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?
006
10.0points
You place two beakers into an evacuated
chamber.
One beaker has 100 mL of a 0.1
M solution of NaCl in water and the other has
100 mL of a 1.0 M sugar solution. When the
system comes to equilibrium,
1.
all of the water will be in the sugar solu-
tion.
2.
the beaker with the sugar solution will
have a larger volume.
correct
3.
the two beakers will have identical vol-
umes.
4.
the beaker with the salt solution will have
a larger volume.
5.
all of the water will be in the salt solu-
tion.
Explanation:

gamboa (gg8745) – Homework #2 – holcombe – (51160)
3
Both liquids will evaporate and condense
from each beaker.
The chamber will come
to equilibrium when the two beakers have
the same concentration (same vapor pres-
sure).
The salt solution will become more
concentrated (lower volume) and the sugar so-
lution will become less concentrated (higher
volume).
007
10.0points
Which of the solutions below will have the
greater boiling point and what will it be?
K
b
= 0
.
512
◦
C
/
m for water.
I) 135 g of glucose (C
6
H
12
O
6
) dissolved in
0.5 kg water
II) 35 g of NaCl dissolved in 0.5 kg of water
1.
Solution
II
with
a
boiling
point
of
100
.
6
◦
C
2.
Solution
I
with
a
boiling
point
of
118
.
4
◦
C
3.
Solution
I
with
a
boiling
point
of
100
.
8
◦
C
4.
Solution
II
with
a
boiling
point
of
101
.
2
◦
C
correct
Explanation:
First calculate the molality of each solution,
which is mol solute per kg solvent.
To do
this, you will need to convert from mass to
mols, using molar mass. Remember also that
i
= 2 for NaCl and
i
= 1 for glucose. Using
Δ
T
=
imK
b
you will find that Δ
T
= 0
.
768
◦
C
for the glucose solution and 1.227 for the salt
solution, which means the boiling point of the
salt solution will be 101.2
◦
C.

#### You've reached the end of your free preview.

Want to read all 7 pages?

- Spring '11
- Vanden Bout
- Chemistry, Partial Pressure, Vapor pressure, Gamboa