Version 094 – Homewrok 3 – Fakhreddine – (53605)
1
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printout
should
have
26
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
1.0 points
Several
interesting
observations
from
the
world around you are listed below.
Which
of these is NOT explained by a colligative
property?
1.
Antifreeze is added to a car radiator to
keep the car from overheating.
2.
A lobster will die when placed in fresh
water.
3.
The freezing point of water is lowered
when salt is added.
4.
At high altitude it takes longer to cook
spaghetti.
correct
5.
Water boils at a higher temperature when
salt is added.
Explanation:
Colligative properties of a solution depend
on the number of solute particles in solution,
not the type.
Boiling point variations due
to pressure changes have nothing to do with
solutions and colligative properties (boiling
point variations due to particles in solution,
etc.).
002
1.0 points
The vapor pressure of pure CH
2
Cl
2
(molecu
lar weight = 85 g/mol) is 133 torr at 0
◦
C and
the vapor pressure of pure CH
2
Br
2
(molecu
lar weight 174 g/mol) is 11 torr at the same
temperature.
What is the total vapor pres
sure of a solution prepared from equal masses
of these two substances?
1.
vapor pressure = 144 torr
2.
vapor pressure = 105 torr
3.
vapor pressure = 93 torr
correct
4.
vapor pressure = 89 torr
5.
vapor pressure = 124 torr
6.
vapor pressure = 72 torr
Explanation:
For CH
2
Cl
2
,
P
0
= 133 torr
MW = 85 g
/
mol
For CH
2
Br
2
,
P
0
= 11 torr
MW = 174 g
/
mol
This is a combination of Raoult’s Law and
Dalton’s Law of Partial Pressures.
The an
swer does not depend on what the masses are,
as long as they are equal. You can choose any
mass you like, but to speed up calculations, it
is convenient to choose the mass the same as
one of the molecular weights given, so that the
number of moles for one of the components is
exactly ONE.
So, for argument’s sake, choose 85 g to be the
mass of each of the components.
That way
you have:
(85 g CH
2
Cl
2
)
parenleftbigg
1 mol CH
2
Cl
2
85 g CH
2
Cl
2
parenrightbigg
= 1
.
0 mol CH
2
Cl
2
Now calculate the moles of the other compo
nent.
(85 g CH
2
Br
2
)
parenleftbigg
1 mol CH
2
Br
2
174 g CH
2
Br
2
parenrightbigg
= 0
.
49 mol CH
2
Br
2
Once you have the two values for moles you
can calculate the mole fraction of each com
ponent.
n
total
= 1
.
0 + 0
.
49 = 1
.
49 mol
X
CH
2
Cl
2
=
1
.
0 mol
1
.
49 mol
= 0
.
67
X
CH
2
Br
2
=
0
.
49 mol
1
.
49 mol
= 0
.
33
Then use those values in Raoult’s Law to
get the vapor pressure for each component.
Raoult’s Law states that:
P
A
=
X
A
P
0
A
P
CH
2
Cl
2
= (0
.
67)(133 torr) = 89 torr
P
CH
2
Br
2
= (0
.
33)(11 torr) = 3
.
6 torr
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Version 094 – Homewrok 3 – Fakhreddine – (53605)
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Add the two together to get the total vapor
pressure (Dalton’s Law).
P
total
=
P
A
+
P
B
+
· · ·
= 89 torr + 3
.
6 torr = 93 torr
You might want to check to see that you
get the same answer no matter what value
you assume as the equal masses of the two
components. As an additional challenge, can
you solve this problem WITHOUT assuming
a definite number of grams, by setting the
mass of each component equal to an algebraic
variable?
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 Spring '09
 Fakhreddine
 Chemistry, Mole, Vapor pressure, Amount of substance

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