Chapter 13
Answers to Questions
1.
Strategy
Mass Na2SO4 → mol Na2SO4
Mol/vol → conc
Mol Na2 SO4 = 40.0 g Na2 SO4 ×
Concentration =
Relationship
1 mol ≡ 142.1 g
c = n/V
1 mol
= 0.281 mol Na2 SO4
142.1 g
0.281 mol
= 0.562 mol ∙ L−1
0.5000 L
2.
Strategy
Mass Fe(NO3)3 → mol Fe(NO3)3
Mol/vol → conc
Relationship
1 mol ≡ 241.8 g
c = n/V
Mol Fe(NO3 )3 = 0.850 g Fe(NO3 )3 ×
1 mol
= 3.52 × 10−3 mol Fe(NO3 )3
241.8 g
Concentration =
3.52 × 10−3 mol
2.50 × 10−2 L
= 0.141 mol ∙ L−1
3.
Strategy
Vol/conc → mol
Mol MgCl2 → mass MgCl2
Relationship
n=cV
1 mol ≡ 95.2 g
Mol = 1.50 mol ∙ L−1 × 0.125 L = 0.188 mol
Mass MgCl2 = 0.188 mol MgCl2 ×
95.2 g
= 17.9 g MgCl2
1 mol
4.
Strategy
Vol/conc → mol
Mol Cu(NO3)2 → mass Cu(NO3)2
Relationship
n=cV
1 mol ≡ 187.5 g
Mol = 2.25 mol ∙ L−1 × 0.125 L = 0.281 mol
Mass Cu(NO3 )2 = 0.281 mol Cu(NO3 )2 ×
187.5 g
= 52.7 g Cu(NO3 )2
1 mol
5.
Strategy
Mass AlCl3 → mol AlCl3
Mol AlCl3 → mol Cl
Mol/vol → conc
Mol AlCl3 = 5.64 g AlCl3 ×
−
Mol Cl = 4.22 × 10
Concentration =
−2
Relationship
1 mol ≡ 133.5 g
1 mol AlCl3 ≡ 3 mol Cl
c = n/V
1 mol
= 4.22 × 10−2 mol AlCl3
133.5 g
3 mol Cl−
mol AlCl3 ×
= 0.127 mol Cl−
1 mol AlCl3
0.127 mol
2.00 L
= 6.34 × 10
−2
mol ∙ L−1
6.
Strategy
Mass K2CO3 → mol K2CO3
Mol K2CO3 → mol K+
Mol/vol → conc
Mol K 2 CO3 = 12.4 g K 2 CO3 ×
Relationship
1 mol ≡ 138.2 g
1 mol K2CO3 ≡ 2 mol K+
c = n/V
1 mol
= 8.97 × 10−2 mol K 2 CO3
138.2 g
Mol K + = 8.97 × 10−2 mol K 2 CO3 ×
Concentration =
0.179 mol
1.25 L
2 mol K +
= 0.179 mol K +
1 mol K 2 CO3
= 0.144 mol ∙ L−1
7.
Strategy
Vol/conc → mol
Mol AgNO3 → mass AgNO3
Relationship
n=cV
1 mol ≡ 169.9 g
Mol = 0.1000 mol ∙ L−1 × 0.2500 L = 2.500 × 10−2 mol
Mass AgNO3 = 2.500 × 10−2 mol AgNO3 ×
169.9 g
= 4.248 g AgNO3
1 mol
8.
Strategy
Vol/conc → mol
Mol KMnO4 → mass KMnO4
Relationship
n=cV
1 mol ≡ 158.0 g
Mol = 0.2000 mol ∙ L−1 × 0.1000 L = 2.000 × 10−2 mol
Mass KMnO4 = 2.000 × 10−2 mol KMnO4 ×
9.
158.0 g
= 3.160 g KMnO4
1 mol
Only for dilution problems, the following short-cut formula can be used:
cconc  vconc = cdil  vdil
cdil =
10.
cconc × vconc
16.0 mol ∙ L−1 × (2.50 × 10−2 L)
=
= 0.400 mol ∙ L−1
vdil
1.00 L
Only for dilution problems, the following short-cut formula can be used:
cconc  vconc = cdil  vdil
cdil =
11.
cconc × vconc
5.00 mol ∙ L−1 × (5.00 × 10−2 L)
=
= 1.00 mol ∙ L−1
vdil
0.2500 L
Only for dilution problems, the following short-cut formula can be used:
cconc  vconc = cdil  vdil
vconc =
12.
cdil × vdil
1.00 mol ∙ L−1 × (0.2500 L)
=
= 4.17 × 10−2 L = 41.7 mL
cconc
6.00 mol ∙ L−1
Only for dilution problems, the following short-cut formula can be used:
cconc  vconc = cdil  vdil
vdil =
cconc × vconc
5.00 mol ∙ L−1 × (5.00 × 10−2 L)
=
= 0.250 L = 250. mL
cdil
1.00 mol ∙ L−1