Chem 116 POGIL Wrksheet - Week 3 Intermlecular Frces, Liquids, Slids, and Slutins Why? Mst substances can exist in either gas, liquid, r slid phase under apprpriate cnditins f temperature and pressure. The phase that we see under rdinary cnditins (rm temperature and nrmal atmspheric pressure) is a result f the frces f attractin between mlecules r ins cmprising the substance. The strength f these attractins als determines what changes in temperature and pressure are needed t effect a phase transitin. The behavir f a pure substance in any f its phases is altered when it is mixed with ther substances t make slutins. Slutins are hmgeneus mixtures, which can ccur in any phase. But mst ften in chemistry we are dealing with slutins that are in the liquid phase. An understanding f the prcesses by which slutins frm and f hw their prperties differ frm their pure-substance cmpnents is useful in many real-life applicatins f materials. Learning Objectives Knw the kinds and relative strengths f intermlecular attractive interactins Understand the cncept f plarizability Understand the effects f strengths f intermlecular frces n phase changes and substance prperties such as viscsity and surface tensin Understand the cncepts f critical pressure, critical temperature, vapr pressure, nrmal biling pint, nrmal melting pint, critical pint, and triple pint Understand the prcesses by which slutins frm Understand the rles f enthalpy and entrpy in determining slubility Understand the slubility f nn-reactive gases Understand the effect f slute cncentratin n clligative prperties Success Criteria Be able t judge relative strengths f intermlecular frces f attractin and their effects n prperties fr a series f cmpunds Be able t interpret heating curves and calculate quantities related t temperature and enthalpies f phase changes Be able t interpret and sketch phase diagrams Be able t identify the frces respnsible fr slubility in particular slutins Be able t predict slubility, based n the intermlecular frces f the cmpnents Be able t d Henry s law calculatins f gas slubility Be able t calculate mlality Be able t calculate the magnitude f clligative changes in vapr pressure, freezing pint, biling pint, and smtic pressure Prerequisite Have read Chapters 11 and 13
Infrmatin (Intermlecular Frces) The tendency f a substance t be fund in ne state r the ther under certain cnditins is largely a result f the kinds frces f attractin that exist between the particles cmprising it. We will cncentrate n the frces between mlecules in mlecular substances, which are called intermlecular frces. (Frces that exist within mlecules, such as chemical bnds, are called intramlecular frces.) The greater the strength f the intermlecular frces, the mre likely the substance is t be fund in a cndensed state; i.e., either a liquid r slid. As we have seen, the mdel f an ideal gas assumes that the gas particles (mlecules r atms) have virtually n frces f attractin between them, are widely separated, and are cnstantly mving with high velcity and kinetic energy. In truth, there are frces f attractin between the particles, but in a gas the kinetic energy is s high that these cannt effectively bring the particles tgether. With strnger intermlecular frces r lwer kinetic energy, thse frces may draw mlecules clser tgether, resulting in a cndensed phase. Ging frm gas t liquid t slid, mlecular velcities and particle separatins diminish prgressively as structural rder increases. In the case f liquids, mlecular attractins give rise t viscsity, a resistance t flw. Als, the absence f intermlecular frces abve the surface f a liquid results in surface tensin, the develpment f a skin n the surface, which causes beading f liquid drplets and als allws light bjects t rest n a liquid surface withut sinking (e.g., water bugs). Slids have strnger intermlecular frces, making them rigid, with essentially n tendency t flw. Althugh the mix f types and strengths f intermlecular frces determines the state f a substance under certain cnditins, in general mst substances can be fund in any f the three states under apprpriate cnditins f temperature and pressure. Changing thse cnditins can induce a change in the state f the substance, called a phase transitin. Key Questins 1. Is the average kinetic energy f mlecules greater r lesser than the energy f intermlecular frces f attractin in (a) slids, (b) liquids, and (c) gases? 2. Why des increasing the temperature cause a substance t change in successin frm a slid t a liquid t a gas? 3. Why d substances with high surface tensin als tend t have high viscsities? 4. Why d surface tensin and viscsity decrease with increasing temperature? Infrmatin (Kinds f Intermlecular Frces) We will cnsider the fllwing types f intermlecular frces: Lndn dispersin, diplediple, and hydrgen bnding. Lndn dispersin frces and diple-diple frces are cllectively knwn as van der Waals frces. Mlecules can have any mix f these three kinds f intermlecular frces, but all substances at least have Lndn dispersin frces.
