1 mol H O 450 g H O 25.0 mol H O. 2 water g H2O


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1 University Chemistry Quiz 1 015/03/19 1. (10%) Hw many grams f urea[(nh ) CO] must be added t 450 g f water t give a slutin with a vapr pressure.50 mmhg less than that f pure water at 30 C? (The vapr pressure f water at 30 C is 31.8 mmhg.) (urea = g ml 1 ) This prblem is very similar t Prblem 9.4. P x urea P water.50 mmhg xurea(31.8 mmhg) xurea The number f mles f water is: n 1 ml H O 450 g H O 5.0 ml H O water 18.0 g HO x urea n urea n water n urea n urea 5.0 n urea nurea.13 ml g urea mass f urea.13 ml urea 1 ml urea 18 g f urea. (10%) A mixture f liquids A and B exhibits ideal behavir. At 84 C, the ttal vapr pressure f a slutin cntaining 1. mles f A and.3 mles f B is 331 mmhg. Upn the additin f anther mle f B t the slutin, the vapr pressure increases t 347 mmhg. Calculate the vapr pressures f pure A and pure B at 84 C. First find the mle fractins f the slutin cmpnents. We will keep an extra
2 significant figure an then rund at the end. x A 1. ml 1. ml.3 ml x B.3 ml 1. ml.3 ml We can nw use Daltn s law and Raults law t derive the fllwing: P Ttal x A P A x B P B 0.343P A 0.657P B 331 mmhg We d the same calculatins fr after an additin mle f B is added. x A x B 1. ml 1. ml 3.3 ml 3.3 ml 1. ml 3.3 ml P Ttal x A P A x B P B 0.67P A 0.733P B 347 mmhg Nw we have tw equatins and tw unknwns. If we slve fr P A in ur tw equatins and then set them equal t each ther we get the fllwing: 965 mmhg 1.915P B 1300 mmhg.745p B Slving fr P B we get: P B 335 mmhg mmhg When we plug this value f P B int either f the tw equatins that we started with, we get P A 190 mmhg 3. (10%) Estimate the mlar heat f vaprizatin f a liquid whse vapr pressure dubles when the temperature is raised frm 85 C t 95 C. Using Equatin 9.4 f the text: P H 1 vap 1 1 ln P R T T1
3 ln 1 H vap J K 1 ml K K H vap J ml 1 Hvap J ml kj ml 1 4. (10%) Explain why reverse smsis is (theretically) mre desirable as a desalinatin methd than distillatin r freezing. What minimum pressure must be applied t seawater at 5 C in rder fr reverse smsis t ccur? (Treat seawater as a 0.70M NaCl slutin.) Reverse smsis uses high pressure t frce water frm a mre cncentrated slutin t a less cncentrated ne thrugh a semipermeable membrane. Desalinatin by reverse smsis is cnsiderably cheaper than by distillatin and avids the technical difficulties assciated with freezing. T reverse the smtic migratin f water acrss a semipermeable membrane, an external pressure exceeding the smtic pressure must be applied. T find the smtic pressure f 0.70 M NaCl slutin, we must use the van t Hff factr given in Table 9.4 in the text, because NaCl is a strng electrlyte (i = 1.9). The smtic pressure f sea water is: icrt (1.9)(0.70 ml L 1 )( L bar ml 1 K 1 )(98 K) 33 bar T cause reverse smsis a pressure in excess f 33 bar must be applied. 5. (10%) The smsis pressure f 0.010M slutins f CaCl and urea at 5 C are and 0.47 bar, respectively. Calculate the van t Hff factr fr the CaCl slutin. The temperature and mlarity f the tw slutins are the same. If we divide Equatin 9.31 f the text fr ne slutin by the same equatin fr the ther, we can find the rati f the van't Hff factrs in terms f the smtic pressures (i 1 fr urea). CaCl urea icrt crt i bar 0.47 bar.48
4 6. (10%) The slubility f N in bld at 37 C and at a partial pressure f 0.80 bar is ml L 1. A deepsea diver breathes cmpressed air with the partial pressure f N equal t 4.0 bar. Assume that the ttal vlume f bld in the diver s bdy is 5.0 L. Calculate the amunt f N gas released (in liters at 37 C and 1 bar) when the diver returns t the surface f the water where the partial pressure f N is 0.80 bar. Strategy: The given slubility allws us t calculate Henry's law cnstant (k), which can then be used t determine the cncentratin f N at 4.0 bar. We can then cmpare the slubilities f N in bld under nrmal pressure (0.80 bar) and under a greater pressure that a deepsea diver might experience (4.0 bar) t determine the mles f N released when the diver returns t the surface. Frm the mles f N released, we can calculate the vlume f N released. Slutin: First, calculate the Henry's law cnstant, k, using the cncentratin f N in bld at 0.80 bar. k = c P k ml L bar ml L 1 bar 1 Next, we can calculate the cncentratin f N in bld at 4.0 bar using k calculated abve. c kp c ( ml L 1 bar 1 )(4.0 bar) ml L 1 Frm each f the cncentratins f N in bld, we can calculate the number f mles f N disslved by multiplying by the ttal bld vlume f 5.0 L. Then, we can calculate the number f mles f N released when the diver returns t the surface. The number f mles f N in 5.0 L f bld at 0.80 bar is: ( ml L 1 )(5.0 L) ml The number f mles f N in 5.0 L f bld at 4.0 bar is: ( ml L 1 )(5.0 L) ml
