Ch. 13 Properties of Solutions (19 problems)
Checked 13.50, 13.54, and 13.58 for calculations.
13.2) (a) Cations that have a high charge and a small radius tend to be strongly
hydrated.
(b) Ionic substances with high lattice energies tend be less soluble in water than
those with lower lattice energies because a large amount of energy must be added
at the beginning of the solution process to separate the ionic solid into ions.
13.4) Increasing magnitude of solute-solvent interactions:
(b)
<
(c)
<
CH2Cl2 in benzene
methanol in H2O
nonpolar in nonpolar
polar in polar
London dispersion forces
hydrogen bonding
(a)
KCl in H2O
ionic in polar
ion-dipole forces
13.5) When NH4NO3 is dissolved in water, energy is added as the hydrogen bonds of
the water are broken and the NH4NO3 dissociates. Energy is then released as iondipole forces are formed. Because the heat of solution of NH4NO3 is positive,
more energy is added overall than released.
13.8) Although the dissolving of KBr is endothermic, its solubility is still relatively
high. This must be due to a large increase in entropy which drives the reaction to
be spontaneous.
13.10) Given: solubility of MnSO4H2O at 20C is 70 g/100 mL of H2O
(a) 1.22 M: saturated, unsaturated, or supersaturated?
70 g MnSO4H2O x 1 mol MnSO4H2O______ = 4 M
0.100 L sol’n
169.016 g MnSO4H2O
Assume no significant change in volume from adding solute
Since 1.22 M < 4 M, the sol’n is unsaturated.
(b) Solubility tests: add crystal of MnSO4H2O
-if it dissolves, then the sol’n is unsaturated
-if it does not dissolve, then the sol’n is saturated
-if it causes crystallization, then the sol’n was supersaturated
13.15) more soluble in water:
(a) CH3CH2CH2OH or CH3CH2OH
shorter hydrocarbon chain and thus more polar
(b) CCl4 or CaCl2
Ionic, not nonpolar covalent
(c) benzene, C6H6, or phenol, C6H5OH
has hydroxyl group for hydrogen bonding
13.18) Pressure affects the solubility of O2 because it is a gas and can be forced into
solution via compression. Pressure does not affect the solubility of NaCl because
it is a solid and cannot be compressed.
13.20) M = ? of O2 if PO2 = 0.21 atm and 20C
Cg = kPg
= (1.38 x 10-3 M/atm)(0.21 atm)
= 2.9 x 10-4 M
13.22) (a) mass % = ? if 0.055 mol I2 in 105 g of CCl4
0.055 mol I2 x (253.82 g I2/1 mol I2) = 13.9601 g I2
mass % = mass solute x 100
mass of sol’n
=_____13.9601 g I2_______ x 100
(13.961 g I2 + 105 g CCl4)
= 12 % (2 SF)
(b)
ppm = mass of solute x 106
mass of sol’n
= ______0.0079 g Sr2+______ x 106
(1000 g H2O + 0.0079 g Sr2+)
= 7.9 ppm (2 SF)
13.26) calculate molarity
(a) 15.0 g CaCl2 in 0.350 L sol’n
15.0 g CaCl2 x (1 mol CaCl2/110.98 g CaCl2) = 0.35159488 mol CaCl2
M = mol solute = 0.35159488 mol CaCl2 = 0.386 M (2 SF)
liters sol’n
0.350 L
(c) 35.0 mL of 6.00 M H2SO4 diluted to 250 mL
M = mol solute
liters sol’n
6.00 M = ___x___
0.0350 L
x = 0.21 mol H2SO4
M = mol solute
liters sol’n
M = 0.21 mol = 0.84 M (2 SF)
0.250 L
OR
M1V1 = M2V2
(6.00 M)(35.0 mL) = M2(250 mL)
M2 = 0.84 M
13.28) (a) m = ? if 1.80 mol KCl in 16.0 mol H2O
16.0 mol H2O x (18.016 g H2O/1 mol H2O) = 288.256 g H2O
m = mol solute
kg solvent
= 1.80 mol KCl
0.288256 kg H2O
= 6.24 m (3 SF)
(b) g S8 = ? if 0.16 m sol’n w/ 100.0 g C10H8
m = mol solute
kg solvent
0.16 m = _____x______
0.1000 kg C10H8
x = 0.016 mol S8 x (256.48 g S8/1 mol S8) = 4.1 g S8
13.30) Given: 80.5 g C6H8O6 in 210 g H2O; d = 1.22 g/mL
(a) mass % = ?
mass % = mass solute x 100
mass of sol’n
= _____80.5 g C6H8O6__________x 100
(80.5 g C6H8O6 + 210 g H2O)
= 28 % (2 SF)
(b) mol fraction = ?
