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5Chem 338
Homework Set #6 solutions
October 17, 2001
From Atkins:
6.8, 6.12, 6.14, 6.16, 6.17, 6.19, 6.22
6.8) The partial molar volumes of propanone and trichloromethane in a mixture in
3
-1
which the mole fraction of CHCl
is 0.4693 are 74.166 and 80.235 cm
mol
,
3
respectively. What is the volume of a solution of total mass 1.000 kg?
Like many problems in Physical Chemistry, there are a couple of ways to tackle this
one. I admit to doing it the hard way the first time around. Here’s the more
straightforward method:
First we find the molecular weight of the solution, convert the 1000g to moles to
obtain the total number of moles of solution, and then use the mole fractions and
partial molar volumes to obtain the total volume.
For 1 mol of solution, there are 0.4693 moles of CHCl
and (1–0.4693)=0.5307
3
moles of propanone. The molecular weights of CHCl
and propanone are 119.37
3
and 58.08 g/mol, respectively, so that the molecular weight of the solution is:
119.37 g
58.08 g
×
+
×
=
0 4693
.
mol
0 5307
.
mol
86 843
.
g / mol
1 mol
1 mol
1 mol
×
=
1000 g of solution corresponds to 1000
g
11 515
.
mol
of solution
86.843 g
Then, (note, c = CHCl
, p = propanone)
3
=
+
V
n V
n V
c c
p p
=
+
x n
V
x n
V
c total c
p total p
=
+
( .
0 4693 11 515 80 235
)( .
)( .
) ( .
0 5307 11 515 74 166
)( .
)( .
)
3
=
887 cm
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