#### Transcript Chap 11 Example and Chem Tour PPT

ChemTour: Lattice Energy Click here to launch this ChemTour Students learn to apply Coulomb’s law to calculate the exact lattice energies of ionic solids. It includes Practice Exercises. © 2012 by W. W. Norton & Company ChemTour: Molecular Motion Click here to launch this ChemTour Students use an interactive graph to explore the relationship between kinetic energy and temperature. It includes Practice Exercises. © 2012 by W. W. Norton & Company ChemTour: Raoult’s Law Click here to launch this ChemTour Students explore the connection between the vapor pressure of a solution and its concentration as a gas above the solution. It includes Practice Exercises. © 2012 by W. W. Norton & Company ChemTour: Fractional Distillation Click here to launch this ChemTour This ChemTour shows how differences in boiling points are used to separate a mixture of volatile molecules, such as crude oil. It includes follow-up questions. © 2012 by W. W. Norton & Company ChemTour: Boiling and Freezing Points Click here to launch this ChemTour Students learn about colligative properties by exploring the relationship between solute concentration and the temperature at which a solution will undergo phase changes. Interactive exercises invite students to practice calculating the boiling and freezing points of different solutions. © 2012 by W. W. Norton & Company ChemTour: Osmotic Pressure Click here to launch this ChemTour Students discover how a solute can build up pressure behind a semipermeable membrane. This tutorial also discusses the osmotic pressure equation and the van’t Hoff factor. © 2012 by W. W. Norton & Company Sample Exercise 11.1 Rank these three ionic compounds in order of increasing lattice energy and increasing melting point: NaF, KF, and RbF. Assume that these compounds have the same solid structure, which means they have the same value of k in Equation 11.4. Collect and Organize: We can determine relative lattice energies from Equation 11.4. To do so we need to establish the charges and radii of the ions in each compound using Figure 10.2. © 2012 by W. W. Norton & Company Sample Exercise 11.1 (cont.) Analyze: All the cations are alkali metal cations and have a charge of 1+; all the anions are fluoride ions, with a charge of 1-. We are told that k is the same for all three solids. Therefore, any differences in lattice energy must be related to differences in the nucleus-to-nucleus distance d between ions. Because the fluoride ion is common to all the salts, variations in d depend only on the size of the cation. © 2012 by W. W. Norton & Company Sample Exercise 11.1 (cont.) Solve: Periodic trends in size predict, and Figure 11.1 confirms, that the cation sizes are Na+< K+< Rb+. Therefore, the compounds in order of increasing value of d are NaF < KF < RbF (Figure 11.1). As d increases, lattice energy decreases, so RbF has the lowest lattice energy, followed by KF, followed by NaF with the highest. The same trend occurs in melting points: RbF < KF < NaF. Think about It: We predicted the order of lattice energies and melting points for three alkali metal fluorides. This predicted order is confirmed by the experimentally measured melting points: 775°C for RbF, 846°C for KF, and 988°C for NaF. © 2012 by W. W. Norton & Company Sample Exercise 11.2 In Sample Exercise 10.1, we predicted that the ion– ion attraction in CaF2 is greater than in NaF. Confirm this prediction by calculating the lattice energies of (a) NaF and (b) CaF2 given the following information: © 2012 by W. W. Norton & Company Sample Exercise 11.2 (cont.) Collect and Organize: We are to calculate the lattice energy of two ionic compounds using enthalpy values for processes that can be summed to describe an overall process that produces a salt from its constituent elements. We can use a Born– Haber cycle to calculate the unknown lattice energy. Table A4.3 in Appendix 4 contains standard enthalpy of formation values for NaF and CaF2: -575.4 kJ/mol and -1228.0 kJ/mol, respectively. These enthalpy changes apply to the formation of one mole of the two compounds by combining their component elements in their standard states. © 2012 by W. W. Norton & Company Sample Exercise 