Version 020 – Exam 3 – fakhreddine – (51010)
1
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32
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before answering.
001
10.0 points
In which of these compounds would you find
ONLY dispersion forces existing between the
molecules?
I. CCl
4
;
II. NH
3
;
III. HBr;
IV. CO
2
.
1.
I and IV only
correct
2.
I and III only
3.
II only
4.
II and IV only
5.
I and II only
6.
II and III only
7.
I only
8.
IV only
9.
III and IV only
10.
III only
Explanation:
A nonpolar covalent molecule would have
only dispersion forces with another nonpolar
covalent molecule.
002
10.0 points
Consider a pure sample of each of these com-
pounds
RbCl
FeCl
3
FeN
SrCl
2
Order from weakest to strongest intermolecu-
lar forces.
1.
RbCl
<
FeN
<
SrCl
2
<
FeCl
3
2.
RbCl
<
SrCl
2
<
FeCl
3
<
FeN
correct
3.
FeCl
3
<
SrCl
2
<
RbCl
<
FeN
4.
FeN
<
RbCl
<
SrCl
2
<
FeCl
3
5.
FeN
<
SrCl
2
<
RbCl
<
FeCl
3
Explanation:
003
10.0 points
What mass of (NH
4
)
2
SO
4
would be required
to prepare 500 mL of a 7.50% solution of am-
monium sulphate ((NH
4
)
2
SO
4
)?
Its density
is 1.04 g/mL.
1.
41.6 g
2.
39.0 g
correct
3.
36.0 g
4.
28.2 g
5.
43.8 g
Explanation:
V
= 500 mL
density
soln
= 1
.
04 g
/
mL
The (NH
4
)
2
SO
4
is 7.50% (NH
4
)
2
SO
4
by
mass.
We
can
write
this
as
the
ratio
7
.
50 g (NH
4
)
2
SO
4
100 g soln
.
The density of the solution is 1.04 g/mL. This
can be more fully written as
1
.
04 g soln
1 mL soln
.
We want to prepare 500 mL of the solution.
We use the solution density to convert from
mL solution to g solution:
? g soln = 500 mL soln
×
1
.
04 g soln
1 mL soln
= 520 g soln
Now we can use the percent by mass to con-
vert from grams to grams (NH
4
)
2
SO
4
:
? g (NH
4
)
2
SO
4
= 520 g soln
×
7
.
50 g (NH
4
)
2
SO
4
100 g soln
= 39 g (NH
4
)
2
SO
4
004
10.0 points

Version 020 – Exam 3 – fakhreddine – (51010)
2
At constant temperature if the volume of a
gas sample is tripled the pressure will
1.
remain the same.
2.
be increased by a factor of 6.
3.
be one third the original.
correct
4.
be cut in half.
5.
be tripled.
Explanation:
3
V
1
=
V
2
Boyle’s Law relates the volume and pressure
of a sample of gas:
P
1
V
1
=
P
2
V
2
P
1
V
1
=
P
2
(3
V
1
)
P
1
= 3
P
2
P
2
=
1
3
P
1
005
10.0 points
What volume of 0
.
135 M K
3
PO
4
is required to
react with 87 mL of 0
.
401 M MgCl
2
according
to the equation
2 K
3
PO
4
+ 3 MgCl
2
→
Mg
3
(PO
4
)
2
+ 6 KCl
1. 54.3243
2. 114.484
3. 32.9169
4. 44.8035
5. 19.5065
6. 172.368
7. 29.8988
8. 146.092
9. 31.9531
10. 31.4894
Correct answer: 172
.
368 mL.
Explanation:
[K
3
PO
4
] = 0
.
135 M
V
= 87 mL
[MgCl
2
] = 0
.
401 M
First, we determine the moles of MgCl
2
:
? mol MgCl
2
= 87 mL soln
×
1 L soln
1000 mL soln
×
0
.
401 mol MgCl
2
1 L soln
= 0
.
034887 mol MgCl
2
Using the mole-to-mole ratio from the bal-
anced
chemical
equation we calculate the
moles of K
3
PO
4
needed to react with this
amount of MgCl
2
:
? mol K
3
PO
4
= 0
.
087 mol MgCl
2
×
2 mol K
3
PO
4
3 mol MgCl
2
= 2
.
3258
×
10
−
5
mol K
3
PO
4
Now we use the molarity of the K
3
PO
4
solu-
tion to convert from moles of K
3
PO
4
to the
volume of K
3
PO
4
:
? mL soln = 2
.
3258
×
10
−
5
mol K
3
PO
4
×
1 L soln
0
.
135 mol K
3
PO
4
×
1000 mL soln
1 L soln
= 172
.
368 mL soln
It would require 172
.
368 mL of the K
3
PO
4
solution to provide enough moles of K
3
PO
4
to
react all of the MgCl
2
.

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- Fall '08
- wandelt
- Chemistry, Mole, Intermolecular force, Van der Waals force