Lndn dispersin frces exist fr all substances, whether cmpsed f plar r nnplar mlecules. They arise frm the frmatin f temprary, instantaneus plarities acrss a mlecule frm circulatins f electrns. An instantaneus plarity in ne mlecule may induce an ppsing plarity in an adjacent mlecule, resulting in a series f attractive frces amng neighbring mlecules. average electrn distributin mmentary distributin (plar) induced attractins Lndn dispersin frces arise frm changing electrn distributins. Because all mlecules have electrns, all mlecular substances have Lndn dispersin frces, regardless f whether they are plar r nn-plar. Mlecules with higher mlecular weights have mre electrns, which are generally mre lsely held. This makes their electrn cluds mre defrmable frm nearby charges, a characteristic called plarizability. As a result, substances with higher mlecular weights have higher Lndn dispersin frces and cnsequently tend t have higher melting pints, biling pints, and enthalpies f vaprizatin. The fllwing data fr the diatmic halgens nicely illustrate these trends. Element F2 Cl2 Br2 I2 m.p. ( C) -220-101 -7.3 114 b.p. ( C) -188-34 58.8 184 At 25 C gas gas liquid slid Diple-diple frces f attractin exist between mlecules that are plar thse that have a permanent diple mment. The plarities f individual mlecules tend t align by ppsites, drawing the mlecules tgether and thereby favring a cndensed phase.
These additinal frces f attractin must be vercme in a transitin t a less-rdered phase (e.g., slid t liquid, liquid t gas), s substances with diple-diple attractins between their mlecules tend t have higher melting pints and biling pints than cmparable cmpunds cmpsed f nnplar mlecules, which nly have Lndn dispersin intermlecular frces. A hydrgen bnd is a nn-cvalent attractin between a hydrgen that is cvalently bnded t a very electrnegative atm (X) and anther very electrnegative atm (Y), mst ften n an adjacent mlecule. (X and Y may be the same r different elements.) Fr the mst part, nly cmpunds in which hydrgen is cvalently bnded t O, N, r F are candidates fr hydrgen bnding. Cvalent bnds with these elements are very plar, resulting in a partial negative charge (ä ) n the O, N, r F. This partial negative charge can be attracted t the partial psitive charge (ä+) f the hydrgen in an X H bnd n an adjacent mlecule. Thus, the H Y hydrgen bnd, unlike the cvalent X H bnd, results mainly frm electrstatic attractin. Hydrgen bnd strengths typically are in the range 4-46 kj/ml, much less than the strengths f typical cvalent bnds. Nnetheless, hydrgen bnd strength is significantly greater than either Lndn dispersin frces r diple-diple frces. Hydrgen bnds in HF(s) and H O(s) (shwn n the next page) are intermediate in strength within this range. 2
Nnplar mlecules nly have Lndn dispersin frces, which tend t be the weakest f the three kinds f intermlecular frces. Plar mlecules add anther kind f frce, beynd their Lndn frces, and s have strnger verall intermlecular frces f attractin. If a mlecule is capable f hydrgen bnding, then it has all three kinds f intermlecular frces and has the strngest verall mix. As the kinds f intermlecular frces increase, substances have a greater tendency t exist in a cndensed phase, have higher melting pints and biling pints, and as liquids have lwer vapr pressure and lwer viscsity. Key Questins 5. Name the kind r kinds f intermlecular frces that must be vercme t cnvert the fllwing frm liquid r slid t gas: (a) Br, (b) CH OH, (c) CO, (d) HCN, (e) NH 2 3 2 3 6. Nrmal alkanes are hydrcarbns with unbranched carbn chains, having a general frmula CnH 2n+2. At rm temperature, ethane, C2H 6, is a gas; hexane, C6H 14, is a liquid; and ctadecane, C18H 38, is a slid. Describe the intermlecular frces present in each substance and explain the differences in their rm-temperature phases. 7. Arrange the fllwing in rder f increasing biling pint:
Infrmatin (Heating Curves) As we cntinuusly heat a slid substance, such as ice, ver time it can pass thrugh all phases, giving a behavir represented by the fllwing heating curve. Fr a given input f heat, q, the temperature rises by different amunts in the slid, liquid, and vapr phases due t the different heat capacities fr each phase. In a single phase, ÄT = q/c, where C is the heat capacity fr the phase in the range f the temperature change. Put anther way, in a single phase the amunt f heat that must be added t raise the temperature a certain amunt, ÄT, is given by q = C ÄT. At bth the melting pint and biling pint, the temperature remains cnstant with heat input s lng as bth phases invlved with the transitin exist. At these pints, all the heat is used t effect the cnversin f ne phase int anther (frm slid t liquid at the melting pint r frm liquid t vapr at the biling pint). Once cnversin is cmplete, the temperature f the substance will rise with additin f heat. At the melting pint, the amunt f heat required t cnvert ne mle f substance frm slid t liquid defines the mlar heat f fusin, ÄH fus, als called the mlar heat f melting, ÄH melt. At the biling pint, the amunt f heat required t cnvert ne mle f substance frm liquid t vapr is called the mlar heat f vaprizatin, ÄH. Key Questin vap 8. Hw much heat is required t heat 10.0 g f ice at -5.00 C t becme liquid water at +7.00 C? In this temperature range, the heat capacity f H2O(s) is 37.7 J/ml K, and the heat capacity f H O(l) is 75.8 J/ml K. The mlar heat f fusin f ice is 6.01 kj/ml. 2
Infrmatin (Vapr Pressure) If a liquid is placed in a clsed cntainer, bth evapratin and cndensatin will ccur simultaneusly. Initially, evapratin predminates, because there are s few mlecules in the vapr phase abve the liquid. As mre and mre mlecules build up in the vapr, sme will strike the liquid surface, be attracted t liquid mlecules, and rejin the liquid phase. At sme pint the rate f mlecules escaping the liquid will equal the rate f mlecules rejining it frm the vapr. This marks the nset f a state f dynamic equilibrium at which the rates f evapratin and cndensatin are equal. At this pint, the number f vapr mlecules in the vlume abve the liquid remains cnstant ver time. The resulting partial pressure f the vapr in equilibrium with the liquid is the vapr pressure. The vapr pressure depends upn the nature f the liquid (ÄH vap), the kinetic energy f the mlecules (increasing with temperature), and the resulting vapr cncentratin. Althugh different substances may have different vapr pressures at the same temperature (higher fr mre vlatile substances), all vapr pressures increase with increasing temperature. When the temperature is high enugh that a liquid s vapr pressure equals the ambient pressure, biling will cmmence. At temperatures where the vapr pressure is lwer than the ambient pressure, n bubbles f vapr can frm in the liquid, because the greater air pressure wuld cause them t cllapse. The higher the ambient pressure, the higher the biling pint will be. Fr this reasn, we define the nrmal biling pint at 1 atm. Cnversely, biling ccurs at a lwer temperature when the ambient pressure is lwer. Fr example, Mt. Washingtn (elevatin 6288 ft) has a typical air pressure f 610 trr. At 94 C, the vapr pressure f water is 610.90 trr, s this is apprximately the biling pint f water atp Mt. Washingtn. Cnsequently, fds tend t require lnger cking times at high altitudes. Key Questins 9. Explain hw each f the fllwing affects the vapr pressure f a liquid: (a) the vlume f the liquid, (b) the vlume f the cntainer, (c) the surface area f the liquid, (d) the temperature, (e) intermlecular frces f attractin, (f) the density f the liquid.
Infrmatin (Phase Diagrams) The relatinships between temperature and pressure and the phase transitins f a substance can be summarized in a phase diagram. The phase diagram fr carbn dixide is shwn belw. Fr any cmbinatin f temperature and pressure that falls entirely within a phase regin, nly that ne phase will exist. Lines n the diagram represent cmbinatins f temperature and pressure under which tw phases cexist (e.g., gas-liquid, liquid-slid, slid-gas). The triple pint is the cmbinatin f temperature and pressure at which all three phases are simultaneusly in equilibrium. The critical pint marks the limit n the ability t cndense a vapr with applied pressure at high temperature. Abve this pint it is nt pssible t cndense the gas regardless f increasing pressure. The critical temperature, T c, is the pint abve which a gas cannt be liquified, regardless f the pressure. The critical pressure, P c, is the minimum pressure needed t cause cndensatin at the critical temperature. Abve the critical pint the substance exists as a supercritical fluid. Key Questins 10. Describe the phases and/r phase transitins experienced by CO 2 under the fllwing cnditins: (a) Heating frm 100 C t 30 C at 1.0 atm (b) Heating frm 100 C t 50 C at 70 atm (c) A sample at 35 C and 100 atm (d) A sample at 50 C and 6.0 atm 11. Des carbn dixide have a nrmal biling pint? Explain. 12. Describe the cnditins under which liquid carbn dixide bils.
Infrmatin (The Slutin Prcess) A slute disslves in a slvent because the attractive frces between slvent and slute particles (i.e., ins and/r mlecules) are similar t thse between the slute particles themselves. As a rule: "Like disslves like." When tw substances have dissimilar intermlecular frces f attractin, their mlecules tend t be mre attracted t themselves and remain in the pure, unmixed state. In such cases, the slute is said t be insluble in the slvent. But the term insluble is relative. Mst slutes have sme (albeit limited) slubility in a given slvent. Thus, "insluble" usually means "negligibly sluble" r "sparingly sluble". Any f the intermlecular frces we have previusly discussed can ccur between slute and slvent mlecules, resulting in slutin frmatin. Lndn dispersin frces, diple-diple attractins, r hydrgen bnding between slute and slvent mlecules can induce the slute mlecules t break away frm the attractins they have with their wn mlecules t enter the slutin. This prcess is called slvatin in general, but when water is the slvent it is called hydratin. In aqueus slutins the slute mlecules r ins are surrunded by waters f hydratin, which prevent their recmbinatin. When an inic crystal disslves in water, it dissciates t frm hydrated ins, which are catins and anins surrunded by water mlecules; e.g., + NaCl(s) + xh O [Na(H O) ] + [Cl(H O) ] 2 2 n 2 m In this case, in-diple attractins and the stability f the resulting hydrated ins vercme the culmbic attractins between the ins in the crystal lattice. The ins in slutin are prevented frm recmbining because the surrunding waters f hydratin d nt allw them t get clse enugh tgether. Sme mlecules with highly plar bnds, such as the binary acids, HX, may disslve in water partially r cmpletely as ins. The prcess is similar t the disslving f inic slutes, but in this case it is the strength f the cvalent bnds within the slute mlecules that must be vercme; e.g., + HCl(g) + H2O H3O (aq) + Cl (aq) The attractin between slvent water mlecules and the slute mlecules is initially diplediple, but the attractin between the dissciated ins and the their waters f hydratin is indiple. These in-diple attractins interfere with recmbinatin t frm undissciated mlecules. If the intermlecular frces in the slute and slvent are very different, the slute will have little r n appreciable slubility. Fr example, a nnplar liquid like benzene, C6H 6, which nly has Lndn dispersin frces between its mlecules, has pr slubility in water, which is a plar slvent with hydrgen bnding. Likewise, benzene is a pr slvent fr inic substances, such as NaCl, because it is incapable f establishing the necessary in-diple attractins that keep ins in slutin.
Key Questins 13. Identify the principal type f slute-slvent interactin that is respnsible fr frming the fllwing slutins: (a) KNO 3 in water; (b) Br 2 in benzene, C6H 6; (c) glycerl, CH 2(OH)CH(OH)CH2OH, in water; (d) HCl in acetnitrile, CH3CN [HCl des nt frm ins in CH CN]. 3 14. Fr the fllwing carbxylic acids, predict whether slubility will be greater in water r carbn tetrachlride, and give yur reasning: (a) acetic acid, CH3CO2H, (b) stearic acid, CH (CH ) CO H. 3 2 16 2 Infrmatin (Enthalpy and Entrpy f Slutin) When a slute disslves in a slvent the enthalpy f the verall prcess, expressed as ÄH be exthermic r endthermic; e.g., 2+ CaCl 2(s) Ca (aq) + 2Cl (aq) + heat ÄH sln < 0 (exthermic) + heat + KNO 3(s) K (aq) + NO 3 (aq) ÄH sln > 0 (endthermic) The enthalpy f slutin can be analyzed as the sum f the fllwing