11.2 (cont.) Analyze: The standard enthalpy of formation of NaF is This value is the enthalpy change for the entire process, ΔHrxn. The fluorine in the equation has a coefficient of ½; this means that the energy needed to break a mole of F―F bonds to make atomic fluorine must be multiplied by ½. Following the same steps for part b: © 2012 by W. W. Norton & Company Sample Exercise 11.2 (cont.) This value is the enthalpy change for the entire process. For calcium, we must include both the first and second ionization energies because Ca loses two electrons when it forms Ca2+ ions. Since two fluorine atoms are needed to react with a single calcium atom, we do not need the factor of ½ in front of the term for energy to break a mole of F―F bonds. However, we need to multiply the electron affinity of F by 2 because two moles of fluorine atoms gain two moles of electrons to form two moles of fluoride ions. Figure 11.3 summarizes the Born–Haber cycles for calculating the lattice energies of NaF and CaF2. Based on the charges on the ions, their ionic radii, and Equation 11.4, the lattice energy of CaF2 should be greater than the lattice energy of NaF. © 2012 by W. W. Norton & Company Sample Exercise 11.2 (cont.) Solve: a. The overall enthalpy change in the reaction producing NaF is © 2012 by W. W. Norton & Company Sample Exercise 11.2 (cont.) b. The overall enthalpy change in the reaction producing CaF2 is Substituting the values given: Solving for U: © 2012 by W. W. Norton & Company Sample Exercise 11.2 (cont.) Think about It: Based on Equation 11.4, we predicted that the lattice energy of CaF2 would be larger than that of NaF. The calculated values of U confirm that © 2012 by W. W. Norton & Company Sample Exercise 11.3 Use the appropriate enthalpy of hydration values in Table 11.3 and ΔHsolution = 0.91 kJ/mol for NaF to calculate the lattice energy of NaF. Collect and Organize: The enthalpies of hydration for Na+ and F- are -418 kJ/mol and -502 kJ/mol, respectively. We can use the enthalpy of solution and Equation 11.5 to calculate the lattice energy. © 2012 by W. W. Norton & Company Sample Exercise 11.3 (cont.) Analyze: Before we can use Equation 11.5 to calculate UNaF, we must add the values of ΔHhydration for the two ions to obtain a value for ΔHhydration for NaF. From our calculations in Sample Exercise 11.2, we expect that the lattice energy of NaF should be close to -927 kJ/mol. Solve: Solving for the enthalpy of hydration of NaF: © 2012 by W. W. Norton & Company Sample Exercise 11.3 (cont.) Substituting ΔHhydration,NaF(aq) and ΔHsolution,NaF into Equation 11.5: Solving for UNaF: © 2012 by W. W. Norton & Company Sample Exercise 11.3 (cont.) Think about It: The value of UNaF calculated from ΔHsolution and ΔHhydration is within 1% of the value calculated in Sample Exercise 11.2 using different kinds of measurements, suggesting that the value calculated this way accurately reflects the lattice energy of NaF. © 2012 by W. W. Norton & Company Sample Exercise 11.4 The liquid used in automobile cooling systems is prepared by dissolving ethylene glycol (HOCH2CH2OH, molar mass 62.07 g/mol) in water. What is the vapor pressure of a solution prepared by mixing 1.000 L of ethylene glycol (density 1.114 g/mL) with 1.000 L of water (density 1.000 g/mL) at 100.0°C? Assume that the mixture obeys Raoult’s law. © 2012 by W. W. Norton & Company Sample Exercise 11.4 (cont.) Collect and Organize: We are asked to calculate the vapor pressure of a solution of ethylene glycol in water. The vapor pressure of a solution is a colligative property that depends on the number of solute particles, and hence the concentration. Volume and density must be used to determine the concentration of ethylene glycol. Ethylene glycol is a nonvolatile solute and its solution obeys Raoult’s law, which means it behaves ideally. The vapor pressure curves in Figure 11.7 show that the vapor pressure of pure water at 100°C (its normal boiling point) is 1.00 atm, whereas that of ethylene glycol is less than 0.05 atm. © 2012 by W. W. Norton & Company Sample Exercise 11.4 (cont.) Analyze: We have a mixture of equal volumes of two liquids. The solvent is the one present in the greater number of moles. We can