Chemistry Worksheets With Answers Page 5
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5Chem 338 Homework #6
For this case, n
is the number of moles of solute in 1 kg of solvent, which is the
A
1000 g
=
definition of molality:
m
x
A
A
18.02 g / mol
−
−
6
4
=
×
=
×
m
( .
9 106 10
)(
1000 18 02
/
. )
5 05 10
.
moles / kg
N
2
−
−
6
4
=
×
=
×
m
( .
4 836 10
)(
1000 18 02
/
. )
2 68 10
.
moles / kg
O
2
Note that the approximation we made above is valid since the number of moles
of N
and O
are both much smaller than the number of moles of water.
2
2
6.22) The vapor pressure of a sample of benzene is 400 Torr at 60.6ºC, but it fell to
386 Torr when 0.125 g of an organic compound was dissolved in 5.00 g of the
solvent. Calculate the molar mass of the compound.
For A = benzene and using Raoult’s law,
p
=
=
386
=
A
x
400 0 965
.
A
*
P
A
From the definition of the mole fraction,
5 00
.
n
A
78 12
.
=
=
x
A
+
5 00
.
0 125
.
n
n
+
A
x
78 12
.
MW
x
0 06400
.
=
0 965
.
0 125
.
+
0 06400
.
MW
x
Solving for the MW of the unknown, MW
= 53.85 g/mol
x
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