Chapter 13 Suggested Homework Answers
Exercises: 14, 28, 36, 38, 42, 50, 60, 62, 66, 68, 72, 74
14. (a) KCl dissolved in water is an example of an electrolyte dissolved in water. The principal type of solute-solvent interaction for an electrolyte dissolved in water is ion-dipole interactions.
(b) CH2Cl2 dissolved in benzene (C6H6) is an example of a slightly polar molecule (CH2Cl2) dissolved in a nonpolar solvent (C6H6). The principal solute-solvent interaction is
(c) Methanol (CH3OH) in water is an example of two hydrogen bonding compounds. The principal solute-solvent interaction is hydrogen bonding.
Since ion-dipole interactions are the strongest, the interactions in part (a) are the strongest. Hydrogen bonding is stronger than
28. Any polar molecule will more likely dissolve in a polar hydrogen bonding solvent such as water.
(a) Cyclohexane (C6H12) is a nonpolar hydrocarbon where glucose contains many –OH bonds which can hydrogen bond with water. Glucose is more likely to be soluble in water than cyclohexane.
(b) Propionic acid is a polar hydrogen bonding solvent but the salt sodium proprionate will involve ion-dipole interactions with water which will make sodium proprionate more soluble.
(c) HCl contains a much more polar H-Cl bond than the C-Cl bond in ethyl chloride. HCl will be more soluble in water since it is much more polar than ethyl chloride.
36. (a) Mole fraction of phenol is defined as follows:
Given that 25.5 g of phenol, C6H5OH, is dissolved in 495 g of ethanol, CH3CH2OH. Calculate the moles of each component:
The mole fraction of phenol is equal to:
(b) Mass percent of phenol is defined as:
Using the given masses, the mass percent of ethanol is equal to:
(c) Molality of ethanol is defined as:
The moles of ethanol has already been calculated in part (a). Calculate the number of kilograms of ethanol:
The molality of ethanol is equal to:
38. Molarity is defined as:
(a) Calculate the number of moles of Al2(SO4)3 in 25.0 g:
The molarity of Al2(SO4)3 is equal to:
(b) Calculate the number of moles of Mn(NO3)2·2H2O in 5.25 g. (Remember to include the two waters of hydration in the molar mass):
Calculate the volume of the solution in liters:
The molarity of the solution is equal to:
(c) Use the dilution formula to calculate the molarity of the diluted solution: McVc = MdVd. From the problem: Vd = 35.0 mL; Mc = 9.00 M; and Vc = 0.500 L. First, convert Vc into units of milliliters:
Solve the dilution formula for Md:
42. Given in the problem: 80.5 g of C6H8O6 dissolved in 210 g of water (H2O), density of solution is 1.22 g/mL
(a) The definition of mass percentage of ascorbic acid is:
(b) The definition of mole fraction of ascorbic acid is:
Calculate the number of moles of each component, ascorbic acid and water:
Using the definition of mole fraction:
(c) The definition of molality is:
The number of moles of ascorbic acid was calculated in part (b). Calculate the number of kilograms of water:
The molality of ascorbic acid is equal to:
(d) The definition of molarity is:
In order to find the volume of the solution, take the total mass of solution (291 g, calculated in part (a)) and divide by the solution’s density to find the volume of solution in mL, then convert it to units of liters:
Using the definition of molarity:
50. Assume that you have exactly one liter (1000 mL) of solution. The mass of 1 L of solution can be calculated by multiplying the density by 1000 mL:
The mass of NH3 contained in 1 L of solution may be calculated by using the mass percent of NH3:
Use the molar mass of NH3 to convert this mass into the number of moles of NH3 contained in 1 L of solution:
This is equal to the molarity of NH3 which is 15 M.
