CHAPTER 12 PHYSICAL PROPERTIES OF SOLUTIONS

Transcription

1 CHATER 1 HYSICAL ROERTIES OF SOLUTIONS roble Categories Biological: 1.56, 1.57, 1.66, 1.76, 1.81, 1.9, 1.1. Conceptual: 1.9, 1.10, 1.11, 1.1, 1.5, 1.6, 1.69, 1.70, 1.71, 1.7, 1.75, 1.8, 1.87, 1.88, 1.89, 1.91, 1.96, 1.97, 1.100, 1.101, 1.10, 1.106, 1.111, 1.11, 1.115, 1.118, 1.119, Descriptive: 1.98, Industrial: Organic: 1.9, 1.10, 1.1, 1.16, 1.17, 1.19, 1.1, 1.4, 1.49, 1.50, 1.51, 1.5, 1.5, 1.54, 1.55, 1.58, 1.59, 1.60, 1.61, 1:, 1.6, 1.64, 1.65, 1.70, 1.88, 1.94, 1.96, 1.104, 1.105, 1.110, 1.111, 1.114, 1.116, 1.1. Difficulty Level Easy: 1.9, 1.10, 1.1, 1.15, 1.17, 1.0, 1., 1.7, 1.7, 1.55, 1.56, 1.6, 1.69, 1.7, 1.75, 1.77, 1.78, 1.8, 1.8, 1.85, 1.96, 1.98, 1.101, Mediu: 1.11, 1.16, 1.18, 1.19, 1.1, 1., 1.4, 1.8, 1.5, 1.6, 1.49, 1.50, 1.51, 1.5, 1.57, 1.58, 1.60, 1.61, 1.6, 1.64, 1.65, 1.66, 1.70, 1.71, 1.7, 1.74, 1.76, 1.84, 1.86, 1.87, 1.88, 1.90, 1.91, 1.9, 1.95, 1.97, 1.100, 1.10, 1.10, 1.106, 1.108, 1.109, 1.110, 1.111, 1.115, 1.116, 1.117, 1.118, 1.1, 1.10, 1.1. Difficult: 1.9, 1.8, 1.5, 1.54, 1.59, 1.81, 1.89, 1.9, 1.94, 1.99, 1.104, 1.105, 1.107, 1.11, 1.114, 1.119, 1.10, 1.11, 1.1, 1.14, 1.15, 1.16, 1.17, 1.18, 1.19, CsF is an ionic solid; the ion ion attractions are too strong to be overcoe in the dissolving process in benzene. The ion induced dipole interaction is too weak to stabilize the ion. Nonpolar naphthalene olecules for a olecular solid in which the only interparticle forces are of the weak dispersion type. The sae forces operate in liquid benzene causing naphthalene to dissolve with relative ease. Like dissolves like Strategy: In predicting solubility, reeber the saying: Like dissolves like. A nonpolar solute will dissolve in a nonpolar solvent; ionic copounds will generally dissolve in polar solvents due to favorable ion-dipole interactions; solutes that can for hydrogen bonds with a solvent will have high solubility in the solvent. Solution: Strong hydrogen bonding (dipole-dipole attraction) is the principal interolecular attraction in liquid ethanol, but in liquid cyclohexane the interolecular forces are dispersion forces because cyclohexane is nonpolar. Cyclohexane cannot for hydrogen bonds with ethanol, and therefore cannot attract ethanol olecules strongly enough to for a solution The order of increasing solubility is: O < Br < LiCl < CH OH. Methanol is iscible with water because of strong hydrogen bonding. LiCl is an ionic solid and is very soluble because of the high polarity of the water olecules. Both oxygen and broine are nonpolar and exert only weak dispersion forces. Broine is a larger olecule and is therefore ore polarizable and susceptible to dipole induced dipole attractions. 1.1 The longer the C C chain, the ore the olecule "looks like" a hydrocarbon and the less iportant the OH group becoes. Hence, as the C C chain length increases, the olecule becoes less polar. Since like dissolves like, as the olecules becoe ore nonpolar, the solubility in polar water decreases. The OH group of the alcohols can for strong hydrogen bonds with water olecules, but this property decreases as the chain length increases.

2 14 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS 1.15 ercent ass equals the ass of solute divided by the ass of the solution (that is, solute plus solvent) ties 100 (to convert to percentage). (a) 5.50 g NaBr 78. g soln 100% 7.0% 1.0 g KCl (1.0 15) g soln 100% 16.9% (c) 4.5 g toluene (4.5 9) g soln 100% 1% 1.16 Strategy: We are given the percent by ass of the solute and the ass of the solute. We can use Equation (1.1) of the text to solve for the ass of the solvent (water). Solution: (a) The percent by ass is defined as ass of solute percent by ass of solute 100% ass of solute + ass of solvent Substituting in the percent by ass of solute and the ass of solute, we can solve for the ass of solvent (water) g urea 16.% 100% 5.00 g urea + ass of water (0.16)(ass of water) 5.00 g (0.16)(5.00g) ass of water 5.9 g Siilar to part (a), 6. g MgCl 1.5% 100% 6. g MgCl + ass of water ass of water g 1.17 (a) The olality is the nuber of oles of sucrose (olar ass 4. g/ol) divided by the ass of the solvent (water) in kg. 1 ol ol sucrose 14. g sucrose ol 4. g sucrose Molality ol sucrose kg HO Molality 7.0 ol ethylene glycol.546 kg HO.0

3 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS olality oles of solute ass of solvent (kg) (a) 1.08 g ass of 1 L soln 1000 L 1080 g 1 L g NaCl ass of water 1080 g.50 ol NaCl 94 g 0.94 kg 1 ol NaCl.50 ol NaCl 0.94 kg HO g of the solution contains 48. g KBr and 51.8 g H O. 1 ol KBr ol of KBr 48. g KBr ol KBr g KBr 1 kg ass of HO (in kg) 51.8 g HO kg HO 1000 g ol KBr kg HO In each case we consider one liter of solution. ass of solution volue density (a) 4. g sugar 1 kg ass of sugar 1. ol sugar 418 g sugar kg sugar 1 ol sugar 1000 g 1.1 g 1 kg ass of soln 1000 L 110 g 1.10 kg 1 L 1000 g olality 1. ol sugar ( ) kg HO g NaOH ass of NaOH 0.87 ol NaOH 5 g NaOH 1 ol NaOH ass solvent (H O) 1040 g 5 g 1005 g kg olality 0.87 ol NaOH kg HO 0.87 (c) g NaHCO ass of NaHCO 5.4 ol NaHCO 440 g NaHCO 1 ol NaHCO ass solvent (H O) 1190 g 440 g 750 g kg olality 5.4 ol NaHCO kg HO 6.99

