rose (kmr2427) – Homework 2 – fakhreddine – (51615)
1
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printout
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have
31
questions.
Multiplechoice questions may continue on
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before answering.
001
1.0points
The mole fraction of a certain nonelectrolyte
compound in a solution containing only that
substance and water is 0.100. The molecular
weight of water is 18.0 g/mol.
What addi
tional information is needed to determine the
molality of the solution?
1.
The mole fraction of water in the solu
tion.
2.
The density of the solution.
3.
No additional information; the molality
can be calculated from the information given.
correct
4.
The density of the solute.
5.
The molecular weight of the compound.
Explanation:
Here we can assume that we have 1 mol
total. (In fact, we can choose any number of
moles, but the math is easier if you choose 1
mol.)
If the mole fraction of the substance
is 0.100, you can then assume that you have
0.100 mol of the substance, and the remaining
0.900 mol is H
2
O. The molality of a solution
is determined by the following formula:
m
=
mol solute
kg solvent
We’ve already assumed that we have 0.100
mol of solute, and we can determine the kg of
H
2
O in the usual way:
0
.
900 mol H
2
O
parenleftbigg
18
.
0 g
1 mol
parenrightbiggparenleftbigg
1 kg
1000 g
parenrightbigg
= 0.0162 kg,
and we can calculate the molality of this solu
tion:
m
=
0
.
100 mol
0
.
0162 kg H
2
O
= 6.17
m
So, it
is
possible to determine the molality of
this solution without any additional informa
tion.
002
1.0points
What additional information, if any, would
enable you to calculate the molality of a 7.35
molar solution of a nonelectrolyte solid dis
solved in water?
1.
None is needed.
2.
Only the density of the solution would be
needed.
3.
Both the density of the solution and
the molecular weight of the solute would be
needed.
correct
4.
Only
the
density
of
water
would
be
needed.
5.
Only the molecular weight of the solute
would be needed.
Explanation:
molarity =
mol solute
L solution
molality =
mol solute
kg solvent
The density of the solution can be used to
convert volume (1 L) of solution into mass of
solution.
Then the molecular weight of the
solute (given or calculated from the formula)
can be used to convert the number of moles
solute in 1 L solution into mass of solute
in grams.
The mass of the solvent is the
difference between the mass of the solution
and the mass of the solute (both of which
have been calculated). Substitute the values
into the molality formula and calculate.
003
1.0points
A solution is 40.0% silver nitrate (AgNO
3
)
by mass. The density of this solution is 1.48
grams/mL. The formula weight of AgNO
3
is
170 grams/mol.
Calculate the molality of
AgNO
3
in this solution.
Correct answer: 3
.
92157
m
.
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rose (kmr2427) – Homework 2 – fakhreddine – (51615)
2
Explanation:
density = 1.48 g/mL
FW
AgNO
3
= 170 g/mol
% AgNO
3
= 40% by mass
molality =
n
AgNO
3
kg water
In 100 g of 40.0% AgNO
3
(aq) there are 40.0
g AgNO
3
and 60.0 g water.
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 Spring '07
 Holcombe
 Mole, Vapor pressure, Molecule, mol, Amount of substance

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