5 The amunt f N released in mles when the diver returns t the surface is: ( ml) ( ml) ml Finally, we can nw calculate the vlume f N released using the ideal gas equatin. The ttal pressure pushing n the N that is released is atmspheric pressure (1 atm). The vlume f N released is: V N = nrt P V N ( ml)(73 37)K (1.0 bar) L bar ml1 K 1 1 = 0.8 L 7. (10%) A slutin is prepared by disslving 35.0 g f hemglbin (Hb) in enugh water t make up 1 L in vlume. If the smtic pressure f the slutin is fund t be 10.0 mmhg at 5 C, Calculate the mlar mass f hemglbin. (R = L atm ml 1 K 1 ) Strategy The steps needed t calculate the mlar mass f Hb are similar t thse utlined in example 9.10, except we use smtic pressure instead f freezingpint depressin. First, we must calculate the mlarity f the slutin frm the smtic pressure f the slutin. Then, frm the mlarity, we can determine the number f mles in 35.0 g f Hb and hence its mlar mass. Because the pressure is given in mmhg, it is mre cnvenient t use R in terms f L atm instead f L bar because the cnversin factr frm mmhg t atm is simpler. Slutin The sequence f cnversins is as fllws: smtic pressure mlarity number f mles mlar mass First, calculate the mlarity using Equatin 9.6: Π = crt c = Π RT = 10.0mmHg 1 atm 760 mmhg (0.081 L atm ml 1 K 1 )(98 K) = M
6 The vlume f the slutin is 1L, s it must cntain ml f Hb. We use this quantity t calculate the mlar mass: mlar mass f Hb = grams f Hb mles f Hb = 35.0 g ml = g ml 1 8. (10%) Hw many liters f the antifreeze ethylene glycl [CH (OH)CH (OH)] wuld yu add t a car radiatr cntaining 6.50 L f water if the cldest winter temperature in yur area is 0 C? Calculate the biling pint f this water/ethylene glycl mixture. (The density f ethylene glycl is 1.11 g ml 1.) (Kf = 1.86 C m 1, Kb = 0.5C m, mlar mass f ethylene glycl = 6.07 g ml 1 ) We want a freezing pint depressin f 0C. m T f K f 0C 1.86C m m The mass f ethylene glycl (EG) in 6.5 L r 6.5 kg f water is: 10.8 ml EG 6.07 g EG 3 mass EG 6.50 kg HO g EG 1 kg H O 1 ml EG The vlume f EG needed is: V 3 1 ml EG 1 L ( g EG) 3.93 L 1.11 g EG 1000 ml Finally, we calculate the biling pint: Tb mkb (10.8 m)(0.5c m 1 ) 5.6C The biling pint f the slutin will be 100.0C 5.6C 105.6C. 9. (10%) Slutin A and B have smtic pressures f.4 and 4.6 bar, respectively, at a certain temperature. What is the smtic pressure f a slutin prepared by mixing equal vlumes f A and B at the same temperature? At cnstant temperature, the smtic pressure f a slutin is prprtinal t
7 the mlarity. When equal vlumes f the tw slutins are mixed, the mlarity will just be the mean f the mlarities f the tw slutins (assuming additive vlumes). Since the smtic pressure is prprtinal t the mlarity, the smtic pressure f the slutin will be the mean f the smtic pressure f the tw slutins..4 bar 4.6 bar 3.5 bar 10. (15%) Liquid A (mlar mass = 100 g ml 1 ) and B (mlar mass = 110 g ml 1 ) frm an ideal slutin. At 55 C, A has a vapr pressure f 95 mmhg and B has a vapr pressure f 4 mmhg. A slutin is prepared by mixing equal masses f A and B. (a) Calculate the mle fractin f each cmpnent in the slutin. (b) Calculate the partial pressures f A and B ver the slutin at 55 C. (c) Suppse that sme f the vapr described in part (b) is cndensed t a liquid. Calculate the mle fractin f each cmpnent in this liquid and the vapr pressure f each cmpnent abve this liquid at 55 C. (a) The slutin is prepared by mixing equal masses f A and B. Let's assume that we have 100 grams f each cmpnent. We can cnvert t mles f each substance and then slve fr the mle fractin f each cmpnent. Since the mlar mass f A is 100 g ml 1, we have 1.00 mle f A. mles f B are: The 1 ml B 100 g B ml B 110 g B The mle fractin f A is: x A n A n A n B Since this is a tw cmpnent slutin, the mle fractin f B is: xb (b) We can use Equatin 9.8 f the text and the mle fractins calculated in part (a) t calculate the partial pressures f A and B ver the slutin.
8 P A x A P A (0.54)(95 mmhg) 50 mmhg P B x B P B (0.476)(4 mmhg) 0 mmhg (c) Recall that pressure f a gas is directly prprtinal t mles f gas (P n). The rati f the partial pressures calculated in part (b) is 50 : 0, and therefre the rati f mles will als be 50 : 0. Let's assume that we have 50 mles f A and 0 mles f B. We can slve fr the mle fractin f each cmpnent and then slve fr the vapr pressures using Equatin 9.8 f the text. The mle fractin f A is: A 50? n A n n A B Since this is a tw cmpnent slutin, the mle fractin f B is: xb The vapr pressures f each cmpnent abve the slutin are: P A x A P A (0.71)(95 mmhg) 67 mmhg P B x B P B (0.9)(4 mmhg) 1 mmhg
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