AP CHEMISTRY WKST KEY: SOLUTIONS—CONCENTRATIONS
1)
2)
Given: 2.255 g NaCl / 1.030 L H2O
m=
g solute
(MM solute)(kg solvent)
m=
2.255 g
= 0.03746 m NaCl
g
(58.44 mol
)(1.030 kg)
Given: 15.00 % C6H12O6 soln / 500.0 mL H2O
% solute
mass solute
=
100
mass solute + mass solvent
15.00
x
=
100
x + 500.0 g
100x = 15.00 x + 7500.
85.00x = 7500.
x = 88.24 g C6H12O6
3)
Given: 20.0 g H3PO4 / 750.0 mL soln / dsoln = 1.005 g/mL
20.0 g H3PO4 = 0.204 mol H3PO4
733.8 g H2O = 40.72 mol H2O
40.92 mol soln
M=
Vsoln = 750.0 mL
msoln = dsolnVsoln = (1.005 g/mL)(750.0 mL) = 753.8 g
mol solute
0.204 mol
=
= 0.272 M H3 PO4
L soln
0.7500 L
N = nM = (3)(0.272 M) = 0.816 N H3PO4
ΧH3PO4 =
4)
mol solute
0.204 mol
=
= 0.00499
mol soln
40.92 mol
Given: 14.0 % Na2CO3
-assume 100 g soln
14.0 g Na2CO3 = 0.132 mol Na2CO3
86.0 g H2O
= 4.77 mol H2O
4.90 mol soln
ΧNa2CO3 =
m=
mol solute
0.132 mol
=
= 0.0269
mol soln
4.90 mol
mol solute
0.132 mol
=
= 1.53 m Na2 CO3
kg solvent
0.0860 kg
1
AP CHEMISTRY WKST KEY: SOLUTIONS—CONCENTRATIONS
5)
Given: 12.4 ppm CHCl3
a)
12.4
6
10
=
x
100
x = 0.00124% CHCl3
b)
assume 106 g soln  12.4 g CHCl3 / 999987.6 g H2O (can assume this will also be the volume of soln)
M=
6)
g solute
12.4 g
=
= 1.04 x 10-4 M CHCl3
g
(MM solute)(L soln)
(119.37 mol
)(999.9876 L)
Given: 5.07 m C2H4(OH)2 / dsoln = 1.045 g/mL
-assume 1 kg solvent
315 g C2H4(OH)2 = 5.07 mol C2H4(OH)2
1000 g H2O
= 55.49 mol H2O
60.56 mol soln
mass soln = 315 g + 1000 g = 1315 g
Vsoln =
1315 g
g = 1258 mL soln
1.045 mL
% solute = (
mass solute
315 g
) 100 = (
) 100 = 24.0% C2 H4 (OH)2
mass soln
1315 g
Χ𝐶2𝐻4(𝑂𝐻)2 =
mol solute
5.07 mol
=
= 0.0837
mol soln
60.56 mol
M=
mol solute
5.07 mol
=
= 4.03 M C2 H4 (OH)2
L soln
1.258 L
N = nM = (2)(4.03 M) = 8.06 N C2H4(OH)2
7)
Given: 0.20% KMnO4 / dsoln = 1.00 g/mL / in KMnO4  Mn7+ ; 4e− + Mn7+ → Mn3+
-assume 100 g soln
0.20 g KMnO4
100 mL soln
M=
g solute
0.20 g
=
= 0.013 M KMnO4
g
(MM solute)(L soln)
(158.04 mol
)(0.100 L)
N = nM = (4)(0.013 M) = 0.052 N KMnO4
2
AP CHEMISTRY WKST KEY: SOLUTIONS—CONCENTRATIONS
8)
Given: ΧMgCl2 = 0.125
-assume 1 mol soln
11.9 g MgCl2 = 0.125 mol MgCl2
15.8 g H2O = 0.875 mol H2O
m=
mol solute
0.125 mol
=
= 7.91 m MgCl2
kg solvent
0.0158 kg
% solute = (
9)
mass soln = 11.9 g + 15.8 g = 27.7 g
mass solute
11.9 g
) 100 = (
) 100 = 43.0% MgCl2
mass soln
27.7 g
Given: 2.37 M H2CO3 / dsoln = 1.10 g/mL
-assume 1 L soln
147 g H2CO3 = 2.37 mol H2CO3
953 g H2O = 52.9 mol H2O
55.3 mol soln
% solute = (
m=
V soln = 1000 mL
mass soln = dV = (1.10 g/mL)(1000 mL) = 110̅ 0 g
mass solute
147 g
) 100 = (
) 100 = 13.4% H2 CO3
mass soln
110̅ 0 g
mol solute
2.37 mol
=
= 2.49 m H2 CO3
kg solvent
0.953 kg
ΧH2CO3 =
mol solute
2.37 mol
=
= 0.0429
mol soln
55.3 mol
N = nM = (2)(2.37 M) = 4.74 N H2CO3
10)
Given: [HCl] = 12.1 M / 3000. mL soln / 0.500 M
M1V1 = M2V2
V1 =
M2 V2
(0.500 M)(3000. mL)
=
= 124 mL conc. HCl
M1
12.1 M
3
AP CHEMISTRY WKST KEY: SOLUTIONS—CONCENTRATIONS
11)
Given: 25 mL C5H12 (d = 0.63 g/mL) / 45 mL C6H14 (d = 0.66 g/mL)
-Since there is less C5H12 than C6H14, the C5H12 is the solute and the C6H14 is the solvent.
mass C5H12 = (0.63 g/mL)(25 mL) = 16 g C5H12 = 0.22 mol C5H12
mass C6H14 = (0.66 g/mL)(45 mL) = 30. g C6H14 = 0.52 mol C6H14
0.74 mol soln
% solute = (
ΧC5H12 =
12)
mass solute
16 g
) 100 = (
) 100 = 35% C5 H12
mass soln
46 g
mol solute
0.22 mol
=
= 0.30
mol soln
0.74 mol
m=
mol solute
0.22 mol
=
= 4.3 m C5 H12
kg solvent
0.030 kg
M=
mol solute
0.22 mol
=
= 3.1 M C5 H12
L soln
0.070 L
a)
2.50 M AgNO3 / 500.0 mL soln
M=
V soln = 25 mL + 45 mL = 70. mL
mass soln = 16 g + 30. g = 46 g
g solute
(MM solute)(L soln)
g solute = (M)(MM solute)(L soln) = (2.50 M)(169.88
g
mol
)(0.5000 L) = 212 g AgNO3
Place 212 g AgNO3 in a 500 mL volumetric flask. Fill  to  way with distilled water. Shake to dissolve. Fill to
the mark with water.
b)
12.5% MgCl2 / 250 g soln
% solute mass solute
=
100
mass soln
12.5
x
=
100
250 g

100x = 3125

x = 31.2 g
Dissolve 31.2 g MgCl2 in 218.8 g H2O. (get the 218.8 by 250 – 31.2)
c)
4.25 m Ba(NO3)2 in 100.0 mL H2O
m=
g solute
(MM solute)(kg solvent)
g solute = (m)(MM solute)(kg solvent) = (4.25 m)(261.35
Dissolve 111 g Ba(NO3)2 in 100.0 mL H2O.
4
g
mol
)(0.1000 kg) = 111 g Ba(NO3)2