4 16 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS 1.0 Let s assue that we have 1.0 L of a M solution. Assuing a solution density of 1.0 g/l, the ass of 1.0 L (1000 L) of the solution is 1000 g or g. The ass of ole of urea is: g urea ol urea 0.60 g urea 1 ol urea The ass of the solvent is: (solution ass) (solute ass) ( g) (0.60 g) g 1.0 kg oles solute ass solvent ol 1.0 kg We find the volue of ethanol in 1.00 L of 75 proof gin. Note that 75 proof eans 75 %. 75 Volue 1.00 L % 0.8 L.8 10 L g Ethanol ass (.8 10 L).0 10 g 1 L 1. (a) Converting ass percent to olality. Strategy: In solving this type of proble, it is convenient to assue that we start with gras of the solution. If the ass of sulfuric acid is 98.0% of g, or 98.0 g, the percent by ass of water ust be 100.0% 98.0%.0%. The ass of water in g of solution would be.0 g. Fro the definition of olality, we need to find oles of solute (sulfuric acid) and kilogras of solvent (water). Solution: Since the definition of olality is olality oles of solute ass of solvent (kg) we first convert 98.0 g H SO 4 to oles of H SO 4 using its olar ass, then we convert.0 g of H O to units of kilogras. 1 ol HSO g HSO ol HSO g H SO 4 1 kg.0 g HO.0 10 kg HO 1000 g Lastly, we divide oles of solute by ass of solvent in kg to calculate the olality of the solution. ol of solute ol kg of solvent.0 10 kg

5 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS 17 Converting olality to olarity. Strategy: Fro part (a), we know the oles of solute (0.999 ole H SO 4 ) and the ass of the solution (100.0 g). To solve for olarity, we need the volue of the solution, which we can calculate fro its ass and density. Solution: First, we use the solution density as a conversion factor to convert to volue of solution. 1 L? volue of solution g 54.6 L L 1.8 g Since we already know oles of solute fro part (a), ole H SO 4, we divide oles of solute by liters of solution to calculate the olarity of the solution. M ol of solute L of soln ol L 18. M 1. 1 ol NH ol NH 0.0 g NH 1.76 ol NH 17.0 g NH 1 L 1 L Volue of the solution g soln 0.10 L 0.98 g 1000 L olarity 1.76 ol NH 0.10 L soln 17. M 1 kg kg of solvent (HO) 70.0 g HO kg HO 1000 g olality 1.76 ol NH kg HO Assue g of solution. (a) The ass of ethanol in the solution is g 10.0 g. The ass of the water is g 10.0 g 90.0 g kg. The aount of ethanol in oles is: 1 ol 10.0 g ethanol 0.17 ol ethanol g ol solute kg solvent 0.17 ol kg.41 The volue of the solution is: 1 L g 10 L 0.10 L g The aount of ethanol in oles is 0.17 ole [part (a)]. M ol solute liters of soln 0.17 ol 0.10 L.1 M

6 18 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS (c) 1L Solution volue 0.15 ol L 58.7 L.1 ol 1.7 The aount of salt dissolved in 100 g of water is:.0 g salt 9.10 g HO 100 g HO 5. g salt Therefore, the solubility of the salt is 5. g salt/100 g H O. 1.8 At 75 C, 155 g of KNO dissolves in 100 g of water to for 55 g of solution. When cooled to 5 C, only 8.0 g of KNO reain dissolved. This eans that ( ) g 117 g of KNO will crystallize. The aount of KNO fored when 100 g of saturated solution at 75 C is cooled to 5 C can be found by a siple unit conversion. 100 g saturated soln 117 g KNO crystallized 55 g saturated soln 45.9 g KNO 1.9 The ass of KCl is 10% of the ass of the whole saple or 5.0 g. The KClO ass is 45 g. If 100 g of water will dissolve 5.5 g of KCl, then the aount of water to dissolve 5.0 g KCl is: The 0 g of water will dissolve: The KClO reaining undissolved will be: 100 g HO 5.0 g KCl 0 g HO 5.5 g KCl 7.1 g KClO 0 g HO 1.4 g KClO 100 g HO (45 1.4) g KClO 44 g KClO 1.5 When a dissolved gas is in dynaic equilibriu with its surroundings, the nuber of gas olecules entering the solution (dissolving) is equal to the nuber of dissolved gas olecules leaving and entering the gas phase. When the surrounding air is replaced by heliu, the nuber of air olecules leaving the solution is greater than the nuber dissolving. As tie passes the concentration of dissolved air becoes very sall or zero, and the concentration of dissolved heliu increases to a axiu. 1.6 According to Henry s law, the solubility of a gas in a liquid increases as the pressure increases (c k). The soft drink tastes flat at the botto of the ine because the carbon dioxide pressure is greater and the dissolved gas is not released fro the solution. As the iner goes up in the elevator, the atospheric carbon dioxide pressure decreases and dissolved gas is released fro his stoach. 1.7 We first find the value of k for Henry's law k c 0.04 ol/l 1 at 0.04 ol/l at For atospheric conditions we write: c k (0.04 ol/l at)( at) ol/l

7 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS Strategy: The given solubility allows us to calculate Henry's law constant (k), which can then be used to deterine the concentration of N at 4.0 at. We can then copare the solubilities of N in blood under noral pressure (0.80 at) and under a greater pressure that a deep-sea diver ight experience (4.0 at) to deterine the oles of N released when the diver returns to the surface. Fro the oles of N released, we can calculate the volue of N released. Solution: First, calculate the Henry's law constant, k, using the concentration of N in blood at 0.80 at. k k = c ol/l 0.80 at ol/l at Next, we can calculate the concentration of N in blood at 4.0 at using k calculated above. c k c ( ol/l at)(4.0 at).8 10 ol/l Fro each of the concentrations of N in blood, we can calculate the nuber of oles of N dissolved by ultiplying by the total blood volue of 5.0 L. Then, we can calculate the nuber of oles of N released when the diver returns to the surface. The nuber of oles of N in 5.0 L of blood at 0.80 at is: ( ol/l )(5.0 L).8 10 ol The nuber of oles of N in 5.0 L of blood at 4.0 at is: (.8 10 ol/l)(5.0 L) ol The aount of N released in oles when the diver returns to the surface is: ( ol) (.8 10 ol) ol Finally, we can now calculate the volue of N released using the ideal gas equation. The total pressure pushing on the N that is released is atospheric pressure (1 at). The volue of N released is: V N = nrt V N ( ol)(7 7)K L at = (1.0 at) ol K 0.8 L 1.49 The first step is to find the nuber of oles of sucrose and of water. 1 ol Moles sucrose 96 g 1.16 ol sucrose 4. g 1 ol Moles water 64 g 4.6 ol water 18.0 g