prcesses: 1. Enthalpy t vercme attractins between slute mlecules. + ÄH > 0 (endthermic) 2. Enthalpy t vercme attractins between slvent mlecules. + ÄH > 0 (endthermic) 3. Enthalpy t frm slvated mlecules thrugh slute-slvent attractins. + ÄH < 0 (exthermic) 1 2 3 The verall enthalpy f slutin is ÄH sln = ÄH 1 + ÄH 2 + ÄH 3. If ÄH3 > {ÄH 1 + ÄH 2}, then ÄH < 0 (exthermic). If ÄH < {ÄH + ÄH }, then ÄH > 0 (endthermic). sln 3 1 2 sln sln, may If a slute is sluble in a slvent, the verall prcess is said t be spntaneus. If a slute des nt disslve in a slvent (insluble) the slutin prcess is said t be nn-spntaneus. A spntaneus prcess r change prceeds in a given directin withut needing t be driven by an external energy input. As we have seen repeatedly, reactins r prcesses that are exthermic, in which the energy cntent f the system ges dwn, tend t be spntaneus (e.g., cmbustin). But sme endthermic reactins r prcesses are als spntaneus. As just nted, bth CaCl 2 and KNO 3 disslve spntaneusly in water. CaCl 2 des s exthermically, but KNO 3 des s endthermically. In bth cases, the slutin prcess disperses the ins frm their highly rdered crystal lattice int the a much mre disrdered state in slutin. The degree f disrder in a system is assciated with the thermdynamic functin called entrpy (abut which we will have much mre t say later). When a slute disslves in a slvent, the dispersal f the slute
particles results in an increase in entrpy, which favrs a spntaneus slutin prcess. In the case f slutes like KNO 3 the entrpy factr favring spntaneity is greater than the enthalpy factr favring nn-spntaneity, and disslving is spntaneus. The prcess is said t be entrpy cntrlled. If the increase in entrpy is nt great enugh, an endthermic prcess will be nnspntaneus and is said t be enthalpy cntrlled. The heat f slutin f an insluble slute is usually endthermic, with t small an entrpy increase t make the verall prcess spntaneus. Key Questin 15. Hexane (C6H 14) and heptane (C7H 16) are miscible in all prprtins with ÄH sln 0. (a) Why are these tw liquids miscible with each ther? (b) Why is ÄH sln 0 fr this pair f liquids? (b) Why d they spntaneusly frm slutins, given that ÄH 0? Infrmatin (Slubility f Nn-reactive Gases) When nn-reactive gases disslve in a slvent, bth pressure and temperature significantly affect slubility. Mst nn-reactive gases have very small slubilities, which increase with increasing partial pressure f the disslving gas. At a given temperature and with mderate partial pressures ( 1 atm), many nn-reactive gases bey Henry's Law: sln P = K g where P g is the partial pressure f the gas ver the slutin, K is a cnstant fr the particular gas and slvent at the given temperature, and g is the mle fractin f the gas in the slutin. At the lw cncentratins typical f disslved nn-reactive gases, mle fractin is prprtinal t cncentratin in the usual units, such as ml/l r g/l. Therefre, Henry's Law can be restated as C = k P g where C g is the cncentratin f the disslved gas, and k is the Henry's Law cnstant fr the particular gas and slvent pair, expressed in apprpriate cncentratin units per unit pressure, usually in atm. Gas slubility is always less at higher temperature (i.e., k is smaller). Key Questin g g -3 16. The slubility f N 2 at p(n 2) = 1 atm is 1.75 x 10 g/100 ml f water. What is the slubility in water at an air pressure f 2.51 atm, the pressure at 50 ft belw the surface f the water? Air is 78.1 vl-% N 2. [Hint: What is the partial pressure f N 2(g) when the air pressure is 2.51 atm?]
