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Revised Spring ’99
Answers to the Solution Chemistry Supplementary Sheet
1.
A solution of sulfuric acid prepared by dissolving 40.0 g of H
2
SO
4
in 60.0 g of H
2
O has a density of
1.303 g/mL.
Calculate:
a)
the percent H
2
SO
4
by weight
b)
the mole fraction of H
2
SO
4
c)
the molarity of the solution
d)
the molality of the solution
e)
the normality of the solution
In problems of this type, where you need to convert one set of concentration units to other concentration units, one
needs to determine the moles and mass of each component present and the volume of the solution.
Then using the
appropriate concentration unit definitions, the desired concentration unit can be determined.
Note: the density
given in the problem is that of the entire solution not the density of the solvent or the solute in the system.
In this problem the grams of
solute, H
2
SO
4
, and the number of grams of
solvent, H
2
O
, are given.
Thus, one first
needs to calculate the total mass of solution, the total volume of solution, and the moles of each component
present.
# g sol’n = # g H
2
SO
4
+ # g H
2
O = 40 g H
2
SO
4
+ 60 g H
2
O = 100 g sol’n
1 mL sol’n
1 L
# L sol’n = 100 g sol’n x

x

0.0767 L sol'n
1.303 g sol’n
1000 mL
1 mole H
2
SO
4
# mole H
2
SO
4
= 40 g H
2
SO
4
x 
= 0.41 mole H
2
SO
4
98 g
H
2
SO
4
1 mole H
2
O
# moles H
2
O = 60 g H
2
O x 
= 3.33 moles H
2
O
18 g H
2
O
total # moles
of sol’n = # moles H
2
SO
4
+ # moles H
2
O = 0.41 moles + 3.33 moles
= 3.74 moles sol’n
# g H
2
SO
4
40.0 g H
2
SO
4
a) % H
2
SO
4
=
 X 100 % = 
X 100 =
40.0 %
# g sol’n
100 g sol’n
# moles H
2
SO
4
0.41 moles H
2
SO
4
b)
X
H
2
SO
4
=

=

=
0.11
# moles sol’n
3.74 moles sol’n
# moles H
2
SO
4
0.41 moles H
2
SO
4
c)
Molarity (M)
=
 =
 =
5.35 M
# L sol’n
0.0767 L sol’n
# moles H
2
SO
4
0.41 moles H
2
SO
4
1000 g
d) Molality (m) =  =
 X  =
6.83 m
# Kg H
2
O
60 g H
2
O
1 Kg
Since H
2
SO
4
contains 2 hydrogen atoms that can react with base, there are 2 equivalents of H per mole of H
2
SO
4
.
# equiv.
2 equiv.
e) Normality (N) = M x  = 5.35 M x  =
10.70 N
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View Full DocumentSolution Chemistry Supplementary Sheet Answers Page
2
1 mole
1 mole
2.
A solution prepared by dissolving 15.6 g of benzene, C
6
H
6
, in enough CCl
4
to give 450 mL of solution
has a density of 1.15 g/mL.
Calculate:
a)
the molarity of the solution
b)
the percent C
6
H
6
by weight
c)
the mole fraction of C
6
H
6
d)
the molality of the solution
This problem is similar to the previous problem, except that the total volume of solution is given rather than
the amount of solvent.
In this problem,
C
6
H
6
is the
solute
, since it is dissolved in the
solvent
,
CCl
4
. The density of
the solution
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 Spring '08
 Martin
 Mole

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