determine the numbers of moles of each by converting their volumes into masses by multiplying by their densities and then calculate the number of moles of each by dividing by their molar masses. From these values we can decide which liquid is the solvent and calculate its mole fraction in the mixture. © 2012 by W. W. Norton & Company Sample Exercise 11.4 (cont.) Solve: For ethylene glycol: For water: © 2012 by W. W. Norton & Company Sample Exercise 11.4 (cont.) The mole fraction for water is Ethylene glycol is essentially nonvolatile, so the vapor pressure of the solution is due only to the solvent, and Psolvent = PH2O = 1.00 atm. Using this value and the calculated mole fraction of water in the mixture yields © 2012 by W. W. Norton & Company Sample Exercise 11.4 (cont.) Think about It: The presence of a nonvolatile solute causes the vapor pressure of the solution to be less than 1 atm, giving us confidence in our result. In addition, our selection of water as the solvent is justified because it is the major component in terms of numbers of moles of substances present: 55.49 moles of water with 17.95 moles of ethylene glycol dissolved in it. © 2012 by W. W. Norton & Company Sample Exercise 11.5 At its normal boiling point of 126°C, n-octane, C8H18, has a vapor pressure of 760 torr. What is its vapor pressure at 25°C? The enthalpy of vaporization of n-octane is 39.07 kJ/mol. Collect and Organize: The vapor pressure of octane at any temperature can be calculated using Equation 11.7, the Clausius–Clapeyron equation, given the enthalpy of vaporization and the vapor pressure of octane at its normal boiling point. © 2012 by W. W. Norton & Company Sample Exercise 11.5 (cont.) Analyze: We can use the Clausius–Clapeyron equation, but to do so we must convert the given temperature to kelvins. Because we will use 8.314 J/(mol · K) for R, we must also convert ΔHvap to joules per mole. Because vapor pressure decreases as temperature decreases, we expect the vapor pressure of octane to be significantly lower at 25°C than at its boiling point, 126°C. Solve: The two temperatures are © 2012 by W. W. Norton & Company Sample Exercise 11.5 (cont.) Inserting these values in Equation 11.7: © 2012 by W. W. Norton & Company Sample Exercise 11.5 (cont.) Think about It: We expect octane to have a low vapor pressure at a temperature much lower than its boiling point, so this number seems reasonable. © 2012 by W. W. Norton & Company Sample Exercise 11.6 How would the graph in Figure 11.11 be different if (a) we started with a mixture of 75 mL of n-octane (C8H18) and 25 mL of n-heptane (C7H16) and (b) we started with a mixture of 75 mL of n-octane (C8H18) and 25 mL of n-nonane (C9H20)? The normal boiling points of n-heptane, n-octane, and n-nonane are 98°C, 126°C, and 151°C, respectively. Collect and Organize: The graph in Figure 11.11 is an idealized plot of the temperature of the solution as a function of the volume of distillate for a 50:50 mixture of C7H16:C8H18. We are asked to describe the changes in the graph if we change the ratio of the components and if we change the identity of one of the components. We are given the normal boiling points of all compounds. © 2012 by W. W. Norton & Company Sample Exercise 11.6 (cont.) Analyze: Fractional distillation separates solutions of volatile liquids into pure substances on the basis of their different boiling points. The components distill in the order of their boiling points, with the lowest-boiling component distilling first. Ideally the total volume of distillate equals the total volume of the solution, and the volume of each fraction reflects the volume of the component in the original solution. © 2012 by W. W. Norton & Company Sample Exercise 11.6 (cont.) Solve: a. The first component that distills is the lower-boiling n-heptane. If our system were perfect, 25 mL of nheptane would distill at 98°C. When the n-heptane was completely removed from the solution, the only remaining component (75 mL of n-octane) would distill at 126°C. The idealized graph (Figure 11.13) shows the boiling points of the two substances along the yaxis and the volume of each distilled along the x-axis. The lengths of the horizontal lines at 98°C and 126°C are