Freezing Point of 40% Isopropyl
Name: Patrick C.
Date: Monday, December 02, 2002
what is the freezing point of
a water and isopropyl alcohol mixture (40% isopropyl)?
It would be great to have a method
of estimating the freezing point of other concentrations too.
The reason I am asking is my fish
samples are preserved and are not in a heated building. I would like to
know if ambient winter temperatures will be a problem.
You can calculate the freezing point of the 40 % isopropyl alcohol/water
solution as follows:
Assume you were to end up with a kilogram (about a litre) of final
solution. This means that 400 grams of the solution would be the alcohol
component and the remainder, 600 grams, would be pure water.
The change in freezing point of the solution is the freezing point
depression constant of water (-1.86 C per mole solute per kilogram
solvent) multiplied by the solution's molality.
The molality of the solution is moles of solute (alcohol) per kilogram of
First, calculate the moles of alcohol present: Using a chemical handbook
or atomic masses from the Periodic Table, the molar mass of isopropyl
alcohol (C3H8O) is 60 g/mol.
Four hundred grams of the alcohol would represent 400g / 60 g/mol = 6.67
moles of alcohol.
The 600 g water component represents 0.6 kilograms of solvent.
Thus, the molality of the solution is: (6.67 mol alcohol / 0.6 kg) = 11.11
As mentioned above, water's freezing point depression constant is -1.86
degrees C per mol of solute per kilogram of solvent. This means that for
every mole of solute dissolved per kilogram of solvent, the freezing point
of water drops by 1.86 C from water's actual freezing point of 0 C (32 F).
Thus, your (40 %) solution would freeze at (11.11)(-1.86) C ==> about -
20.7 C (- 5.2 F).
Is this sufficiently low for conditions inside your unheated building?
Unless the temperature inside falls to less than - 5.2 F, there is little
danger of your specimens freezing.
To do the calculation for other concentrations: Assume you are preparing
1000 grams of solution. Express the % alcohol and water in grams. For
example, a 50:50 solution would contain 500 grams of alcohol and 500 grams
As outlined above, use the formula mass of the alcohol -- in the case of
isopropyl alcohol it is 60 g/mol -- to calculate the number of moles
present in the mass of alcohol used.
For 500 g alcohol, the moles present are 500 g / 60 g/mol = 8.33 moles.
Calculate the mass of the water component in kilograms -- in this case, 0.5 kg
Solution molality = 8.33 moles alcohol per 0.5 kilogram of (water) solvent
= 16.66 molal
Next, multiply the molality of the solution by -1.86 C to find the drop in
the freezing point. As already determined immediately above, in the case
of 500 g alcohol, the molality is 16.66.
Thus, the solution freezing point is: (16.66)(-1.86C) = - 30.99 C or about
- 24 F.
The above assumes that the alcohol being used is 100 % pure. Actually, it
is not easy to purchase 100 % isopropyl alcohol because what's available
in the drugstore contains water that must be factored into the calculation
of solution molality. Even so, within limits, this procedure can serve as
The problem here is that 40% isopropanol in water may not have a simple
melting point. You would have to know the phase diagram of the system. This
is probably in the literature, but I could not find it on the Internet.
Further details: As the 40% solution is cooled to its "melting point"
one of several things could occur:
1. Water freezes to form ice and the
higher concentration of isopropanol in the remaining liquid lowers the
freezing point further, and more water ice forms. At some point either
isopropanol itself, or some solid phase containing both water and
isopropanol begins to form.
2. In principle (although I do not think this route will occur) isopropanol
could freeze out as a solid pure isopropanol, or as a hydrated molecule of
3. The solution forms a viscous isopropanol / water glass. This would be
"unstable" but the time required to achieve a stable composition of
isopropanol and water is very long because of the high viscosity -- the
molecules cannot move. This can happen if the cooling rate is rapid -- for
example, if you "quench" the solution by immersing it in liquid nitrogen.
If you cannot find the data in the literature, you may just have to try
the experiment using liquid nitrogen, or a dry ice bath to cool the sample.
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