8 0 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS The ole fraction of water is: HO 4.6 ol 4.6 ol 1.16 ol The vapor pressure of the solution is found as follows: solution H O H O (0.968)(1.8 Hg) 0.8 Hg 1.50 Strategy: Fro the vapor pressure of water at 0 C and the change in vapor pressure for the solution (.0 Hg), we can solve for the ole fraction of sucrose using Equation (1.5) of the text. Fro the ole fraction of sucrose, we can solve for oles of sucrose. Lastly, we convert for oles to gras of sucrose. Solution: Using Equation (1.5) of the text, we can calculate the ole fraction of sucrose that causes a.0 Hg drop in vapor pressure. 1 sucrose water.0 Hg sucrose = = = Hg water Fro the definition of ole fraction, we can calculate oles of sucrose. sucrose = nsucrose nwater nsucrose 1 ol oles of water 55 g 0.6 ol HO 18.0 g sucrose n sucrose nsucrose = 0.11 = 0.6 nsucrose.8 ol sucrose Using the olar ass of sucrose as a conversion factor, we can calculate the ass of sucrose. 4. g sucrose ass of sucrose.8 ol sucrose g sucrose 1 ol sucrose 1.51 Let us call benzene coponent 1 and caphor coponent. n n1 n n 1 n 1 ol 98.5 g benzene 1.6 ol benzene g 1 ol 4.6 g caphor 0.16 ol caphor 15. g 1.6 ol Hg 88.6 Hg ( ) ol

9 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS For any solution the su of the ole fractions of the coponents is always 1.00, so the ole fraction of 1 propanol is The partial pressures are: ethanol ethanol ethanol (0.00)(100 Hg) 0.0 Hg 1 propanol 1 propanol 1 propanol (0.700)(7.6 Hg) 6. Hg Is the vapor phase richer in one of the coponents than the solution? Which coponent? Should this always be true for ideal solutions? 1.5 (a) First find the ole fractions of the solution coponents. Moles ethanol 0.0 g 1 ol.04 g 0.96 ol CHOH Moles ethanol 45.0 g 1 ol g ol CH5OH ethanol ethanol 0.96 ol 0.96 ol ol 1 ethanol The vapor pressures of the ethanol and ethanol are: ethanol (0.489)(94 Hg) 46 Hg ethanol (0.511)(44 Hg) Hg Since n V/RT and V and T are the sae for both vapors, the nuber of oles of each substance is proportional to the partial pressure. We can then write for the ole fractions: ethanol 46 Hg ethanol ethanol ethanol 46 Hg Hg 0.68 X ethanol 1 ethanol 0. (c) The two coponents could be separated by fractional distillation. See Section 1.6 of the text This proble is very siilar to roble urea water.50 Hg X urea (1.8 Hg) X urea The nuber of oles of water is: 1 ol HO nwater 450 g HO 5.0 ol HO 18.0 g HO

10 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS urea n urea nurea nwater nurea nurea 5.0 nurea.1 ol ass of urea.1 ol urea g urea 1 ol urea 18 g of urea 1.55 T b K b (.5 C/)(.47 ) 6.5 C The new boiling point is 80.1 C 6.5 C 86.4 C T f K f (5.1 C/)(.47 ) 1.6 C The new freezing point is 5.5 C 1.6 C 7.1 C 1.56 Tf Kf 1.1 C = C/ 1.57 METHOD 1: The epirical forula can be found fro the percent by ass data assuing a g saple. 1 ol Moles C g 6.76 ol C 1.01 g 1 ol Moles H 1.56 g 1.45 ol H g 1 ol Moles O 5.66 g 0.54 ol O g This gives the forula: C 6.76 H 1.45 O Dividing through by the sallest subscript (0.54) gives the epirical forula, C 19 H 8 O. The freezing point depression is T f 5.5 C.7 C.1 C. This iplies a solution olality of: Tf Kf.1 C 5.1 C/ 0.41 Since the solvent ass is 8.50 g or kg, the aount of solute is: 0.41 ol 1 kg benzene kg benzene.5 10 ol Since 1.00 g of the saple represents.5 10 ol, the olar ass is: olar ass 1.00 g.5 10 ol 86 g/ol The ass of the epirical forula is 8 g/ol, so the olecular forula is the sae as the epirical forula, C 19 H 8 O.

11 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS METHOD : Use the freezing point data as above to deterine the olar ass. olar ass 86 g/ol Multiply the ass % (converted to a decial) of each eleent by the olar ass to convert to gras of each eleent. Then, use the olar ass to convert to oles of each eleent. n C (0.8078) (86 g) 1 ol C 1.01 g C 19. ol C n H (0.156) (86 g) 1 ol H g H 8.5 ol H n O (0.0566) (86 g) 1 ol O g O 1.01 ol O Since we used the olar ass to calculate the oles of each eleent present in the copound, this ethod directly gives the olecular forula. The forula is C 19 H 8 O METHOD 1: Strategy: First, we can deterine the epirical forula fro ass percent data. Then, we can deterine the olar ass fro the freezing-point depression. Finally, fro the epirical forula and the olar ass, we can find the olecular forula. Solution: If we assue that we have 100 g of the copound, then each percentage can be converted directly to gras. In this saple, there will be 40.0 g of C, 6.7 g of H, and 5. g of O. Because the subscripts in the forula represent a ole ratio, we need to convert the gras of each eleent to oles. The conversion factor needed is the olar ass of each eleent. Let n represent the nuber of oles of each eleent so that n C n H n O 1 ol C 40.0 g C. ol C 1.01 g C 1 ol H 6.7 g H 6.6 ol H g H 1 ol O 5. g O. ol O g O Thus, we arrive at the forula C. H 6.6 O., which gives the identity and the ratios of atos present. However, cheical forulas are written with whole nubers. Try to convert to whole nubers by dividing all the subscripts by the sallest subscript.. C : = This gives us the epirical, CH O. 6.6 H : =.0.. O: = Now, we can use the freezing point data to deterine the olar ass. First, calculate the olality of the solution. Tf 1.56 C K 8.00 C/ f Multiplying the olality by the ass of solvent (in kg) gives oles of unknown solute. Then, dividing the ass of solute (in g) by the oles of solute, gives the olar ass of the unknown solute.

12 4 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS ol solute? ol of unknown solute kg diphenyl 1 kg diphenyl ol solute g olar ass of unknown g/ol ol Finally, we copare the epirical olar ass to the olar ass above. epirical olar ass 1.01 g (1.008 g) g 0.0 g/ol The nuber of (CH O) units present in the olecular forula is: olar ass g epirical olar ass 0.0 g 4.00 Thus, there are four CH O units in each olecule of the copound, so the olecular forula is (CH O) 4, or C 4 H 8 O 4. METHOD : Strategy: As in Method 1, we deterine the olar ass of the unknown fro the freezing point data. Once the olar ass is known, we can ultiply the ass % of each eleent (converted to a decial) by the olar ass to convert to gras of each eleent. Fro the gras of each eleent, the oles of each eleent can be deterined and hence the ole ratio in which the eleents cobine. Solution: We use the freezing point data to deterine the olar ass. First, calculate the olality of the solution. Tf 1.56 C K 8.00 C/ f Multiplying the olality by the ass of solvent (in kg) gives oles of unknown solute. Then, dividing the ass of solute (in g) by the oles of solute, gives the olar ass of the unknown solute ol solute? ol of unknown solute kg diphenyl 1 kg diphenyl ol solute g olar ass of unknown g/ol ol Next, we ultiply the ass % (converted to a decial) of each eleent by the olar ass to convert to gras of each eleent. Then, we use the olar ass to convert to oles of each eleent. n C n H n O (0.400) (1.0 1 ol C 10 g) 1.01 g C 4.00 ol C (0.067) (1.0 1 ol H 10 g) g H 7.98 ol H (0.5) (1.0 1 ol O 10 g) g O 4.00 ol O Since we used the olar ass to calculate the oles of each eleent present in the copound, this ethod directly gives the olecular forula. The forula is C 4 H 8 O 4.

13 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS We want a freezing point depression of 0 C. Tf Kf 0 C 1.86 C/ 10.8 The ass of ethylene glycol (EG) in 6.5 L or 6.5 kg of water is: 10.8 ol EG 6.07 g EG ass EG 6.50 kg HO g EG 1 kg HO 1 ol EG The volue of EG needed is: V 1 L EG 1 L ( g EG) 1.11 g EG 1000 L.9 L Finally, we calculate the boiling point: T b K b (10.8 )(0.5 C/) 5.6 C The boiling point of the solution will be C 5.6 C C We first find the nuber of oles of gas using the ideal gas equation. n V RT 1 at 748 Hg (4.00 L) 760 Hg ol K (7 + 7) K L at ol ol olality kg benzene T f K f (5.1 C/)(.76 ) 14.1 C freezing point 5.5 C 14.1 C 8.6 C 1.61 The experiental data indicate that the benzoic acid olecules are associated together in pairs in solution due to hydrogen bonding. O H O C C O H O 1.6 First, fro the freezing point depression we can calculate the olality of the solution. See Table 1. of the text for the noral freezing point and K f value for benzene. T f (5.5 4.) C 1. C Tf Kf 1. C 5.1 C/ 0.

14 6 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS Multiplying the olality by the ass of solvent (in kg) gives oles of unknown solute. Then, dividing the ass of solute (in g) by the oles of solute, gives the olar ass of the unknown solute. 0. ol solute? ol of unknown solute kg benzene 1 kg benzene ol solute.50 g olar ass of unknown g/ol ol The epirical olar ass of C 6 H 5 is g/ol. Therefore, the olecular forula is (C 6 H 5 ) 4 or C 4 H MRT (1.6 ol/l)(0.081 L at/k ol)(.0 7)K.9 at 1.64 Strategy: We are asked to calculate the olar ass of the polyer. Gras of the polyer are given in the proble, so we need to solve for oles of polyer. want to calculate given olar ass of polyer gras of polyer oles of polyer need to find Fro the osotic pressure of the solution, we can calculate the olarity of the solution. Then, fro the olarity, we can deterine the nuber of oles in 0.80 g of the polyer. What units should we use for and teperature? Solution: First, we calculate the olarity using Equation (1.8) of the text. MRT M RT 1 at 5.0 Hg 760 Hg ol K 98 K L at M Multiplying the olarity by the volue of solution (in L) gives oles of solute (polyer).? ol of polyer ( ol/l)(0.170 L) ol polyer Lastly, dividing the ass of polyer (in g) by the oles of polyer, gives the olar ass of the polyer g polyer 4 olar ass of polyer g/ol ol polyer

15 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS Method 1: First, find the concentration of the solution, then work out the olar ass. The concentration is: Molarity RT 1.4 at (0.081 L at/k ol)(00 K) ol/l The solution volue is L so the nuber of oles of solute is: ol 1L L ol The olar ass is then: g ol 40 g/ol The epirical forula can be found ost easily by assuing a g saple of the substance. 1 ol Moles C 41.8 g.48 ol C 1.01 g 1 ol Moles H 4.7 g 4.7 ol H g 1 ol Moles O 7. g. ol O g 1 ol Moles N 16. g 1.16 ol N g The gives the forula: C.48 H 4.7 O. N Dividing through by the sallest subscript (1.16) gives the epirical forula, C H 4 O N, which has a ass of 86.0 g per forula unit. The olar ass is five ties this aount ( ), so the olecular forula is (C H 4 O N) 5 or C 15 H 0 O 10 N 5. METHOD : Use the olarity data as above to deterine the olar ass. olar ass 40 g/ol Multiply the ass % (converted to a decial) of each eleent by the olar ass to convert to gras of each eleent. Then, use the olar ass to convert to oles of each eleent. n C (0.418) (40 g) 1 ol C 1.01 g C 15.0 ol C n H (0.047) (40 g) 1 ol H g H 0 ol H n O (0.7) (40 g) 1 ol O g O 10.0 ol O n N (0.16) (40 g) 1 ol N g N 5.00 ol N Since we used the olar ass to calculate the oles of each eleent present in the copound, this ethod directly gives the olecular forula. The forula is C 15 H 0 O 10 N 5.

