Chapter 10: Solutions
10. Refer to Section 10.1 and Examples 10.4 and 10.5.
Molality
Mass Percent
of Solvent
Ppm Solute
Mole Fraction
of Solvent
2.577
86.58%
1.340 x 105
0.9556
a.
b.
20.4
45.0
5.50 x 10
c.
0.07977
99.5232%
4768
d.
12.6
57.0%
4.30 x 10
5
0.731
0.9986
5
0.815
Since these are aqueous solutions, the solvent is water. The solute is urea, CO(NH2)2.
Molar mass of urea: (12.01) + 16.00 + 2(14.01 + 2(1.008)) = 60.06 g/mol.
a. 2.577 m indicates that there are 2.577 mol. of solute per 1000 g of solvent and the
calculations are based on that ratio.
2.577 mol. x
1 kg x
60.06 g urea
= 154.8 g urea
1 mol. urea
1000 g 1 mol. H 2 O
x
= 55.49 mol. H 2 O
1 kg
18.02 g H 2 O
mass % =
1000 g
1000 g
x 100% =
= 86.58%
(1000 g + 154.8 g)
1155 g
ppm solute =
X H 2O =
154.8 g
x 10 6 = 1.340 x 10 5
(1000 g + 154.8 g)
55.49 mol.
= 0.9556
(55.49 mol. + 2.577 mol.)
b. 45.0 mass % of solvent means that 45.0 g of solvent are present for each 100.0 g of
solution. Calculations are then based on this ratio.
45.0 g x
1 mol. H 2 O
= 2.50 mol. H 2 O
18.02 g
mass of solute = 100.0 g (total) – 45.0 g (water) = 55.0 g CO(NH2)2
55.0 g urea x
molality =
0.916 mol. urea
= 20.4 m (remember to convert mass of solvent to kg)
0.0450 kg H 2 O
ppm solute =
c.
1 mol. urea
= 0.916 mol. urea
60.06 g urea
55.0 g urea
x 106 = 5.50 x 105
100 g
X H 2O =
2.50 mol.
= 0.731
(2.50 mol. + 0.916 mol.)
ppm =
mass of solute
x 10 6 = 4768 g
mass of solution
If we assume a total mass of 106 g, then the mass of solute = 4768 g by definition. Any
assumption for mass is valid here, this one was chosen for simplicity.
4768 g urea x
1 mol. urea
= 79.39 mol. urea
60.06 g urea
10 6 g - 4768 g = 995232 g; 995232 g H 2 O x
X H 2O =
1 mol.
= 55244.0 mol. H 2 O
18.0152 g
55244.0 mol.
= 0.9986
(55244.0 mol. + 79.39 mol.)
mass % =
9.95232 x 10 5 g
x 100% = 99.5232%
10 6 g
molality =
79.39 mol.
= 0.07977 m
995.232 kg
d. Xsolvent = 0.815 indicates that there are 0.815 moles H2O per 1 mole of solvent and solute
combined. Consequently, there must be 0.185 mol. (1-0.815) urea.
0.185 mol. urea x
60.06 g urea
= 11.1 g urea
1 mol. urea
0.815 mol. H 2 O x
18.02 g
= 14.7 g H 2 O
1 mol.
0.185 mol. urea
= 12.6 m (remember to convert mass of solvent to kg)
0.0147 kg H 2 O
Mass % =
ppm =
mass of solvent
14.7 g
x 100% =
x 100% = 57.0%
total mass
(11.1 g + 14.7 g)
mass of solute
11.1 g urea
x 10 6 =
x 10 6 = 4.30 x 10 5
mass of solution
(11.1 g + 14.7 g)
20. Refer to Sections 10.2 and 9.3.
The compound which exhibits intermolecular forces most similar to water will be the more
soluble in water (like dissolves like). Recall that water has dispersion, dipole, and H-bonding
forces.
dispersion forces
a. CH3Cl:
CH3OH: dispersion, dipole, and H-bonding forces
CH3OH would be more soluble since it shares H-bonding with water.
dispersion and dipole forces
b. NI3:
KI:
ionic
KI would be more soluble because it is ionic and ionic compounds generally exhibit high
solubility in water.
c. LiCl:
ionic
dispersion and dipole forces
C2H5Cl:
LiCl would be more soluble because it is ionic and ionic compounds generally exhibit
high solubility in water.
dispersion, dipole, and H-bonding forces
d. NH3:
dispersion and dipole forces
CH4:
NH3 would be more soluble because of the H-bonding.
30. Refer to Section 10.3.
The first step is to calculate the mole fraction of oxalic acid.
∆ P = PHo 2 O - Psolution = 22.38 mm Hg - 21.97 mm Hg = 0.41 mm Hg
∆ P = ( X H 2 C 2 O 4 )( PHo 2 O ) ⇒ 0.41 mm Hg = ( X H 2 C 2 O 4 )( 22.38 mm Hg)
X H 2C 2O 4 = 0.018
Assuming 1 mole total, this means we have 0.018 mol. H2C2O4 and 0.982 mol. water
(1.00 - 0.018 = 0.982). The next step is to calculate the masses associated with these
quantities, and from that the mass of solution and volume of solution.
0.982 mol. H 2 O x
18.02 g H 2 O
= 17.7 g H 2 O
1 mol. H 2 O
0.018 mol. H 2 C 2 O 4 x
90.04 g H 2 C 2 O 4
= 1.6 g H 2 C 2 O 4
1 mol. H 2 C 2 O 4
(17.7 + 1.6) g solution x
1 mL
1L
x
= 0.0184 L
1.05 g 1000 mL
Now one can either:
1. Calculate molarity (mol./L) and convert mol./L to grams/L (using molecular mass) or
2. Directly calculate grams of H2C2O4 in one liter (as shown below)
1.00 L solution x
1.6 g H 2 C 2 O 4
= 87 g H 2 C 2 O 4
0.0184 L solution
Thus, to prepare the prescribed solution, one must dissolve 87 g H2C2O4 in enough water to
make 1.00 L of solution.
40. Refer to Sections 10.3 and 3.3, and Example 10.7.
Use the freezing point depression to calculate molality, and from that, moles of the compound
and the compound's molecular mass.
∆Tf = Tf° -Tf = 178.40°C – 173.44°C = 4.96°C
∆Tf = kfm
4.96°C = (40.0°C/m)(m)
m = 0.124 m
50.00 g camphor x
m=
1 kg
= 0.0500 kg camphor
1000 g
mol. solute
x mol.
⇒ 0.124 m =
kg solvent
0.05000 kg camphor
x = 0.00620 mol.
2.50 g
= 403 g/mol.
0.00620 mol.