Chem 116 POGIL Worksheet - Week 4 Properties of Solutions

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1 Chem 116 POGIL Wrksheet - Week 4 Prperties f Slutins Key Questins 1. 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 (a) in-diple (b) Lndn dispersin (c) hydrgen bnding (d) diple-diple 2. 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 (a) Acetic acid s -OH grups make hydrgen bnding pssible, which is cmpatible with slvent water. Carbn tetrachlride has nly Lndn dispersin frces, which are less cmpatible. Therefre, acetic acid is mre sluble in water than carbn tetrachlride. (b) Stearic acid has a very lng chain and much higher Lndn dispersin frces than acetic acid. This makes it mre cmpatible with carbn tetrachlride, despite the ptential fr hydrgen bnding (which is largely mitigated by the lng chain getting in the way). Therefre, stearic acid is mre sluble in carbn tetrachlride than water. 3. Hexane (C6H 14) and heptane (C7H 16) are miscible in all prprtins with H sln 0. (a) Why are these tw liquids miscible with each ther? They have very similar Lndn dispersin frces, wing t their similar mlar masses. (b) Why is H sln 0 fr this pair f liquids? The intermlecular frces f attractin in the neat liquids are s similar t each ther that little change ccurs n mixing. It is the change in intermlecular attractin strength that is principally respnsible fr the sense and magnitude f H. (b) Why d they spntaneusly frm slutins, given that H sln 0? Mixing is a mre disrdered state than exists in the separate neat liquids. The increase in entrpy is the driving factr in making slutin frmatin spntaneus in this case. sln 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?]

2 Frm Daltn's Law f Partial Pressures, at air pressure f 2.51 atm, the partial pressure f N 2 is = (0.781)(2.51 atm) = 1.96 atm Frm Henry's Law, the slubility is 5. Calculate the mlality f ethanl, C2H5OH (m.w. = 46.06) in a slutin prepared by disslving 5.00 g f ethanl in g f water. 6. 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] (NH ) SO (s) 2NH (aq) + SO (aq) 7. Cnsider a 2.00 m slutin f sugar in water at 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 = u) 2 T calculate mle fractin, we need the numbers f mles f water and f sugar. If we assume that exactly 1000 g f water were used, then we already knw that the slutin cntains 2.00 mles f sugar. All we need is the number f mles in 1000 g f water, and then we can calculate (H O). 2

3 ml sugar = 2.00 ml sugar ml H2O = (1000 g H2O)(1 mle H2O/18.02 g H2O) = ml H2O (H2O) = ml/( ) ml = 55.49/57.49 = (b) Calculate the vapr pressure abve a 2.00 m slutin f sugar in water at C, given that the vapr pressure f pure water at this temperature is mm Hg. P sln = (H2O) P (H2O) = mm Hg = mm Hg 8. Calculate the expected vapr pressure abve a 2.00 m slutin f Na2SO 4 in water at C. Cmpare this result t what yu fund in part a. We must take accunt f the dissciatin f Na SO (s): Na SO (s) 2Na (aq) + SO (aq) Again, assume a slutin prepared with exactly 1000 g f water. ml ins = 3 ml Na2SO 4 = ml = 6.00 ml (H2O) = ml/( ) ml = 55.49/61.49 = P sln = (H2O) P (H2O) = mm Hg = mm Hg The amunt by which the vapr pressure f water has been lwered is almst three times greater in the Na SO slutin, cmpared t the sugar slutin 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] ml C6H 6 = (12.5 g C6H 6)(ml C6H 6/78.11 g C6H 6) = ml C6H6 ml C6H5CH 3 = (44.2 g C6H5CH 3)(ml C6H5CH 3/92.14 g C6H5CH 3) = ml C6H5CH3 ttal mles = ml ml = ml (C6H 6, sln) = ml/0.640 ml = (C6H5CH 3, sln) = = P(C6H 6) = (0.250)(74.5 trr) = 18.7 trr P(C6H5CH 3) = (0.750)(22.3 trr) = 16.7 trr

4 P t = ( ) trr = 35.4 trr 10. In terms f mle fractins, what is the cmpsitin f the vapr in the previusly described benzene-tluene mixture? We previusly fund P(C6H 6) = (0.250)(74.5 tr) = 18.7 trr P(C6H5CH 3) = (0.750)(22.3 trr) = 16.7 trr P t = ( ) trr = 35.4 trr Using these values, we can calculate the mle fractins in the vapr as fllws: (C6H 6, vap) = 18.7 trr/35.4 trr = (C6H5CH 3, vap) = 16.7 trr/35.4 trr = = Ntice that mre vlatile benzene has an increased mle fractin in the vapr. 11. 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 sln T f = (5.12 C/m)(0.15 m) = 0.77 C T f = T f T f = ( ) C = 4.7 C sln T b = (2.53 C/m)(0.15 m) = 0.38 C T b = T b + T b = ( ) C = 80.5 C b 12. When 45.0 g f an unknwn nnvlatile nnelectrlyte is disslved in g f water, the resulting slutin freezes at C. What is the mlar mass f the unknwn substance? K = 1.86 C/m fr water. f T f = C m = T f/k f = C/1.86 C/m = m = ml X/kg H2O 13. What is the smtic pressure f a M glucse slutin in trr at 25.0 C? = MRT = (0.100 ml/l)( L atm/k ml)(298 K) = 2.45 atm (760 trr atm ) -1 = 1860 trr

5 14. 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? We must use the mlarity f ins, nt the stated mlarity f NaCl (analytical cncentratin f NaCl). + NaCl(s) Na (aq) + Cl (aq) mlarity f ins = 2 C NaCl = M = 1.20 M = MRT = (1.20 ml/l)( L atm/k ml)(298 K) = atm = 29.4 atm

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