16 8 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS 1.66 We use the osotic pressure data to deterine the olarity. M RT 4.61 at ol K (0 7) K L at 0.19 ol/l Next we use the density and the solution ass to find the volue of the solution. ass of soln 6.85 g g g soln 1 L volue of soln g soln L L 1.04 g Multiplying the olarity by the volue (in L) gives oles of solute (carbohydrate). ol of solute M L (0.19 ol/l)( L) ol solute Finally, dividing ass of carbohydrate by oles of carbohydrate gives the olar ass of the carbohydrate. olar ass 6.85 g carbohydrate ol carbohydrate 4 g/ol 1.69 CaCl is an ionic copound (why?) and is therefore an electrolyte in water. Assuing that CaCl is a strong electrolyte and copletely dissociates (no ion pairs, van't Hoff factor i ), the total ion concentration will be , which is larger than the urea (nonelectrolyte) concentration of (a) (c) The CaCl solution will show a larger boiling point elevation. The CaCl solution will show a larger freezing point depression. The freezing point of the urea solution will be higher. The CaCl solution will have a larger vapor pressure lowering Boiling point, vapor pressure, and osotic pressure all depend on particle concentration. Therefore, these solutions also have the sae boiling point, osotic pressure, and vapor pressure Assue that all the salts are copletely dissociated. Calculate the olality of the ions in the solutions. (a) 0.10 Na O 4 : ions/unit NaCl: 0.5 ions/unit 0.70 (c) 0.0 MgCl : 0.0 ions/unit 0.60 (d) 0.15 C 6 H 1 O 6 : nonelectrolyte, 0.15 (e) 0.15 CH COOH: weak electrolyte, slightly greater than 0.15 The solution with the lowest olality will have the highest freezing point (sallest freezing point depression): (d) > (e) > (a) > (c) >. 1.7 The freezing point will be depressed ost by the solution that contains the ost solute particles. You should try to classify each solute as a strong electrolyte, a weak electrolyte, or a nonelectrolyte. All three solutions have the sae concentration, so coparing the solutions is straightforward. HCl is a strong electrolyte, so under ideal conditions it will copletely dissociate into two particles per olecule. The concentration of particles will be Acetic acid is a weak electrolyte, so it will only dissociate to a sall extent. The concentration of particles will be greater than 0.50, but less than Glucose is a nonelectrolyte, so glucose olecules reain as glucose olecules in solution. The concentration of particles will be For these solutions, the order in which the freezing points becoe lower is: 0.50 glucose > 0.50 acetic acid > 0.50 HCl In other words, the HCl solution will have the lowest freezing point (greatest freezing point depression).

17 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS (a) NaCl is a strong electrolyte. The concentration of particles (ions) is double the concentration of NaCl. Note that 15 L of water has a ass of 15 g (why?). The nuber of oles of NaCl is: 1 ol 1. g NaCl 0.6 ol NaCl g Next, we can find the changes in boiling and freezing points (i ) 0.6 ol 0.15 kg.70 T b ik b (0.5 C/)(.70 ).8 C T f ik f (1.86 C/)(.70 ) 10.0 C The boiling point is 10.8 C; the freezing point is 10.0 C. Urea is a nonelectrolyte. The particle concentration is just equal to the urea concentration. The olality of the urea solution is: 1 ol urea oles urea 15.4 g urea 0.56 ol urea g urea 0.56 ol urea kg HO.84 T b ik b 1(0.5 C/)(.84 ).0 C T f ik f 1(1.86 C/)(.84 ) 7.14 C The boiling point is 10.0 C; the freezing point is 7.14 C Using Equation (1.5) of the text, we can find the ole fraction of the NaCl. We use subscript 1 for H O and subscript for NaCl Hg.98 Hg = = Hg Let s assue that we have 1000 g (1 kg) of water as the solvent, because the definition of olality is oles of solute per kg of solvent. We can find the nuber of oles of particles dissolved in the water using the definition of ole fraction. = n n1 n 1 ol HO n 1 = 1000 g HO ol HO 18.0 g HO

18 0 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS n n = 0.08 n ol Since NaCl dissociates to for two particles (ions), the nuber of oles of NaCl is half of the above result. 1 ol NaCl Moles NaCl ol particles ol ol particles The olality of the solution is: ol kg Both NaCl and CaCl are strong electrolytes. Urea and sucrose are nonelectrolytes. The NaCl or CaCl will yield ore particles per ole of the solid dissolved, resulting in greater freezing point depression. Also, sucrose and urea would ake a ess when the ice elts Strategy: We want to calculate the osotic pressure of a NaCl solution. Since NaCl is a strong electrolyte, i in the van't Hoff equation is. imrt Since, R is a constant and T is given, we need to first solve for the olarity of the solution in order to calculate the osotic pressure ( ). If we assue a given volue of solution, we can then use the density of the solution to deterine the ass of the solution. The solution is 0.86% by ass NaCl, so we can find gras of NaCl in the solution. Solution: To calculate olarity, let s assue that we have L of solution ( L). We can use the solution density as a conversion factor to calculate the ass of L of solution g soln ( L soln) 1005 g of soln 1 L soln Since the solution is 0.86% by ass NaCl, the ass of NaCl in the solution is: 0.86% 1005 g 8.6 g NaCl 100% The olarity of the solution is: 8.6 g NaCl 1 ol NaCl L g NaCl 0.15 M Since NaCl is a strong electrolyte, we assue that the van't Hoff factor is. Substituting i, M, R, and T into the equation for osotic pressure gives: imrt 0.15 ol L at () (10 K) L ol K 7.6 at

19 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS The teperature and olarity of the two solutions are the sae. If we divide Equation (1.1) of the text for one solution by the sae equation for the other, we can find the ratio of the van't Hoff factors in ters of the osotic pressures (i 1 for urea). CaCl urea imrt MRT i at 0.45 at Fro Table 1. of the text, i 1. imrt ol L at (1.) (98 K) L ol K 1.6 at 1.81 For this proble we ust find the solution ole fractions, the olality, and the olarity. For olarity, we can assue the solution to be so dilute that its density is 1.00 g/l. We first find the nuber of oles of lysozye and of water. n lysozye n water 1 ol g ol 190 g 1 ol 150 g 8. ol 18.0 g Vapor pressure lowering: lysozye water nlysozye (.76 Hg) nlysozye nwater ol (.76 Hg) 6 [( ) 8.]ol Hg Freezing point depression: Boiling point elevation: T T ol 5 f Kf (1.86 C/ ) C kg ol 5 b Kb (0.5 C/ ).5 10 C kg Osotic pressure: As stated above, we assue the density of the solution is 1.00 g/l. The volue of the solution will be 150 L. MRT ol L (0.081 L at/ol K)(98 K) at Hg Note that only the osotic pressure is large enough to easure. 1.8 At constant teperature, the osotic pressure of a solution is proportional to the olarity. When equal volues of the two solutions are ixed, the olarity will just be the ean of the olarities of the two solutions (assuing additive volues). Since the osotic pressure is proportional to the olarity, the osotic pressure of the solution will be the ean of the osotic pressure of the two solutions..4 at 4.6 at.5 at

