solution. mass 011,011 =1000.0 g solnx W
1044.2 g soln mass CH3OH ='1.00 kg H20 x = 44.2 g_CH3OH = 42.3 g (311,011 Thus, 1.0000 kg of this original solution contains (1000.0 —42.3 x) 957.7 g H20) Now, a 1.00 m CH3OH solution contains 32.04 g CH 3011 {one mole) for every 1000.0 g
H20. We can compute the mass of H20 associated with 42.3 g CH3OH in such a solution. 1000.0 g H20 mass H20=42.3 g CH3OH. x ——
32.04 g CHaOH =1320 g H20 (We have temporarily retained an extra signiﬁcant ﬁgure in the calcuIation.) Since we already
have 957.7 g H20 in the original solution, the mass of water that must be added equals 1320 g—957.7 g=362 gH,o=3.6x102 gH,0. 32. 46. 1;. (M) _
1 mol (32115011 ': _ 1 jHOHx uﬁ—:0.471molCHOH
(3) moles ofCZHSOH 21 7 g, (2 5 46.07 g CZHfFOH 2 5 t
0 471 mol C H OH
lmolH O _ :;_#_u_=0.0977
[11110 2 783 g Hzox $8.03 g M 4'35 m0] H20 Z“‘h*‘”‘°' (0471+ 4.35) total moles
0.684 lurea
(b) 11,, O = 1000. g X JWmOI H204 : 55.49 mol H20 2 m0 = 0.0122 18.02 g H20 (55.49+0.684) total moles (M) We ﬁrst compute the amount of O2 dissolved in 515 mL : 0.515 L at each temperature. e73
2.18M?! mol 02 :1_12x10—3 mol 02 at 0 °C amount 02 = 0.515 L x ﬁll—“191$ = 6.49 x 10*1 mol 02 at 25 "c amount 02 = 0.515 L x gas expelled. 0.08206 L atm 11.2 ‘— 6.49)><10“‘ molO *4 x 298 K
2 VO = “101 1K =1.2><102 L :12 mL expelled
_ 2 1 atm
g (M) First we determine the number of moles of each component, its mole fraction in the
solution, the partial pressure due to that component above the solution, and ﬁnally the total
pressure. 7
amount benzene = ab = 35.8 g C6116 x M = 0.458 mol C6H6
78.11 g CGH,5
amount toluene = n, = 56.7 g CTHB x M = 0.615 mol C7H3
, 1 92.14 g C7H8 .
. . ‘ 1
2b : 0458 mol CéH6 _0.427 It :_ 0615 mo C7H8 =0_573 (0.458 + 0.615) 101211111312; — ( 0.458 + 0.615) total moles PL = 0.427x 95.1 mian = 40.6 mmHg R = 0.573x 28.4 1111’an = 16.3 1111an
total pressure T406 man + 16.3 mmHg "—f 56.9 1111an ._‘_. 66. (M) We need the molar concentration of ions in the solution to determine its osmotic
pressure. We assume that the solution has a density of 1.00 g/mL. [ions]: ' ' g ‘‘ L 1:10 3 =0.32 M
100.0 mL soln x "—439—
1000 mL
_ 12 mol ,1 _1
‘ a" — FRT = 0.32—L—>< 0.08206 L atm mol K x (37.0 +2732) K = 8.1 atm
(M) (a) First we determine the molality of the solution, then the value of the freezing—point
depression constant.
1 mol (36116 m: ‘ (1.—f’—=O,]60m Kf 24:44:20oC/m
80.00 g solvent x uﬁEr ﬁm —0.I60m
1000 g (b) For benzene, K. = 5.12 “Cm—I. Cyclohexane is the better solvent for freezing—point
depression determinations of molar mass, because a less concentrated solution will still
give a substantial freezing—point depression. For the same concentration, cyclohexane
solutions will show a freezing—point depression approximately four times that of
benzene. Also, one should steer clear of benzene because it is a known carcinogen. EB 0) “ Co Od'oms um um} dell 4Q l \w') * \Cd'ﬁca «Poms O
a) w ‘
;, I C05 (30" :1: = E:
3r  '2‘, 2.
r
m >‘ " «Fe r Ania WP 413? 1de =(.:2.r)(ﬁr\ I 243 r2 VO‘UHR 0? Ce” 3 r%)< r) :l f3 d) ‘1 CO Oxth lCﬁH 2,; (4/gvrn‘57 : r _ \Jm od'omS o _ 3/31Tf5
PGC'VUﬂs cixf‘g‘iClCnC‘j ' Vbi can l i100‘0 ‘1 "LUCY5 3 .
en mew} : mos/cw * “2‘an “‘— 2 5.55%8 u/ﬁd mass 0? WM cw: moiecww Hunks) 57.: 1(6IBNIB3’ZUL 3 1 \\’+O‘ofo4ua Venome eeu ' 85—7: <3 “10°94”: 5);)5q58 u/ﬂs :5 V3 3 L061? —‘> i<= L145? i
85 r‘=
. ,, I; \
at) Wm / cg xnU’L/m— “ti“ﬁg ~71; )
m... ‘anu%»/1 I \
I 1 2.0 .; 3??\<Smd A; 1
SOL“) 'l /’/ 1 HA 8.314003 \CS/moW. 3200p r cans l
'l t T1 = 4058K
‘Tﬂkh hail > ...
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This note was uploaded on 06/10/2011 for the course CHEM 2090 taught by Professor Zax,d during the Spring '07 term at Cornell University (Engineering School).
 Spring '07
 ZAX,D
 Chemistry

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