Department of Energy Argonne National Laboratory Office of Science NEWTON's Homepage NEWTON's Homepage
NEWTON, Ask A Scientist!
NEWTON Home Page NEWTON Teachers Visit Our Archives Ask A Question How To Ask A Question Question of the Week Our Expert Scientists Volunteer at NEWTON! Frequently Asked Questions Referencing NEWTON About NEWTON About Ask A Scientist Education At Argonne Freezing Point of 40% Isopropyl
Name: Patrick C.
Status: other
Age: 40s
Location: N/A
Country: N/A
Date: Monday, December 02, 2002



Question:
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.


Replies:
Patrick,

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 solvent.

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 molal.

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 of water.

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 a guideline.

Regards,
ProfHoff 523


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 isopropanol.

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.

Vince Calder



Click here to return to the Chemistry Archives

NEWTON is an electronic community for Science, Math, and Computer Science K-12 Educators, sponsored and operated by Argonne National Laboratory's Educational Programs, Andrew Skipor, Ph.D., Head of Educational Programs.

For assistance with NEWTON contact a System Operator (help@newton.dep.anl.gov), or at Argonne's Educational Programs

NEWTON AND ASK A SCIENTIST
Educational Programs
Building 360
9700 S. Cass Ave.
Argonne, Illinois
60439-4845, USA
Update: June 2012
Weclome To Newton

Argonne National Laboratory