in a 1:3 ratio reflecting the composition of the mixture, 25 mL of C7H16 and 75 mL of C8H18. © 2012 by W. W. Norton & Company Sample Exercise 11.6 (cont.) b. The first component that distills in the second mixture is the lower boiling n-octane. If our system were perfect, 75 mL of n-octane would distill at 126°C. Once the n-octane is completely removed from the solution, the only remaining component (25 mL of nnonane) would distill at 151°C. The idealized graph (Figure 11.14) shows the boiling points of the two substances along the y-axis and the volume of distillate along the x-axis. © 2012 by W. W. Norton & Company Sample Exercise 11.6 (cont.) Think about It: Our graphs show the best results we could achieve. In a real world system, a small fraction that is a mixture of two hydrocarbons distills at temperatures between the boiling points of the two pure components. © 2012 by W. W. Norton & Company Sample Exercise 11.7 Calculate the vapor pressure of a solution prepared by dissolving 13 g of n-heptane (C7H16) in 87 g of n-octane (C8H18) at 25°C. By what factor does the concentration of the more volatile component in the vapor exceed the concentration of this component in the liquid? The vapor pressures of n-octane and n-heptane at 25°C are 11 torr and 31 torr, respectively. Collect and Organize: We can use Equation 11.8 to determine the total vapor pressure of the solution from the vapor pressures of the components after we determine the composition of the solution in terms of mole fractions. © 2012 by W. W. Norton & Company Sample Exercise 11.7 (cont.) Analyze: To calculate mole fractions, we need the molar masses of n-heptane and n-octane. The mole fraction is then equal to the number of moles of each component divided by the total number of moles of material in the solution. The vapor pressure of the solution must lie between the vapor pressures of the pure compounds. © 2012 by W. W. Norton & Company Sample Exercise 11.7 (cont.) Solve: The number of moles of each component is The mole fraction of each component in the mixture is © 2012 by W. W. Norton & Company Sample Exercise 11.7 (cont.) Using these mole fraction values and the vapor pressures of the two hydrocarbons in Equation 11.8, we have To calculate how enriched the vapor phase is in the more volatile component (the component with the greater vapor pressure, nheptane), we need to recall Dalton’s law of partial pressures (Section 6.7) and the concept that the partial pressure of a gas in a mixture of gases is proportional to its mole fraction in the mixture. Therefore, the ratio of the mole fraction of n-heptane to that of noctane is the ratio of their two vapor pressures: © 2012 by W. W. Norton & Company Sample Exercise 11.7 (cont.) The mole ratio of n-heptane to n-octane in the liquid mixture is Therefore, the vapor phase is enriched in n-heptane by a factor of © 2012 by W. W. Norton & Company Sample Exercise 11.7 (cont.) Think about It: As expected, the vapor pressure of the mixture (13.9 torr) is between the vapor pressures of the separate components. This result illustrates how fractional distillation works. The vapor is enriched in the lower-boiling component. © 2012 by W. W. Norton & Company Sample Exercise 11.8 How many grams of Na2SO4 should be added to 275 mL of water to prepare a 0.750 m solution of Na2SO4? Assume the density of water is 1.000 g/mL. Collect and Organize: We are asked to determine the mass of a solute, sodium sulfate, needed to prepare a 0.750 m solution. Molality is moles of solute per kilogram of solvent. The volume of water, the solvent, is 275 mL. © 2012 by W. W. Norton & Company Sample Exercise 11.8 (cont.) Analyze: Following the procedure shown in Figure 11.18, we use the density of water to convert 275 mL of water to kilograms. We can convert molality into a number of moles of Na2SO4 needed, and then convert that number into grams using the molar mass of Na2SO4. Our goal is to prepare a relatively small volume (~ ⅓ L) of a dilute solution (< 1 m), so we predict that the mass of solute needed will probably be less than mol of solute, which has a molar mass of 142 g/mol, or less than 50 g. © 2012 by W. W. Norton & Company Sample Exercise 11.8 (cont.) Solve: Multiplying 275 mL of water by the density of water, we find that the mass of water is We can rearrange Equation 11.9 to determine the number of moles of solute needed: © 2012 by W. W. Norton & Company Sample Exercise 11.8 (cont.) The molar mass of Na2SO4 is 142.04 g/mol, so the number of grams of Na2SO4 needed is Dissolving 29.3 g of Na2SO4 in 275 mL of water produces a 0.750 m solution. © 2012 by W. W. Norton & Company Sample Exercise 11.8 (cont.) Think about It: The calculated value of 29 g Na2SO4 is consistent with our prediction that less than 50 g of solute would be required. © 2012 by W. W. Norton & Company Sample Exercise 11.9 What is the boiling point of seawater if the concentration of ions in seawater is 1.15 m? Collect and Organize: We are asked to calculate the elevation in the boiling point of an aqueous solution in which the ion concentration is 1.15 m. The boiling-point-elevation constant of water is Kb = 0.52°C/m, and the normal boiling point of water is 100.0°C. © 2012 by W. W. Norton & Company Sample Exercise 11.9 (cont.) Analyze: The boiling point elevation is the product of the boilingpoint-elevation constant,0.52°C/m,and the solute concentration. Since the concentration is close to 1 m, we predict that the boiling point elevation will be close to 0.5°C. Solve: The temperature at which this seawater boils is 0.60°C higher than the normal boiling point of pure water: 100.0°C + 0.60°C = 100.6°C. © 2012 by W. W. Norton & Company Sample Exercise 11.9 (cont.) Think about It: As predicted, the boiling point of the solution is only slightly higher than the boiling point of the pure solvent. © 2012 by W. W. Norton & Company Sample Exercise 11.10 What is the freezing point of radiator fluid prepared by mixing 1.00 L of ethylene glycol (HOCH2CH2OH, density 1.114 g/mL) with 1.00 L of water (density 1.000 g/mL)? The freezing-point-depression constant of water, Kf, is 1.86°C/m. Collect and Organize: We are asked to determine the freezing point of a solution of ethylene glycol in water. We are given the volumes of the two liquids, their densities, and Kf for water, and we have Equation 11.11. © 2012 by W. W. Norton & Company Sample Exercise 11.10 (cont.) Analyze: To determine the freezing point, we need to calculate ΔTf for the solution and then subtract that value from water’s normal freezing point. Because using Equation 11.11 requires us to know m, the molality of the solution, we need to convert volumes of solute and solvent into moles of solute and kilograms of solvent; knowing their densities allows us to make both conversions, and we use ethylene glycol’s formula to calculate its molar mass. Based on Sample Exercise 11.4, we choose water as the solvent. It is a logical choice, because ethylene glycol has a much higher molar mass and a density very similar to water, so we can assume that 1 L of ethylene glycol contains fewer moles of material than 1 L of water. The relatively large volume (103 mL) and a density close to 1 g/mL for the solute allow us to predict that the mass of solute will be on the order of 103 g. With a solute molar mass of 62 g/mol, the molality of the solution should be between 10 and 20 m. Considering Kf = 1.86°C/m, we predict ΔTf to be in the range of 20° to 40°C. © 2012 by W. W. Norton & Company Sample Exercise 11.10 (cont.) Solve: The solvent mass is After calculating the ethylene glycol molar mass to be 62.07 g/mol, we have for the solute © 2012 by W. W. Norton & Company Sample Exercise 11.10 (cont.) Therefore the molal concentration is Using Equation 11.11 gives Subtracting this temperature change from the normal freezing point of water, 0.0°C, we have Freezing point of radiator fluid = 0.0°C - 33.3°C = -33.3°C © 2012 by W. W. Norton & Company Sample Exercise 11.10 (cont.) Think about It: The answer makes sense because the reason we add antifreeze to radiators is to depress the freezing point of water, and the solution we evaluated here certainly has a freezing point lower than that of pure water. To be an effective radiator fluid, the solution must have a freezing point sufficiently below the coldest expected temperatures, so our answer of -33°C is reasonable and matches our prediction. © 2012 by W. W. Norton & Company Sample Exercise 11.11 The salt lithium perchlorate (LiClO4) is one of the most water-soluble salts known. At what temperature does a 0.130 m solution of LiClO4 freeze? The Kf of water is 1.86°C/m; assume i = 2 for LiClO4 and the freezing point of pure water is 0.00°C. Collect and Organize: We are asked to determine the freezing point of a salt solution. We know the formula of the solute, its molal concentration, and the Kf of the solvent. We know that the freezing point of the pure solvent is 0.00°C. Equation 11.13 relates freezing point depression to solute concentration when we take the van’t Hoff factor into consideration. We are given the value of the van’t Hoff factor as i = 2. © 2012 by W. W. Norton & Company Sample Exercise 11.11 (cont.) Analyze: We need Equation 11.13 to solve for the freezing point depression. The value of Kf is close to 2, the value of i is 2, and the concentration of solute is close to 0.1 m, so we predict that the freezing point of the solution will be about 0.4°C lower than that of pure water. © 2012 by W. W. Norton & Company Sample Exercise 11.11 (cont.) Solve: The freezing point of the solution is (0.00 - 0.48)°C = -0.5°C. Think about It: Lithium perchlorate dissolves in aqueous solution to form 2 moles of ions for every mole of solute that dissolves: 1 mole of Li+ cations and 1 mole of ClO4- anions, consistent with i = 2. The calculated value is consistent with our prediction. © 2012 by W. W. Norton & Company Sample Exercise 11.12 The experimentally measured freezing point of a 1.90 m aqueous solution of NaCl is -6.57°C. What is the value of the van’t Hoff factor for this solution? Is the solution behaving ideally, or is there evidence that solute particles are interacting with one another? The freezingpoint-depression constant of water is Kf = 1.86°C/m, and the freezing point of pure water is 0.00°C. Collect and Organize: We are asked to calculate the i factor for a solution of known molality. We are given the measured freezing point and the Kf value. Equation 11.13 relates ΔTf, Kf, the molality of the solution, and the value of i. We are also asked whether the solution is behaving ideally. © 2012 by W. W. Norton & Company Sample Exercise 11.12 (cont.) Analyze: Equation 11.13 tells us that We can rearrange the equation to solve for i before substituting the values for ΔTf, Kf, and m. If the solution behaves ideally, we would expect i = 2 for the strong electrolyte NaCl; however, given the concentration of NaCl (1.90 m), we predict a value less than 2. © 2012 by W. W. Norton & Company Sample Exercise 11.12 (cont.) Solve: Rearranging Equation 11.13 gives us That i is not an integer tells us that the solution is not behaving ideally. Ion pairs must be forming in solution. © 2012 by W. W. Norton & Company Sample Exercise 11.12 (cont.) Think about It: As predicted, the value of i for this solution is less than the theoretical value of 2. © 2012 by W. W. Norton & Company Sample Exercise 11.13 At the beginning of this section, we mentioned that the concentration of solutes in a red blood cell is about a third of that of seawater—more precisely, about 0.30 M. If red blood cells are bathed in pure water, they swell, as shown in Figure 11.22(c). Calculate the osmotic pressure at 25°C of red blood cells across the cell membrane from pure water. Collect and Organize: We are to calculate an osmotic pressure across a membrane that separates pure water from red blood cells. The osmotic pressure of a solution is related to the total particle concentration (0.30 M) and temperature (25°C) of the solution by Equation 11.14. We must convert the temperature to kelvins. © 2012 by W. W. Norton & Company Sample Exercise 11.13 (cont.) Analyze: Osmotic pressure is related to the total concentration of all particles in solution and the absolute temperature. The total particle concentration, 0.30 M, represents the iM term in Equation 11.14. Solve: First we need to convert the temperature in degrees Celsius to kelvins: T(°C) + 273 = T(K). Inserting the values of iM, R, and T in Equation 11.14 we have © 2012 by W. W. Norton & Company Sample Exercise 11.13 (cont.) Think about It: The calculated pressure across the membranes of red blood cells in pure water is over 