Chemistry Reference
In-Depth Information
The number of grams of solute per kilogram of solvent is
15.103
From the boiling-point elevation, the molality of the solution
can be calculated:
14.5 g solute
0.0500 kg solvent
1.000 kg solvent
a
b
290 g solute
m
¢
t
2.60°C
0.512°C/m
5.07
8
m
k
b
The molar mass is the number of grams in a kilogram of
solvent divided by the number of moles in a kilogram of
solvent:
From that molality, the freezing-point depression can be cal-
culated:
290 g
1.34
4
mol
216 g/mol
¢
t
k
f
m
(1.86°C/m) (5.07
8
m)
9.45°C
Molar mass
The freezing point is
15.104
The number of moles per
liter of solution
is known, and
the number of moles per
kilogram of solvent
is to be found.
Assume that there is 1.000 L of solution. That volume of
solution contains 0.503 mol of
9.45°C.
15.91
KNO
3
changes solubility the most with increasing tem-
perature, so it would be purified most effectively by this
technique.
Sn(NO
3
)
2
and has the fol-
lowing mass:
a
1.13 mol
3.00 mol
15.93
P
benzene
b
105 torr
39.5
5
torr
a
1.083 g
1 mL
a
1000 mL
1 L
1.000 L solution
b
b
1083 g solution
a
1.87 mol
3.00 mol
P
toluene
b
34.0 torr
21.1
9
torr
The mass of the
Sn(NO
3
)
2
is
The total pressure is
39.5
5
torr
21.1
9
torr
60.7 torr
a
242.7 g
1 mol
0.503 mol Sn(NO
3
)
2
b
122.
1
g Sn(NO
3
)
2
15.96
(a)
P
solvent
P
solvent
P
solvent
The mass of the solvent is
P
solvent
X
solvent
P
solvent
P
solvent
(1
X
solvent
)
1083 g solution
122.1 g Sn(NO
3
)
2
960.
9
g solvent
P
solvent
X
solute
Therefore, the molality is
(b)
P
solvent
P
solvent
X
solute
0.503 mol Sn(NO
3
)
2
0.960
9
kg solvent
solvent
X
solvent
The mole fractions of different components are used in
the two equations.
P
solvent
P
m
0.523 m
15.105
This compound is a hydrocarbon, which is nonionic. The
empirical formula is determined as in Section 7.4:
V
nRT
15.97
a
1 mol C
12.01 g C
n
V
93.71 g C
b
7.803 mol C
RT
[ (27.1 torr)
(760 torr/1 atm) ] (0.0600 L)
(0.0821 L
#
atm/mol
#
K) (298 K)
a
1 mol H
1.008 g H
n
6.29 g H
b
6.24 mol H
8.74
5
10
5
mol
The ratio is 1.25 mol of carbon to 1 mol of hydrogen or 5 mol
of carbon to 4 mol of hydrogen. The empirical formula is
The freezing-point depression is
and therefore the molality of the solution is
(8.74
5
10
5
mol) (78.0 g/mol)
mass
solute
10
3
g
6.82
1
C
5
H
4
.
5.5°
(
2.90°C)
8.4°C,
mass
solution
(60.0 mL) (1.01 g/mL)
60.6 g
m
¢
t
8.4°C
5.12°C/m
1.6
4
m
60.6 g
6.82
1
10
3
g
60.6 g
mass
solvent
k
f
0.0606 kg
(8.74
5
10
5
mol)/ (0.0606 kg)
molality
The number of grams in 1.00 kg of solvent is
10
3
m
1.44
3
18.6 g solute
0.0866 kg solvent
1.00 kg solvent
a
b
215 g solute
10
3
m)
¢
t
k
b
m
(1.86°/m) (1.44
3
0.00268°C
The molar mass is the number of grams in a kilogram of
solvent divided by the number of moles in a kilogram of
solvent:
The example illustrates that the osmotic pressure is much
more sensitive and can be used with substances of high mo-
lar mass. There is practically no change of freezing point, but
the osmotic pressure is easily determined at 27.1 torr.
215 g solute
1.6
4
mol solute
Molar mass
13
1
g/mol
15.99
(c)
(This compound contains 3 mol of ions per mole
of compound, therefore its concentration of solute particles
is greatest.)
CaCl
2
The empirical formula mass for is 64 amu. The num-
ber of moles of empirical formula units per mole of mole-
cules is
C
5
H
4
15.101
Assume 100.00 g of solution, containing
13
1
g/mol
64 g/mol empirical formula units
a
1 mol AlCl
3
133.4 g AlCl
3
b
0.0749
6
mol AlCl
3
10.0 g AlCl
3
2 empirical formula units
1 molecule
3 mol Cl
1 mol AlCl
3
0.224
9
mol Cl
0.0749
6
mol AlCl
3
a
b
The molecular formula is
C
10
H
8
.
1 kg
1000 g
m
¢
t
1.151°C
1.86°C/m
0.618
8
m
0.618
8
mol
1 kg water
90.0 g H
2
O
a
b
0.0900 kg
15.107
k
f
6.11 g solute
0.2500 kg water
24.4
4
g solute
1 kg water
0.224
9
mol
0.0900 kg
2.50 m Cl
