Nov. 13
th
,
17
th
Tutorial Notes
1) Atkins & dePaula, Exercise 5.7 a)
We have been asked to calculate the freezing point of an aqueous solution, given that its
osmotic pressure at 300 K is 120 kPa.
Normally the freezing point of pure water is 273 K.
However, the fact that this solution has
osmotic pressure means that other colligative properties, including freezing point depression
will also be present.
To calculate the new freezing point of this solution we must use the
equation for freezing point depression:
solute
solvent
mass
f
f
T
K
m
n
K
!
=
"
=
"
The cryoscopic constant for water can be obtained from Table 7.2 in Atkins & dePaula and is
1.86 K kg mol
1
.
However, we do not know the molality of this solution.
Instead we have
been given the magnitude of another colligative property which is also dependent on the
amount of solute present, osmotic pressure:
solute
solvent
cRT
n
RT
V
!
=
=
Solving for
n
solute
:
solvent
solute
V
n
RT
!
=
We do not know the volume of the solvent.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 GIORGI
 Colligative properties, Solvent, Freezingpoint depression, Atkins

Click to edit the document details