20 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS 1.8 Water igrates through the seiperiable cell walls of the cucuber into the concentrated salt solution. When we go swiing in the ocean, why don't we shrivel up like a cucuber? When we swi in fresh water pool, why don't we swell up and burst? 1.84 (a) We use Equation (1.4) of the text to calculate the vapor pressure of each coponent First, you ust calculate the ole fraction of each coponent. Siilarly, na 1.00 ol A na + nb 1.00 ol ol B Substitute the ole fraction calculated above and the vapor pressure of the pure solvent into Equation (1.4) to calculate the vapor pressure of each coponent of the solution. A A A (0.500)(76 Hg) 8 Hg B B B (0.500)(1 Hg) 66 Hg The total vapor pressure is the su of the vapor pressures of the two coponents. Total A B 8 Hg 66 Hg 104 Hg This proble is solved siilarly to part (a). Siilarly, na.00 ol A na + nb.00 ol ol B A A A (0.86)(76 Hg) Hg B B B (0.714)(1 Hg) 94 Hg Total A B Hg 94 Hg 116 Hg 1.85 T f ik f i Tf K f.6 (1.86)(0.40) Fro the osotic pressure, you can calculate the olarity of the solution. M RT 1 at 0. Hg 760 Hg ol K 08 K L at ol/l

21 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS Multiplying olarity by the volue of solution in liters gives the oles of solute. ( ol solute/l soln) (0.6 L soln) ol solute Divide the gras of solute by the oles of solute to calculate the olar ass. 1. g olar ass of solute g/ol ol 1.87 One anoeter has pure water over the ercury, one anoeter has a 1.0 M solution of NaCl and the other anoeter has a 1.0 M solution of urea. The pure water will have the highest vapor pressure and will thus force the ercury colun down the ost; colun X. Both the salt and the urea will lower the overall pressure of the water. However, the salt dissociates into sodiu and chloride ions (van't Hoff factor i ), whereas urea is a olecular copound with a van't Hoff factor of 1. Therefore the urea solution will lower the pressure only half as uch as the salt solution. Y is the NaCl solution and Z is the urea solution. Assuing that you knew the teperature, could you actually calculate the distance fro the top of the solution to the top of the anoeter? 1.88 Solve Equation (1.7) of the text algebraically for olality (), then substitute T f and K f into the equation to calculate the olality. You can find the noral freezing point for benzene and K f for benzene in Table 1. of the text. T f 5.5 C.9 C 1.6 C Tf Kf 1.6 C = C/ Multiplying the olality by the ass of solvent (in kg) gives oles of unknown solute. Then, dividing the ass of solute (in g) by the oles of solute, gives the olar ass of the unknown solute. 0.1 ol solute? ol of unknown solute ( kg benzene) 1 kg benzene.5 10 ol solute 0.50 g olar ass of unknown.0 10 g/ol.5 10 ol The olar ass of cocaine C 17 H 1 NO 4 0 g/ol, so the copound is not cocaine. We assue in our analysis that the copound is a pure, onoeric, nonelectrolyte The pill is in a hypotonic solution. Consequently, by ososis, water oves across the seipereable ebrane into the pill. The increase in pressure pushes the elastic ebrane to the right, causing the drug to exit through the sall holes at a constant rate The olality of the solution assuing AlCl to be a nonelectrolyte is: 1 ol AlCl ol AlCl 1.00 g AlCl ol AlCl 1. g AlCl ol kg 0.150

22 4 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS The olality calculated with Equation (1.7) of the text is: Tf Kf 1.11 C 1.86 C/ The ratio is 4. Thus each AlCl dissociates as follows: AlCl (s) Al (aq) Cl (aq) 1.91 Reverse ososis uses high pressure to force water fro a ore concentrated solution to a less concentrated one through a seipereable ebrane. Desalination by reverse ososis is considerably cheaper than by distillation and avoids the technical difficulties associated with freezing. To reverse the osotic igration of water across a seipereable ebrane, an external pressure exceeding the osotic pressure ust be applied. To find the osotic pressure of 0.70 M NaCl solution, we ust use the van t Hoff factor because NaCl is a strong electrolyte and the total ion concentration becoes (0.70 M) 1.4 M. The osotic pressure of sea water is: imrt (0.70 ol/l)(0.081 L at/ol K)(98 K) 4 at To cause reverse ososis a pressure in excess of 4 at ust be applied. 1.9 First, we tabulate the concentration of all of the ions. Notice that the chloride concentration coes fro ore than one source. MgCl : If [MgCl ] M, [Mg ] M [Cl ] M Na SO 4 : if [Na SO 4 ] M, [Na ] M [SO 4 ] M CaCl : if [CaCl ] M, [Ca ] M [Cl ] M NaHCO : if [NaHCO ] M [Na ] M [HCO ] M KCl: if [KCl] M [K ] M [Cl ] M The subtotal of chloride ion concentration is: [Cl ] ( ) ( 0.010) (0.0090) 0.17 M Since the required [Cl ] is.60 M, the difference ( M) ust coe fro NaCl. The subtotal of sodiu ion concentration is: [Na ] ( 0.051) (0.000) M Since the required [Na ] is.56 M, the difference ( M) ust coe fro NaCl. Now, calculating the ass of the copounds required: NaCl: MgCl : Na SO 4 :.46 ol ol ol g NaCl 1 ol NaCl 14.8 g 95.1g MgCl 1 ol MgCl 5.14 g 14.1 g NaSO4 1 ol NaSO4 7.5 g

23 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS 5 CaCl : KCl: NaHCO : ol ol ol g CaCl 1 ol CaCl 1.11 g g KCl 1 ol KCl 0.67 g g NaHCO 1 ol NaHCO 0.17 g 1.9 (a) Using Equation (1.8) of the text, we find the olarity of the solution. M RT 0.57 at (0.081 L at/ol K)(98 K) ol/l This is the cobined concentrations of all the ions. The aount dissolved in 10.0 L ( L) is ol 4? oles L ol 1L Since the ass of this aount of protein is 0.5 g, the apparent olar ass is 0.5 g ol g/ol We need to use a van t Hoff factor to take into account the fact that the protein is a strong electrolyte. The van t Hoff factor will be i 1 (why?). M irt 0.57 at (1)(0.081 L at/ol K)(98 K) ol/l This is the actual concentration of the protein. The aount in 10.0 L ( L) is ol 1L L ol 6 Therefore the actual olar ass is: 0.5 g ol g/ol 1.94 Solution A: Let olar ass be M. A A ( ) A(760) A n ass olar ass A na na nwater 5.00 / M 5.00 / M 100 / M 14 g/ol

