58.08 g 1 mol. 1 mol

Transcription

1 Chem 338 Homework Set #6 solutios October 17, 001 From Atkis: 6.8, 6.1, 6.14, 6.16, 6.17, 6.19, ) The partial molar volumes of propaoe ad trichloromethae i a mixture i which the mole fractio of CHCl 3 is 4693 are ad 835 cm 3 mol -1, respectively. What is the volume of a solutio of total mass kg? Like may problems i Physical Chemistry, there are a couple of ways to tackle this oe. I admit to doig it the hard way the first time aroud. Here s the more straightforward method: First we fid the molecular weight of the solutio, covert the 1000g to moles to obtai the total umber of moles of solutio, ad the use the mole fractios ad partial molar volumes to obtai the total volume. For of solutio, there are 4693 moles of CHCl 3 ad (1 4693)=5307 moles of propaoe. The molecular weights of CHCl 3 ad propaoe are ad g/mol, respectively, so that the molecular weight of the solutio is: g g 4693 mol mol = g / mol 1000 g of solutio correspods to 1000 g = mol of solutio g The, (ote, c = CHCl 3, p = propaoe) V = cvc + pvp = x c totalvc + x p totalvp = ( 4693)( )( 8 35) + ( 5307)( )( ) = 887 cm 3

2 6.1) Calculate (a) the (molar) Gibbs eergy of mixig, (b) the (molar) etropy of mixig whe the two major compoets of air (itroge ad oxyge) are mixed to form air at 98 K. The mole fractios of N ad O are 78 ad, respectively. Is the mixig spotaeous? a) For oe mole, ( ) Gmix = RT xa l xa + xb l xb = ( )( 98)( 78 l 78 + l ) = J = 13. kj / mol ; spotaeous b) For oe mole, ( ) Smix = R xa l xa + xb l xb G = mix T = 1305 = +4.4 J / mol 98 ; spotaeous 6.14) A solutio is prepared by dissolvig 1.3 g of C 60 (buckmisterfulleree) i 100 g of toluee (methylbezee). Give that the vapor pressure of pure toluee is 5.00 kpa at 30ºC, what is the vapor pressure of toluee over the solutio? 13. g C g = mol 100 g toluee = mol 9.14 g total = , so that x(c60) = x 10-3, x(toluee) = Usig Raoult s law, p toluee = x(toluee) p* toluee p toluee = (99843)(5.00) = 4.99 kpa

3 6.16) At 300 K, the vapor pressure of dilute solutios of HCl i liquid GeCl4 as as follows: x(hcl) p/kpa Show that the solutio obeys Hery s law i this rage of mole fractios ad calculate the Hery s law costat at 300 K. Hery s law is p B = x B K B, where B = HCl i this case. If this solutio obeys Hery s law i this mole fractio rage, a plot of mole fractio versus vapor pressure should yield a straight lie with a slope give by the Hery s law costat. A more accurate method is to fit the data to a quadratic polyomial i x ad evaluate the first derivative at x=0 (this yields the taget to the curve). Both methods are show i the plot below ad yield essetially the same result. From the liear fit, oe obtais K=6411 kpa or 6.4 MPa, while the 1 st derivative of the quadratic yields 640 kpa or also 6.4 MPa. Certaily Hery s law is obeyed for these cocetratio rages. 140 y = x R= Y = M0 + M1*x +... M8*x 8 + M9*x 9 M M1 64 M R

4 A eve simpler test i this case would be to calculate the ratios of p B /x B for each set of data poits ad determie if the resultig K B are cosistet with each other. I this case all three K B s calculated i this way yield 6.4 MPa. Note that this is the least accurate way to fid a Hery s law costat. 6.17) Calculate the cocetratio of carbo dioxide i fat give that the Hery s law costat is 8.6 x 10 4 Torr ad the partial pressure of carbo dioxide is 55 kpa. 1 Torr p = 55 kpa = Torr 1333 kpa From Hery s law, x = p K = = ) The mole fractios of N ad O i air at sea level are approximately 78 ad 1. Calculate the molalities of the solutio formed i a ope flask of water at 5ºC. For a total pressure of 760 Torr, p N = (78)(760) = 59.8 Torr p O = (1)(760) = Torr From Hery s law (H.L. costats from the text): xn xo pn = = = K 7 N po = = = K 7 O The defiitio of molality: For 1kg of water, xa ma = A kg HO = A A 1000 g 1000 g A g / mol 18.0 g / mol

5 For this case, A is the umber of moles of solute i 1 kg of solvet, which is the 1000 g defiitio of molality: ma = xa 18.0 g / mol mn mo 6 4 = ( )( 1000 / 18. 0) = moles / kg 6 4 = ( )( 1000 / 18. 0) = moles / kg Note that the approximatio we made above is valid sice the umber of moles of N ad O are both much smaller tha the umber of moles of water. 6.) The vapor pressure of a sample of bezee is 400 Torr at 66ºC, but it fell to 386 Torr whe 15 g of a orgaic compoud was dissolved i 5.00 g of the solvet. Calculate the molar mass of the compoud. For A = bezee ad usig Raoult s law, xa = pa = 386 * =. P A From the defiitio of the mole fractio, xa = A = A + x MWx = MWx Solvig for the MW of the ukow, MW x = g/mol