– Exam 1 – holcombe – (51055)
1
This
printout
should
have
31
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0 points
The vapor pressure of benzene at 25
◦
C is
94.6 torr and its enthalpy of vaporization is
30.8 kJ
·
mol
−
1
.
Estimate the normal boiling
point of benzene.
Assume the enthalpy of
vaporization is independent of temperature.
1.
640 K
2.
Not enough information is given.
3.
358 K
correct
4.
624 K
5.
470 K
Explanation:
T
1
= 25
◦
C + 273
.
15 = 298
.
15 K
P
1
= 94.6 torr
P
2
= 1
.
0 atm = 760 torr
At ANY liquid’s
normal
boiling point, the
vapor pressure = 1.0 atm.
Using the ClausiusClapeyron equation,
ln
parenleftbigg
P
2
P
1
parenrightbigg
=
Δ
H
◦
vap
R
parenleftbigg
1
T
1

1
T
2
parenrightbigg
R
Δ
H
◦
vap
ln
parenleftbigg
P
2
P
1
parenrightbigg
=
1
T
1

1
T
2
1
T
2
=
1
T
1

R
Δ
H
◦
vap
ln
parenleftbigg
P
2
P
1
parenrightbigg
1
T
2
=
1
298
.
15 K

8
.
314 J
/
(mol
·
K)
30
.
8
×
10
3
J
/
mol
×
ln
parenleftbigg
760 torr
94
.
6 torr
parenrightbigg
= 0
.
00279156
T
2
=
1
0
.
00279156
= 358
.
222 K
002
10.0 points
Solution A contains 0.5 grams of solute A
and solution B contains 0.5 grams of solute
B (both A and B are nonelectrolytes). Other
than the solutes, the solutions are identical
(volume, temperature,
etc
.).
Now you mea
sure the osmotic pressure of each solution and
find that the osmotic pressure of solution B is
twice
that of solution A. What is the relation
ship between the molecular weights of solutes
A and B?
1.
The molecular weight of solute B is four
times that of solute A.
2.
The molecular weight of solute A is twice
that of solute B.
correct
3.
The molecular weight of solute A is four
times that of solute B.
4.
The molecular weight of solute B is twice
that of solute A.
5.
The molecular weight ratio cannot be
determined from this experiment.
Explanation:
Because the solution with B has twice the
osmotic pressure,
it must also have twice
the number of dissolved particles (number of
moles of B). The only way to have more moles
for B (knowing the same mass of A and B was
used) is that B has a smaller molecular weight
than A. For the osmotic pressure to be ex
actly twice that of A, the molecular weight of
B must be half that of A which is the same as
A being twice that of B.
003
10.0 points
Consider the following system at equilibrium.
H
2
(g) + I
2
(g)
⇀
↽
2 HI(g) + heat
Which response includes all the following
that will shift the equilibrium to the left, and
no others?
I) increasing the temperature
II) decreasing the temperature
III) increasing the pressure
IV) decreasing the pressure
V) removing some HI
VI) adding some HI
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
– Exam 1 – holcombe – (51055)
2
VII) removing some I
2
VIII) adding some I
2
1.
I, VI, and VII only
correct
2.
II, V, and VIII only
3.
II only
4.
II, IV, VII, and VIII only
5.
I, III, V, and VII only
Explanation:
Increasing temperature adds heat and will
shift equilibrium to the left. Changing pres
sure will not affect equilibrium because the
number of gas molecules is the same on both
sides of the equation. Removing a reactant or
adding a product will shift equilibrium to the
left.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '07
 Holcombe
 Chemistry, Thermodynamics, Enthalpy, Partial Pressure, Nitric oxide, Vapor pressure, sodium nitrate, Henry

Click to edit the document details