80.5 g C6H8O6 X (1 mol C6H8O6/176.124 g C6H8O6) = 0.457064 mol C6H8O6
210 g H2O x (1 mol H2O/18.016 H2O) = 11.6563 mol H2O
X = mol solute
mol sol’n
= 0.457064 mol C6H8O6
(0.457064 mol + 11.6563 mol)
= 0.038 (2 SF)
(c) molality = ?
m = mol solute
kg solvent
= 0.457064 mol C6H8O6
0.210 kg H2O
= 2.2 m (2 SF)
(d) molarity = ?
80.5 g C6H8O6 + 210 g H2O = 290.5 g sol’n x (1 mL sol’n/1.22 g sol’n) =
238.11475 mL sol’n
M = mol solute
liters sol’n
= 0.457064 mol C6H8O6
0.23811475 L sol’n
= 1.9 M (2 SF)
13.40) Given: 0.0850 m C8H10N4O2 in CHCl3
(a) % mass = ?
0.0850 m = 0.0850 mol C8H10N4O2/1 kg CHCl3
0.0850 mol C8H10N4O2 x (194.2 g C8H10N4O2/1 mol C8H10N4O2) = 16.507 g C8H10N4O2
mass % = mass solute x 100
mass of sol’n
=_____16.507 g C8H10N4O2_______ x 100
(16.507 g + 1000 g)
= 1.62 % (3 SF)
(b) mole fraction = ?
1000 g CHCl3 x (1 mol CHCl3/119.368 g CHCl3) = 8.37745 mol CHCl3
X = mol solute
mol sol’n
= 0.0850 mol C6H8O6
(0.0850 mol + 8.37745 mol)
= 0.0100 (3 SF)
13.42) Increasing concentration of nonvolatile solute to water:
(a) vapor pressure decreases
(b) freezing point decreases
(c) boiling point increases
(d) osmotic pressure increases
13.46) (a) 32.5 g C3H8O3 x (1 mol C3H8O3/92.094 g C3H8O3) = 0.352900 mol C3H8O3
140 g H2O x (1 mol H2O/18.016 g H2O) = 7.770887 mol H2O
PH2O = XH2OPH2O = (7.770887 mol/(7.770887 mol + 0.352900 mol))((233.7 torr)
= 220 torr (2 SF)
13.50) (b) f.p. = ?, b.p. = ? if 18.0 g C10H22 (solute) in 425 g CHCl3 (solvent)
18.0 g C10H22 x (1 mol C10H22/142.276 g C10H22) = 0.126515 mol C10H22
m = mol solute
kg solvent
= 0.126515 mol C10H22
0.425 kg CHCl3
= 0.29768 m
Tf = iKfm = (1)(4.68 C/m)(0.29768 m) = 1.3931 C
f.p. = -63.5 C - 1.3931 C = -64.9 C
Tb = iKbm = (1)(3.63 C/m)( 0.29768 m) = 1.080578 C
b.p. = 61.2 C + 1.080578 C = 62.3 C
13.52) decreasing freezing point: (fewest particles) 0.075 M glucose (van’t Hoff = 1) >
0.030 M Zn(NO3)2 (van’t Hoff = 3) > 0.075 M LiBr (van’t Hoff = 2) (most
particles)
13.54)  = ? at 20 C if 3.4 g NaCl/1 L sol’n
3.4 g NaCl x (1 mol NaCl/58.44 g NaCl) = 0.058179 mol NaCl
 = iMRT
= (2)(0.058179 mol NaCl/1 L sol’n)(0.0821 Latm/molK)(293 K)
= 2.8 atm (2 SF)
13.58) molar mass = ? if 22.3 g of nonelectrolyte solute in 0.250 L sol’n,  = 2.12 atm at 25 C
 = iMRT
2.12 atm = (1)(0.0821 Latm/molK)(298 K)
M = 0.08665157 M
M = mol sol
L sol’n
0.08665157 M = x/0.250 L
x = 0.02166289 mol
molar mass = 22.3 g/ 0.02166289 mol = 1030 g/mol (3 SF)