Homework Chapter 14 14.4 Compound Molality Weight percent Mole fraction KNO3 1.10 10.0 0.0194 CH3CO2H 0.0183 0.110 3.30 × 10–4 HOC2H4OH 3.54 18.0 0.0599 KNO3: 10.0 g KNO3 dissolved in 90.0 g H2O 330989.01.10110.10 KNOmolgKNOmolg =⎟⎟⎠⎞⎜⎜⎝⎛ OHmolgOHmolg2299.402.1810.90 =⎟⎟⎠⎞⎜⎜⎝⎛ mOHkgKNOmolsolvenofkgsoluteofamountm 10.10900.00989.023=== XKNO3 = 0.0989 mol0.0989 mol + 4.99 mol = 0.0194 CH3CO2H: 0.0183 mol CH3CO2H dissolved in 1.00 kg H2O HCOCHgHCOCHmolgmol232310.1105.600183.0 =⎟⎟⎠⎞⎜⎜⎝⎛ OHmolgOHmolg2235.5502.1811000.1 =⎟⎟⎠⎞⎜⎜⎝⎛× %110.0100)1000.1(10.110.1%3=⎟⎟⎠⎞⎜⎜⎝⎛×+=gggWeight 41030.35.550183.00183.023−×=⎟⎟⎠⎞⎜⎜⎝⎛+=ΧmolmolmolHCOCH HOCH2CH2OH: 18.0 g HOCH2CH2OH dissolved in 82.0 g H2O OHCHHOCHmolgOHCHHOCHmolg2222290.007.6210.18 =⎟⎟⎠⎞⎜⎜⎝⎛ OHmolgOHmolg2255.402.1810.82 =⎟⎟⎠⎞⎜⎜⎝⎛m = amount of solutekg of solvent = 0.290 mol HOCH2CH2OH0.0820 kg H2O = 3.54 m XHOCH2CH2OH = 0.290 mol0.290 mol + 4.55 mol = 0.0599 14.8 0.125 = x mol HOCH2CH2OHx mol HOCH2CH2OH + 955 g H2O ⋅ 1 mol H2O18.02 g⎛⎝⎜⎞⎠⎟ x = 7.57 mol HOCH2CH2OH 7.57 mol HOCH2CH2OH ⋅ 62.07 g1 mol HOCH2CH2OH = 470. g HOCH2CH2OHm = 7.57 mol HOCH2CH2OH0.955 kg H2O = 7.93 m 14.23 )49.1(1130/1048.40506.0522atmormmHgmmHgMMkSPHCOCO=×==− 14.28 442211.28.1531325414.08.2531105CClmolgCClmolgImolgImolg=⎟⎟⎠⎞⎜⎜⎝⎛=⎟⎟⎠⎞⎜⎜⎝⎛ ()( )mmHgmmHgPXPmolmolmoloCClCClsolutionCCl444531836.0836.0414.011.211.2444====+=Χ14.38 11 g · 1mol C2H5OH46.1 g = 0.24 mol C2H5OH mC2H5OH = 0.24 mol C2H5OH0.089 kg H2O = 2.7 m ΔTfp = (–1.86 ºC/m)(2.7 m) = –5.0 ºC Tfp = –5.0 ºC The solution will begin freeze if it is chilled to –20 ºC. 14.42 ΔTbp = 80.34 ºC – 80.10 ºC = 0.24 ºC 2/89/180/180500.00028.00028.00300.0095.0095.0/53.224.0=========Δ=molgmolgmolgmassformulamassformulagmolmassformulamasssamplemolesmolanthracenemoleskganthracenemolesmbenzeneKganthracenemolesmmmCCKTmanthaceneoobpbpanthrcene The molecular formula is (C7H5)2 or C14H1014.52 Concentration of ions in solution = (0.16 M)(1.9) = 0.30 M Π = cRT = (0.30 mol/L)(0.082057 L·atm/K·mol)(310. K) = 7.7