60. (a) The vapor pressure of water over a solution containing a nonvolatile solute is given by Raoult’s Law:
343 K is equal to 70°C. The vapor pressure of pure water at 70°C is equal to 233.7 torr (from Appendix B). In order to calculate the mole fraction of water, calculate the moles of glycerin and water in the solution:
Calculate the mole fraction of water in this solution:
The vapor pressure of water over this solution is equal to:
(b) If the vapor pressure of the solution is reduced by 10.0 torr at 35°C from the vapor pressure of pure ethanol, the vapor pressure over the solution will be equal to:
Using Raoult’s Law, calculate the mole fraction of ethanol in a solution that has a vapor pressure of ethanol equal to 90. torr:
In order to calculate how many grams of ethylene glycol need to be added to 1.00 kg of ethanol, convert 1.00 kg of ethanol to moles of ethanol and then calculate the number of moles of ethylene glycol needed to get a mole fraction of ethanol equal to 0.90:
Using the definition of mole fraction:
Convert to grams of ethylene glycol (C2H6O2):
62. (a) For a solution that contains two volatile components:
As given in the problem at 20°C, 
. If a solution has P = 35 torr at 20°C, the following two equations relate the mole fractions of each component:
The second equation is the result of the mole fractions in a mixture adding up to one. From the second equation, Xtoluene = 1.00–Xbenzene. Make this substitution for Xtoluene in the 1st equation and solve for Xbenzene:
The mole fraction of toluene is equal to: Xtoluene = 1.00 – Xbenzene = 1.00 – 0.25 = 0.75.
(b) The partial pressure of each component over the solution is given by Raoult’s Law:
The vapor phase is governed by
66. For electrolytes in water solution, the molality of the solute is multiplied by the number of ions that one mole of the electrolyte forms in water to find the effect of the solute on a colligative property. Glycerin (C3H8O3) contains only nonmetal atoms and is not an acid so it is a nonelectrolyte. Therefore, 0.040 m glycerin has the effect of a 0.040 m solution on colligative properties. KBr is an electrolyte that dissolves in water to form 2 ions per formula unit. Therefore, 0.020 m KBr will have the effect a little bi lower than a 0.040 m solution on colligative properties. Phenol (C6H5OH) is a nonelectrolyte. Therefore, a 0.030 m solution will have the effect of a 0.030 m solution. The greater the molality of solute, the larger the freezing point depression or the lower the freezing point of solution. Therefore, 0.040 m glycerin will probably have the lowest freezing point closely followed by 0.020 m KBr. Phenol will have the highest freezing point.
68. (a) For ethanol, the normal boiling point is 78.4°C and the normal freezing point is –114.6°C. The Kb value is 1.22°C/m and the Kf value is 1.99°C/m. The freezing point depression and the boiling point elevation for a solution that is 0.40 m in glucose is:
The boiling point of the solution is equal to: 78.4°C + 0.49°C = 78.9°C.
The freezing point of the solution is equal to: –114.6°C – 0.80°C = –115.4°C.
(b) For CHCl3, the normal boiling point is 61.2°C and the normal freezing point is –63.5°C. The Kb value is 3.63°C/m and the Kf value is equal to 4.68°C/m. First, calculate the molality of a solution prepared by dissolving 20.0 g of C10H22 in 45.5 g of CHCl3:
The freezing point depression and the boiling point elevation for the solution is:
The boiling point of the solution is equal to: 61.2°C + 11.2°C = 72.4°C.
The freezing point of the solution is equal to: –63.5°C – 14.5°C = –78.0°C.
(c) For water, the boiling point is 100.0°C and the freezing point is 0.0°C. The Kb value is equal to 0.51°C/m and the Kf value is equal to 1.86°C/m. Calculate the molalities of both solutes (150 g of water is equal to 0.150 kg):
Since KBr is an electrolyte that dissolves to produce 2 ions, it will have twice the effect so it will have an apparent molality of 2.0 m. The total molality of solute is equal to 5.0 m. The freezing point depression and the boiling point elevation for the solution is:
The boiling point of the solution is equal to: 100.0°C + 2.6°C = 102.6°C.
The freezing point of the solution is equal to: 0.0°C – 9.3°C = –9.3°C.
72. Calculate the freezing point depression for the solution:
DTf = 5.5°C – 4.1°C = 1.4°C
Using the Kf value for benzene, calculate the molality of lauryl alcohol in this solution:
Multiply the molality of lauryl alcohol by the number of kilograms of bezene to obtain the number of moles of lauryl alcohol in the solution:
Divide the number of grams of lauryl alcohol, given in the problem, by the number of moles of lauryl alcohol to calculate the molar mass:
74. Calculate the molarity of the solute by solving the van’t Hoff equation for the molarity. (T = 25°C +273 = 298 K):
Multiply the molarity by the volume of the solution in liters to obtain the number of moles of solute:
Divide the number of grams of compound, given in the problem, by the number of moles to get the molar mass:
