Solutions_Ch16

Solutions_Ch16 - MW 18. As the strengths of the...

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Unformatted text preview: MW 18. As the strengths of the interparticle forces increase, boiling point, freezing point, viscosity, 4H,,,,, and OHM, increase while vapor pressure decreases. ’ CHAPTER 16 LIQUIDS AND SOL.le a. HBr; HBr is polar, whereas Kr and Cl; are nonpolar. HBr has dipole forces unlike Kr and Cig. b. NaCl; ionic forces are much stronger-than the intermolecular forces for molecular solids. . _ _____ _ ..__ _ _.._fi,___H c. 12; All are nonpolar, so the largest molecule (1;) will have the strongest LD forces and the lowest vapor pressure. (1. N2; nonpolar and smallest, so it has weakest intermolecular forces. e. H202; HMH structure produces stronger l—l-bonding interactions than [-1— F, so H302 has the greatest viscosity. f. CH3CH20H; can form H—bonding interactions unlike the other covalent compounds. g. 12; [2 has only LD forces, whereas CsBr and CaO have much stronger ionic forces. 12 has the weakest intermolecular forces, so it has the smallest HMO“. 22. NaCl, MgCIZ, NaF, Mng, and AlF; all have very high melting points indicative of strong intermolecular forces. They are all ionic solids. SiCl4, SiF4, F2, C12, PPS, and SF, are nonpolar covalent molecules. Only LD forces are present. PC13 and SC]; are polar molecules. LD forces and dipole forces are present. In these eight molecular substances the intermolecular forces are weak and the melting points low. AlCl; doesn't seem to fit in as Well. From the melting point, there are much stronger forces present than in the nonmetal halides, but they aren't as strong as we would expect for an ionic solid. AlCl; illustrates a gradual transition from ionic to covalent bonding, from an ionic solid to discrete molecules. 32. C02 is a gas at room temperature. As melting point and boiling point increase, the strength of the intermolecular forces also increases. Therefore, the strength of forces is C02 < C52 < C862. From a structural standpoint, this is expected. All three are linear, nonpolar molecules. Thus only London dispersion forces are present. Because the molecules increase in size from C02 < C82 < C862, th e strength of the intermolecular forces will increase in the same order. H 84. mi :4 “'1’ i—i P, R T, T, P1 = 760. torr, T; = 565°C + 273.2 = 329.7 K; P: = 630. torr, T2 = ? 3 In fl = 321W“) Elm], i- 1 0.188 23.85 x103(i—3.033 no“) 630. 8.314SJK mol T, 329.7K T, Tifisms x 10‘3 = 4.88 x 10‘5, 1—= 3.082 x 10‘“, T2 = 324.5 K = 513°C 2 2 3 In 630. torr : 32.0 X10 I.II’Inoll ] __ l 3 111630. __In P2 =1’05 P2 8.3145 J K— mol' 298.2 K 324. 5 in P2 = 5.40, P, = as“ x 221 torr 7 90. H20(g, 125°C)—-> H20(g, 100°C), q, : 2.0 J “C ' grl x 75.0 g x (-25°C) =-—3800 J = "-3.8 k] we 1009C) amen, 100.0(2), q2 = 75.0 g x “‘1‘” x M =-169 k1 18.02 g mol H200, 3009a).; H200, 0°C), q3 = 4.2 J °c ‘ g 1 x 75.0 g x («-100.°C)= -32,000 J =—32 id To convert H20(g) at 125°C to H200) at 0°C requires 6-3.8 k1 - 169 kJ .- 32 kJ =) -205 k] of heat removed. To convert from H200) at 0°C to H2005) at 0°C requires: lmol x —6.01kJ =__25.0 kJ 18.02g mol (l4 g X This amount of energy puts us over the --215 kJ limit (-205 kJ -—25.0 kJ =-230. k3). Therefore, a mixture of 1120(5) and H200) wili be present at 0°C when 215 Id of heat are removed from the gas sample. ...
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Solutions_Ch16 - MW 18. As the strengths of the...

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