Infrmatin (Mlality and Clligative Prperties) The presence f slute mlecules changes certain prperties in the slutin frm what they are in the pure slvent. Slutin prperties such as these, which are dependent n cncentratin, are called clligative prperties. Examples f clligative prperties include vapr pressure, biling pint, freezing pint, and smtic pressure. Fr discussins f clligative prperties it is ften cnvenient t express slutin cncentratin in terms f mlality: The magnitude f any clligative effect depends n the ttal cncentratin f slute particles in the slutin, nt n the number f mles f slute added t make up the slutin. In shrt, yu have t ask yurself What are the cncentratins f all slute particles in this slutin? Fr a strng electrlyte at mderate cncentratin, we can assume that all f the inic slid has brken up int its cmpnent ins n disslving. Fr example, a 0.100 m slutin f K3PO 4 has fur times the cncentratin f slute particles as a 0.100 m slutin f sugar, because each frmula + 3 unit f K PO breaks up n disslving int three K ins and ne PO in: 3 4 4 K PO (s) 3 K (aq) + PO + 3 3 4 4 + 3 In ther wrds, the mlality f K ins is (3)(0.100 m) = 0.300 m, the mlality f PO 4 ins is (1)(0.100 m), and the ttal mlality fr all ins is 0.300 m + 0.100 m = 0.400 m. It is the ttal cncentratin f ins that causes the change in slutin prperties. If calculating a clligative effect n the basis f mlality, we wuld use 0.400 m fr such a slutin f K3PO 4, rather than 0.100 m, as we wuld fr a sugar slutin. (aq) Key Questins 17. Calculate the mlality f ethanl, C2H5OH (m.w. = 46.06) in a slutin prepared by disslving 5.00 g f ethanl in 25.00 g f water. 18. Calculate the ttal mlality f all ins in a slutin prepared by disslving 20.0 g f (NH ) SO in 95.0 g f water. [f.w. (NH ) SO = 132 u] 4 2 4 4 2 4 Infrmatin (Vapr Pressure abve Slutins with Nnvlatile Slutes) If we add a nnvlatile slute (ne that desn t have a vapr pressure f its wn) t a vlatile slvent, the slute particles (mlecules, ins, r mixture f bth) will blck the escape f the slvent mlecules int the vapr. This will reduce the vapr pressure ver the slutin, cmpared t what it wuld be if the slvent had nthing disslved in it. The slvent s vapr pressure abve such a slutin is reasnably well predicted by a special case f Rault's Law:
P sln = slvp slv where P sln is the vapr pressure abve the slutin at equilibrium, slv is the mle fractin f the slvent (nt the slute) in the slutin, and P slv is the vapr pressure the pure slvent wuld have at the temperature f the slutin. Key Questins 19. Cnsider a 2.00 m slutin f sugar in water at 25.00 C. (a) What is the value f the mle fractin f water in this slutin? [Hint: Imagine that the slutin was made up with exactly 1 Kg f water.] (m.w. H O = 18.02 u) 2 (b) Calculate the vapr pressure abve a 2.00 m slutin f sugar in water at 25.00 C, given that the vapr pressure f pure water at this temperature is 23.76 mm Hg. 20. Calculate the expected vapr pressure abve a 2.00 m slutin f Na2SO 4 in water at 25.00 C. Cmpare this result t what yu fund in part a f the preceding Key Questin. Infrmatin (Vapr Pressure abve Slutins f Vlatile Cmpnents) If a slutin is made f tw r mre vlatile cmpnents, the ttal vapr pressure abve the slutin shuld be the sum f the vapr pressures (partial pressures) arising frm each f the cmpnents in the mixture, as predicted by Daltn's Law f Partial Pressures. If the cmpnent liquids d nt interact with each ther appreciably, they are said t frm an ideal slutin. In this case the partial vapr pressure abve the slutin frm each cmpnent liquid is given by sln P i = i Pi sln where P i is the vapr pressure frm an individual cmpnent, i is the mle fractin f the cmpnent in the slutin, and P i is the vapr pressure the pure liquid wuld have at the given temperature. Thus, the general frm f Rault's Law can be stated as sln sln sln sln P t = i P i = 1 P 1 + 2 P 2 +... + n Pn Fr a tw-cmpnent system, this is P = P + P sln sln t A A B B It shuld be emphasized that Rault's Law applies t ideal slutins, in which interactins between slvent and slute mlecules are virtually identical t thse in the neat liquids. N real slutin shws this behavir ver a significant range f cncentratins. A few carefully chsen mixtures, particularly with ne cmpnent in very small mle fractin, apprximate Rault's Law. The cmpsitin f the vapr abve a mixture f tw r mre vlatile liquids is nt the same as in the slutin. In general, mre vlatile cmpnents (as indicated by their higher vapr pressure
