017
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 = 89 torr
2.
vapor pressure = 105 torr
3.
vapor pressure = 144 torr
4.
vapor pressure = 93 torr
correct
5.
vapor pressure = 72 torr
6.
vapor pressure = 124 torr
Explanation:
For CH
2
Cl
2
,
P
0
= 133 torr
MW = 85 g
/
mol

spano velasco (ds28282) – hw 02 – Fakhreddine – (52395)
6
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
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?
018
1.0 points
Consider two liquids A and B. The vapor
pressure of pure A (molecular weight = 50
g/mol) is 225 torr at 25
◦
C and the vapor
pressure of pure B (molecular weight = 75
g/mol) is 90 torr at the same temperature.
What is the total vapor pressure at 25
◦
C of a
solution that is 25% A and 75% B by weight?
1.
108 torr
2.
225 torr
3.
76 torr
4.
191 torr
5.
115 torr
6.
335 torr
7.
124 torr
8.
135 torr
correct
9.
203 torr
Explanation:
For A,
P
0
= 255 torr
MW = 50 g
/
mol
For B,
P
0
= 90 torr
MW = 75 g
/
mol
The mole fractions are
1
3
for A and
2
3
for B.
parenleftBig
1
3
parenrightBig
(225) +
parenleftBig
2
3
parenrightBig
(90) = 135 torr
019
1.0 points
At a given temperature, pure A has a vapor
pressure of 140 torr and pure B has a vapor
pressure of 60 torr. At this temperature, we

spano velasco (ds28282) – hw 02 – Fakhreddine – (52395)
7
make a solution in which the mole fractions
of A and B are equal.
We then allow this
solution to reach equilibrium with its vapor.

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- Fall '07
- Holcombe
- Chemistry, Mole, Vapor pressure, Molecule, Torr, spano velasco