24 6 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS Solution B: Let olar ass be M B B B n ass olar ass B nb nb nbenzene.1/ M.1/ M 100 / M 48 g/ol The olar ass in benzene is about twice that in water. This suggests soe sort of dierization is occurring in a nonpolar solvent such as benzene H O H O O.0 g HO 1 ol HO 1 ol O 10 L ol O 100 L 4.0 g HO ol HO (a) Using the ideal gas law: V nrt ( ol O )(0.081 L at/ol K)(7 K) 1.0 at 99 L The ratio of the volues: 99 L 10 L 9.9 Could we have ade the calculation in part (a) sipler if we used the fact that 1 ole of all ideal gases at ST occupies a volue of.4 L? 1.96 As the chain becoes longer, the alcohols becoe ore like hydrocarbons (nonpolar) in their properties. The alcohol with five carbons (n-pentanol) would be the best solvent for iodine (a) and n-pentane (c) (why?). Methanol (CH OH) is the ost water like and is the best solvent for an ionic solid like KBr (a) Boiling under reduced pressure. CO boils off, expands and cools, condensing water vapor to for fog I H O: Dipole - induced dipole. I H O: Ion - dipole. Stronger interaction causes ore I to be converted to I Let the 1.0 M solution be solution 1 and the.0 M solution be solution. Due to the higher vapor pressure of solution 1, there will be a net transfer of water fro beaker 1 to beaker until the vapor pressures of the two solutions are equal. In other words, at equilibriu, the concentration in the two beakers is equal. At equilibriu, M 1 = M Initially, there is ole glucose in solution 1 and 0.10 ole glucose in solution, and the volue of both solutions is L. The volue of solution 1 will decrease, and the volue of solution will increase by the sae volue. Let x be the change in volue.

25 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS 7 The final volues are: ol 0.10 ol (0.050 x) L (0.050 x) L x x 0.15x x L = 16.7 L solution 1: ( ) L. L solution : ( ) L 66.7 L (a) If the ebrane is pereable to all the ions and to the water, the result will be the sae as just reoving the ebrane. You will have two solutions of equal NaCl concentration. (c) This part is tricky. The oveent of one ion but not the other would result in one side of the apparatus acquiring a positive electric charge and the other side becoing equally negative. This has never been known to happen, so we ust conclude that igrating ions always drag other ions of the opposite charge with the. In this hypothetical situation only water would ove through the ebrane fro the dilute to the ore concentrated side. This is the classic ososis situation. Water would ove through the ebrane fro the dilute to the concentrated side To protect the red blood cells and other cells fro shrinking (in a hypertonic solution) or expanding (in a hypotonic solution) First, we calculate the nuber of oles of HCl in 100 g of solution. nhcl 7.7 g HCl 1 ol HCl 100 g soln 1.0 ol HCl 100 g soln 6.46 g HCl Next, we calculate the volue of 100 g of solution. V 1 L 1 L 100 g L 1.19 g 1000 L Finally, the olarity of the solution is: 1.0 ol L 1. M 1.10 (a) Seawater has a larger nuber of ionic copounds dissolved in it; thus the boiling point is elevated. (c) (d) (e) Carbon dioxide escapes fro an opened soft drink bottle because gases are less soluble in liquids at lower pressure (Henry s law). As you proved in roble 1.0, at dilute concentrations olality and olarity are alost the sae because the density of the solution is alost equal to that of the pure solvent. For colligative properties we are concerned with the nuber of solute particles in solution relative to the nuber of solvent particles. Since in colligative particle easureents we frequently are dealing with changes in teperature (and since density varies with teperature), we need a concentration unit that is teperature invariant. We use units of oles per kilogra of ass (olality) rather than oles per liter of solution (olarity). Methanol is very water soluble (why?) and effectively lowers the freezing point of water. However in the suer, the teperatures are sufficiently high so that ost of the ethanol would be lost to vaporization.

26 8 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS Let the ass of NaCl be x g. Then, the ass of sucrose is (10. x)g. We know that the equation representing the osotic pressure is: MRT, R, and T are given. Using this equation and the definition of olarity, we can calculate the percentage of NaCl in the ixture. olarity ol solute L soln Reeber that NaCl dissociates into two ions in solution; therefore, we ultiply the oles of NaCl by two. 1 ol NaCl 1 ol sucrose ol solute x g NaCl (10. x )g sucrose g NaCl 4. g sucrose ol solute 0.04x x ol solute 0.010x Molarity of solution ol solute (0.010 x ) ol L soln 0.50 L Substitute olarity into the equation for osotic pressure to solve for x. MRT ( x ) ol L at 7. at (96 K) 0.50 L ol K x x 1.45 g ass of NaCl 1.45 g Mass % NaCl 100% 14.% 10. g T f C C 10 H 8 : 18. g/ol Tf Kf C 6 H 1 : g/ol Let x ass of C 6 H 1 (in gras). Using, ol solute kg solvent and ol ass olar ass x 1. x kg 18.x x (84.16)(18.) x 0.47 g

27 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS 9 %C6 H % 6% %C10 H % 65% The percentages don t add up to 100% because of rounding procedures (a) Solubility decreases with increasing lattice energy. (c) Ionic copounds are ore soluble in a polar solvent. Solubility increases with enthalpy of hydration of the cation and anion The copleted table is shown below: Attractive Forces Deviation fro Raoult s H solution A A, B B A B ositive ositive (endotheric) A A, B B A B Negative Negative (exotheric) A A, B B A B Zero Zero The first row represents a Case 1 situation in which A s attract A s and B s attract B s ore strongly than A s attract B s. As described in Section 1.6 of the text, this results in positive deviation fro Raoult s law (higher vapor pressure than calculated) and positive heat of solution (endotheric). In the second row a negative deviation fro Raoult s law (lower than calculated vapor pressure) eans A s attract B s better than A s attract A s and B s attract B s. This causes a negative (exotheric) heat of solution. In the third row a zero heat of solution eans that A A, B B, and A B interparticle attractions are all the sae. This corresponds to an ideal solution which obeys Raoult s law exactly. What sorts of substances for ideal solutions with each other? olality 1 ol HSO 98.0 g H 4 SO g HSO4 1 kg HO.0 g HO 1000 g H O We can calculate the density of sulfuric acid fro the olarity. olarity 18 M 18 ol HSO4 1 L soln The 18 ol of H SO 4 has a ass of: 1 L 1000 L 98.0 g HSO4 18 ol HSO g HSO4 1 ol HSO4 density ass HSO g volue 1000 L 1.80 g/l