as a neat liquid) will have a higher mle fractin in the vapr than in the slutin. Frm Daltn s Law f Partial Pressures, the mle fractin f a cmpnent in the vapr mixture is its partial vap pressure divided by the ttal pressure; i.e., = P /P. Key Questins i i t 21. What are the partial pressures and ttal vapr pressure abve a slutin at 20.0 C made by mixing 12.5 g benzene (C6H 6) with 44.2 g tluene (C6H5CH 3). At 20.0 C, P (C6H 6) = 74.7 trr and P (C H CH ) = 22.3 trr. [m.w. C H = 78.11; m.w. C H CH = 92.14] 6 5 3 6 6 6 5 3 22. In terms f mle fractins, what is the cmpsitin f the vapr abve the previusly described benzene-tluene mixture? Infrmatin (Biling Pint Elevatin and Freezing Pint Depressin) The presence f nnvlatile slute particles in slutin interferes with the slvent mlecules escape frm the liquid phase. This is the cause f the vapr pressure lwering. This als means that a higher temperature must be used t achieve a vapr pressure f 1 atm, which defines the nrmal biling pint. Thus, a slutin f a nnvlatile slute and a vlatile slvent has a higher biling pint (biling pint elevatin). Similarly, slute particles interfere with the frmatin f the slid n cling, s freezing ccurs at a lwer temperature (freezing pint depressin). The temperature shift due t biling pint elevatin r freezing pint depressin can be calculated by the frmula ÄT = Km where K is either the biling pint elevatin cnstant (K b) r the freezing pint depressin cnstant (K f). The tw cnstants have different values fr a particular liquid, and K f is generally a bigger number. Dn t wrry abut the sign n the ÄT. Just knw that the value must be added t the nrmal biling pint and subtracted frm the nrmal freezing pint f the pure slvent. Biling pint elevatins and freezing pint depressins can be used t determine mlecular weights f slutes. Because K f > K b fr a given slvent and biling pints are sensitive t pressure cnditins, freezing pint depressin (r, equivalently, melting pint depressin) is generally the preferred methd. This is called cryscpic mlecular weight determinatin. A weighed amunt f substance whse mlar mass is t be determined (the unknwn) is disslved in a knwn mass f slvent, and the new freezing pint (r melting pint) f the mixture is bserved t determine ÄT b. Using this and the slvent s K b value, the mlality f the slutin is calculated. With this knwn number f mles f slute per kilgram f slvent (mlality) and the mass cmpsitin f the slutin as it was made up, it is pssible t calculate the slute s number f grams per mle (its mlar mass).
Key Questins 23. Pure benzene has a freezing pint f 5.5 C and a biling pint f 80.1 C. What are the expected freezing pint and biling pint fr a 0.15 m slutin f a nnvlatile slute in benzene? Fr benzene, K = 5.12 C/m and K = 2.53 C/m. f 24. When 45.0 g f an unknwn nnvlatile nnelectrlyte is disslved in 500.0 g f water, the resulting slutin freezes at -0.930 C. What is the mlar mass f the unknwn substance? K = 1.86 C/m fr water. f Infrmatin (Osmtic Pressure) Osmtic pressure is assciated with the situatin in which a slutin and the pure slvent (r a mre dilute slutin) are separated by a semipermeable membrane, which has pres nly big enugh t allw slvent mlecules t pass thrugh, but nt slute mlecules. Because the cncentratin f slvent mlecules is greater n the pure slvent side, mre slvent mlecules n that side will strike the membrane, hit a hle, and pass thrugh. This results in net flw f slvent frm the pure slvent side t the slutin side, resulting in dilutin ver time. Many tissues in living rganisms serve as semipermeable membranes, and smtic pressure is very imprtant in understanding physilgy. In 1887 Jacbus van't Hff discvered that the flw culd be stpped r even reversed by applying pressure t the slutin side. The necessary minimum pressure t stp the diffusin f slvent int the slutin is called the smtic pressure, ð, given by ð = MRT where M is the mlarity f the slutin, T is temperature in Kelvin, and R is the gas cnstant, 0.08206 L atm/k ml. If a pressure greater than ð is applied, reverse smsis will ccur, reducing the amunt f slvent n the slutin side as it is transferred t the pure slvent side. This is used as ne way f purifying drinking water. Key Questins 25. What is the smtic pressure f a 0.100 M glucse slutin in trr at 25.0 C? 26. Sea water is apprximately 0.60 M NaCl. What is the minimum applied pressure that must be exceeded t achieve water purificatin by reverse smsis at 25 C? b