Chapter 13 In Class Problems
Period 1
13.3
In CH3(CH2)nCOOH, as n increases, the hydrophobic (CH) portion of the carboxylic acid increases and the
hydrophilic part of the molecule stays the same, with a resulting decrease in water solubility.
13.12
a) dispersion forces
b) hydrogen bonding
c) dispersion forces
13.14
a) CH3CH2-O-CH3(g), due to its smaller size (smaller molar mass).
b) CH2Cl2, because it is more polar than CCl4.
c) Tetrahydropyran is more water soluble due to hydrogen bonding between the oxygen atom and water
molecules.
13.28
a) Endothermic
b) The lattice energy term is much larger than the combined ionic heats of hydration.
c) The increase in entropy outweighs the increase in enthalpy, so ammonium chloride dissolves.
13.39
a) Entropy increases
b) Entropy decreases
c) Entropy increases
Period 2
13.43
An increase in temperature produces an increase in kinetic energy; the solute molecules overcome the weak
intermolecular forces, which results in a decrease in solubility of any gas in water. In nearly all cases, gases
dissolve exothermically (ΔHsoln < 0).
13.49
Solubility for a gas is calculated from Henry’s law: S gas = kH × Pgas.
Sgas = (3.7 x 10–2 mol/L•atm) (5.5 atm) = 0.2035 = 0.20 mol/L
13.51
Molarity is defined as the number of moles of solute dissolved in one liter of solution. Molality is defined
as the number of moles of solute dissolved in 1000 g (1 kg) of solvent. Molal solutions are prepared by
measuring masses of solute and solvent, which are additive and not changed by temperature, so the
concentration in molality does not change with temperature.
13.56 b
13.69
Period 3
13.86
Univalent ions behave more ideally than divalent ions. Ionic strength (which affects “activity”
concentration) is greater for divalent ions. Thus, 0.01 m NaBr has a freezing point that is closer to its
predicted value.
13.87
Cyclohexane, with a freezing point depression constant of 20.1°C/m, would make calculation of molar
mass of a substance easier, since ΔT f would be greater.
13.91
13.103