7 atmospheres: enough to rupture the membranes. © 2012 by W. W. Norton & Company Sample Exercise 11.14 Red blood cells placed in seawater shrivel as shown in Figure 11.22(b). Calculate the osmotic pressure across the semipermeable cell membrane, separating the solution inside a red blood cell from seawater at 25°C if the total concentration of all the particles inside the cell is 0.30 M and the total concentration of ions in seawater is 1.15 M. Compare the result with the answer from Sample Exercise 11.13. Collect and Organize: We are to calculate an osmotic pressure across a membrane that separates two solutions with different total concentrations of particles at a constant temperature of 25°C. We can use Equation 11.15 to calculate the difference in osmotic pressure between the two solutions. As in Sample Exercise 11.13, we must convert the temperature to units of kelvins. © 2012 by W. W. Norton & Company Sample Exercise 11.14 (cont.) Analyze: We know that the concentration of particles in seawater is greater than inside a red blood cell, so solvent will flow from the cell to the seawater. Thus, the osmotic pressure in the cell (πcell) will be less than the osmotic pressure of the seawater (πseawater) and the osmotic pressure across the membrane will be the difference between πseawater and πcell, or Δπ. The value of Δπ should be greater than the π calculated in Sample Exercise 11.13, where red blood cells were bathed in pure water because the difference in solute concentrations is greater. © 2012 by W. W. Norton & Company Sample Exercise 11.14 (cont.) Solve: First we need to convert the temperature in degrees Celsius to kelvins: T(°C) + 273 = T(K). Inserting the values of iM, R, and T in Equation 11.15 for seawater and cells we have © 2012 by W. W. Norton & Company Sample Exercise 11.14 (cont.) Think about It: As predicted, the calculated pressure across a semipermeable membrane separating seawater from red blood cells, 21 atm, is greater than theosmotic pressure across a semipermeable membrane separating red blood cells from pure water (7.3 atm). © 2012 by W. W. Norton & Company Sample Exercise 11.15 What is the reverse osmotic pressure required at 20°C to purify brackish well water containing 0.355 M dissolved particles if the product water is to contain no more than 87 mg of dissolved solids (as NaCl) per liter? Collect and Organize: We are asked to calculate the reverse osmotic pressure needed to purify water to a stated solute concentration. We are given the temperature, the solute concentration in the water to be purified, and the amount of solute (expressed as NaCl) tolerable in the product water. We can adapt Equation 11.15 to calculate the difference in osmotic pressure between the two solutions. © 2012 by W. W. Norton & Company Sample Exercise 11.15 (cont.) Analyze: We need the molarity of the less concentrated solution, the drinkable water, to calculate the osmotic pressure exerted by the drinkable water and we can calculate that molarity from the information given. We can also calculate the osmotic pressure once we know the difference in the molarities of the solutions on the two sides of the semipermeable membrane. © 2012 by W. W. Norton & Company Sample Exercise 11.15 (cont.) Solve: First we convert 87 mg NaCl/L to molarity: The total ion concentration for a 1.5 ×10-3 M NaCl solution is 2(1.5 ×10-3 M) = 0.0030 M. Using Equation 11.15: © 2012 by W. W. Norton & Company Sample Exercise 11.15 (cont.) If we maintain an osmotic pressure of exactly 8.47 atm on the side of the membrane with the brackish well water, no net flow of solvent will occur between the two solutions. Pressures greater than 8.47 atm will force water molecules from the well water through the membrane to give water containing less than 87 mg NaCl per liter. (The reason we can never obtain absolutely pure water in this process is that, in practice, some Na+ and Cl- ions pass through the membrane from the well water to the drinkable water.) © 2012 by W. W. Norton & Company Sample Exercise 11.15 (cont.) Think about It: In the absence of any external pressure, solvent will flow from the product water to the well water, driven by the concentration difference between the two solutions