# hw1 soln - CHAPTER 13 THE PROPERTIES OF MIXTURES: SOLUTIONS...

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13-1 CHAPTER 13 THE PROPERTIES OF MIXTURES: SOLUTIONS AND COLLOIDS FOLLOW-UP PROBLEMS 13.1 Plan: Compare the intermolecular forces in the solutes with the intermolecular forces in the solvent. The intermolecular forces for the more soluble solute will be more similar to the intermolecular forces in the solvent than the forces in the less soluble solute. Solution: a) 1,4-Butanediol is more soluble in water than butanol. Intermolecular forces in water are primarily hydrogen bonding. The intermolecular forces in both solutes also involve hydrogen bonding with the hydroxyl groups. Compared to butanol, each 1,4-butanediol molecule will form more hydrogen bonds with water because 1,4- butanediol contains two hydroxyl groups in each molecule, whereas butanol contains only one –OH group. Since 1,4-butanediol has more hydrogen bonds to water than butanol, it will be more soluble than butanol. b) Chloroform is more soluble in water than carbon tetrachloride because chloroform is a polar molecule and carbon tetrachloride is nonpolar. Polar molecules are more soluble in water, a polar solvent. 13.2 Plan: Solubility of a gas can be found from Henry’s law: S gas = k H × P gas . The problem gives k H for N 2 but not its partial pressure. To calculate the partial pressure, use the relationship from Chapter 5: P gas = X gas x P total where X represents the mole fraction of the gas. Solution: To find partial pressure use the 78% N 2 given for the mole fraction: 2 N P = 0.78 x 1 atm = 0.78 atm Use Henry’s law to find solubility at this partial pressure: 2 N S = 7 x 10 -4 mol/L•atm x 0.78 atm = 5.46 x 10 -4 mol/L = 5 x 10 -4 mol/L Check: The solubility should be close to the value of k H since the partial pressure is close to 1 atm. 13.3 Plan: Molality (m) is defined as amount (mol) of solute per kg of solvent. Use the molality and the mass of solvent given to calculate the amount of glucose. Then convert amount (mol) of glucose to mass ( g ) of glucose. Solution: Mass of solvent must be converted to kg: 563 g x (1 kg/1000 g) = 0.563 kg () 2 61 26 2.40 x 10 mol C H O 180.16 C H O 0.563 kg 1kg 1molC H O    = 2.4343 = 2.43 g C 6 H 12 O 6 Check: To check these results estimate the molality from the calculated mass of glucose and given mass of solvent. The number of moles of glucose is approximately (2.4 g) (1 mol / 180 g) = 0.013 mol glucose. Using the calculated value for amount of glucose and the given mass of solvent, calculate the molality: (0.013) / (0.56 kg) = 0.023 m, which is close to the given value of 0.0240 m. 13.4 Plan: Mass percent is the mass ( g ) of solute per 100 g of solution. For each alcohol, divide the mass of the alcohol by the total mass. Multiply this number by 100 to obtain mass percent. To find mole fraction, first find the amount (mol) of each alcohol, then divide by the total moles. Solution: Mass percent: Mass% propanol = 35.0 g x 100% 35.0 150. g + = 18.9189 = 18.9% propanol Mass% ethanol = 150. g x 100% 35.0 150. g + = 81.08108 = 81.1% ethanol

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13-2 Mole fraction: Moles propanol = () 37 1molC H OH 35.0 g C H OH 60.09 g C H OH    = 0.5824596 mol propanol (unrounded) Moles ethanol = 25 150. g C H OH 46.07 g C H OH = 3.2559149 mol ethanol (unrounded) X propanol = (0.5824596 mol propanol) / (0.5824596 + 3.2559149)mol = 0.151746 = 0.152 X ethanol = (3.2559149 mol ethanol) / (0.5824596 + 3.2559149)mol = 0.84825 = 0.848
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hw1 soln - CHAPTER 13 THE PROPERTIES OF MIXTURES: SOLUTIONS...

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