28 40 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS Let's assue we have 100 g of solution. The 100 g of solution will contain 70.0 g of HNO and 0.0 g of H O. 1 ol HNO ol solute (HNO ) 70.0 g HNO 1.11 ol HNO 6.0 g HNO 1 kg kg solvent (HO) 0.0 g HO kg HO 1000 g olality 1.11 ol HNO kg HO 7.0 To calculate the density, let's again assue we have 100 g of solution. Since, d ass volue we know the ass (100 g) and therefore need to calculate the volue of the solution. We know fro the olarity that 15.9 ol of HNO are dissolved in a solution volue of 1000 L. In 100 g of solution, there are 1.11 oles HNO (calculated above). What volue will 1.11 oles of HNO occupy? 1000 L soln 1.11 ol HNO 69.8 L soln 15.9 ol HNO Dividing the ass by the volue gives the density. d 100 g 69.8 L 1.4 g/l = A A A ethanol (0.6)(108 Hg) 67.0 Hg 1-propanol (0.8)(40.0 Hg) 15. Hg In the vapor phase: 67.0 ethanol Since the total volue is less than the su of the two volues, the ethanol and water ust have an interolecular attraction that results in an overall saller volue NH can for hydrogen bonds with water; NCl cannot. (Like dissolves like.) 1.11 In solution, the Al(H O) 6 ions neutralize the charge on the hydrophobic colloidal soil particles, leading to their precipitation fro water We can calculate the olality of the solution fro the freezing point depression. T f K f

29 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS 41 The olality of the original solution was Soe of the solution has ionized to H and CH COO. CH COOH CH COO H Initial Change x x x Equil x x x At equilibriu, the total concentration of species in solution is (0.106 x) x x 0.00 The percentage of acid that has undergone ionization is: % % Egg yolk contains lecithins which solubilize oil in water (See Figure 1.0 of the text). The nonpolar oil becoes soluble in water because the nonpolar tails of lecithin dissolve in the oil, and the polar heads of the lecithin olecules dissolve in polar water (like dissolves like) First, we can calculate the olality of the solution fro the freezing point depression. T f (5.1) (5.5.5) (5.1) 0.9 Next, fro the definition of olality, we can calculate the oles of solute. 0.9 ol solute ol solute kg solvent The olar ass (M) of the solute is: ol solute kg benzene 0.01 ol.8 g 0.01 ol g/ol The olar ass of CH COOH is g/ol. Since the olar ass of the solute calculated fro the freezing point depression is twice this value, the structure of the solute ost likely is a dier that is held together by hydrogen bonds. O H O H C C C CH A dier O H O g g or g g 5 ass of lead/l g/l.6 L 4

30 4 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS Safety liit: pp iplies a ass of g b per g g of water. 1 liter of water has a ass of g b 5 ass of lead 1000 g H 6 O g/l 1 10 g H O The concentration of lead calculated above ( g/l) exceeds the safety liit of g/l. Don t drink the water! (a) T f K f (1.86)() olality 1.1 This concentration is too high and is not a reasonable physiological concentration. Although the protein is present in low concentrations, it can prevent the foration of ice crystals If the can is tapped with a etal object, the vibration releases the bubbles and they ove to the top of the can where they join up to for bigger bubbles or ix with the gas at the top of the can. When the can is opened, the gas escapes without dragging the liquid out of the can with it. If the can is not tapped, the bubbles expand when the pressure is released and push the liquid out ahead of the As the water freezes, dissolved inerals in the water precipitate fro solution. The inerals refract light and create an opaque appearance At equilibriu, the vapor pressure of benzene over each beaker ust be the sae. Assuing ideal solutions, this eans that the ole fraction of benzene in each beaker ust be identical at equilibriu. Consequently, the ole fraction of solute is also the sae in each beaker, even though the solutes are different in the two solutions. Assuing the solute to be non-volatile, equilibriu is reached by the transfer of benzene, via the vapor phase, fro beaker A to beaker B. The ole fraction of naphthalene in beaker A at equilibriu can be deterined fro the data given. The nuber of oles of naphthalene is given, and the oles of benzene can be calculated using its olar ass and knowing that 100 g 7.0 g 9.0 g of benzene reain in the beaker. C10H ol 0.15 ol 9.0 g benzene 1 ol benzene g benzene 0.11 Now, let the nuber of oles of unknown copound be n. Assuing all the benzene lost fro beaker A is transferred to beaker B, there are 100 g 7.0 g 107 g of benzene in the beaker. Also, recall that the ole fraction of solute in beaker B is equal to that in beaker A at equilibriu (0.11). The ole fraction of the unknown copound is: unknown n 107 g benzene n 1 ol benzene g benzene 0.11 n n 1.70 n 0.17 ol

31 CHATER 1: HYSICAL ROERTIES OF SOLUTIONS 4 There are 1 gras of the unknown copound dissolved in benzene. The olar ass of the unknown is: 1 g 0.17 ol g/ol Teperature is assued constant and ideal behavior is also assued. 1.1 To solve for the olality of the solution, we need the oles of solute (urea) and the kilogras of solvent (water). If we assue that we have 1 ole of water, we know the ass of water. Using the change in vapor pressure, we can solve for the ole fraction of urea and then the oles of urea. Using Equation (1.5) of the text, we solve for the ole fraction of urea..76 Hg.98 Hg 0.78 Hg 1 urea water urea water 0.78 Hg.76 Hg 0.0 Assuing that we have 1 ole of water, we can now solve for oles of urea. urea ol urea ol urea ol water 0.0 nurea nurea 1 0.0n urea 0.0 n urea n urea n urea 0.04 ol 1 ole of water has a ass of 18.0 g or kg. We now know the oles of solute (urea) and the kilogras of solvent (water), so we can solve for the olality of the solution. ol solute kg solvent 0.04 ol kg 1.9 H O H 1.1 (a) H C C C H S C S H H Acetone is a polar olecule and carbon disulfide is a nonpolar olecule. The interolecular attractions between acetone and CS will be weaker than those between acetone olecules and those between CS olecules. Because of the weak attractions between acetone and CS, there is a greater tendency for these olecules to leave the solution copared to an ideal solution. Consequently, the vapor pressure of the solution is greater than the su of the vapor pressures as predicted by Raoult's law for the sae concentration. Let acetone be coponent A of the solution and carbon disulfide coponent B. For an ideal solution, A A A, B B B, and T A B.