as shown in Figure 11.23. However, if we apply sufficient external pressure we can reverse the flow of water. The value of the external pressure, about 8.5 atm, represents the minimum external pressure that must be applied to make this device function. © 2012 by W. W. Norton & Company Sample Exercise 11.16 Eicosene is a molecular compound and nonelectrolyte with the empirical formula CH2. The freezing point of a solution prepared by dissolving 100 mg of eicosene in 1.00 g of benzene was 1.75°C lower than the freezing point of pure benzene. What is the molar mass of eicosene? (Kf for benzene is 4.90°C/m.) Collect and Organize: We are asked to determine the molar mass of a compound. We are given the mass of the compound that lowers the freezing point of a solvent by a known amount, and we have the Kf of the solvent. Because the solute is a nonelectrolyte, i = 1. Equation 11.13 relates concentration to the change in freezing point. © 2012 by W. W. Norton & Company Sample Exercise 11.16 (cont.) Analyze: The molar mass of a compound is expressed in units of grams per mole. Our sample consists of 100 mg, or 0.100 g. To find the molar mass we need to determine how many moles of eicosene are contained in the sample. The concentration term in Equation 11.13 is molality, or moles of solute per kilogram of solvent, so we can use the experimental data to first calculate molality. When we know the molality of the eicosene, we know the number of moles of eicosene in 1 kg of benzene; we are given the mass of benzene used, so we can calculate the number of moles of eicosene in the sample, from which we may calculate the molecular mass of eicosene. The empirical formula of eicosene enables us to check the validity of our answer as the molar mass of eicosene must be a whole-number multiple of the mass of CH2 (14 g/mol). © 2012 by W. W. Norton & Company Sample Exercise 11.16 (cont.) Solve: The molality of the eicosene solution is which means 0.357 mol of eicosene is dissolved per kilogram of solvent. Only 1.00 g (1.00 × 10-3 kg) was used in this sample. Calculating the moles of eicosene in the sample: © 2012 by W. W. Norton & Company Sample Exercise 11.16 (cont.) Because the molar mass is the mass of one mole of eicosene, and 100 mg of eicosene was used to prepare the solution, Think about It: The molar mass of eicosene, 280 g/mol, is reasonable since it corresponds to (CH2)n where n = 20. The molecular formula of eicosene is therefore C20H40. © 2012 by W. W. Norton & Company Sample Exercise 11.17 A molecular compound that is a nonelectrolyte was isolated from a South African tree. A 47 mg sample was dissolved in water to make 2.50 mL of solution at 25°C, and the osmotic pressure of the solution was 0.489 atm. Calculate the molar mass of the compound. Collect and Organize: We are given the mass of a substance, the volume of its aqueous solution, the temperature, and its osmotic pressure. We can relate these parameters using Equation 11.14, using the value i = 1 because this is a nonelectrolyte. © 2012 by W. W. Norton & Company Sample Exercise 11.17 (cont.) Analyze: We can calculate the molar concentration of the solution from the information given and Equation 11.14. Since we know the solution volume, we can calculate the number of moles of the solute from the molarity. Because we know the mass of this number of moles, we can calculate the molar mass of the solute. © 2012 by W. W. Norton & Company Sample Exercise 11.17 (cont.) Solve: Rearranging Equation 11.14 to isolate M and substituting the given values: Next we solve the defining equation for molarity, M = n/V, for n, the number of moles of solute: © 2012 by W. W. Norton & Company Sample Exercise 11.17 (cont.) We know that this number of moles of solute has a mass of 47 mg. The molar mass of the solute is therefore © 2012 by W. W. Norton & Company Sample Exercise 11.17 (cont.) Think about It: One of the advantages of determining molar mass by osmotic pressure is that only a small amount of material is required. With only a 47 mg sample of a rather large molecule (molar mass 940 g/mol), the osmotic pressure is sufficiently large (0.489 atm) to enable us to calculate the molar mass accurately. © 2012 by W. W. Norton & Company