Expansion of ice From Solution
Most recent answer: 12/14/2015
- Anonymous (age 13)
Texas
This is a very thought-provoking question.
Let's say that the solute is completely left out of the ice when the water freezes.That's usually a very good approximation. Let's also say that the volume of the liquid changes linearly with the amount of solute in it. That's a good approximation for low concentrations of solute. We can write the liquid volume as
V0+sv, where V0 is the volume of the pure water, s is the number of solute molecules, and v is the volume change per added solute molecule. (As an aside, there are cases where v is negative!)
Then the total effect of the solute on the liquid volume doesn't change when a bit of ice forms. The new volume when frction x of the liquid has frozen is:
V0(1-x) +xVI+sv, where is the volume of ice you'd get if all the water froze. So the change is still just x(VI-V0), the same as you'd get without the solute.
But here's the complication: the solute lowers the equilibrium freezing point. Below 4°C liquid water expands as it cools. So V0 at the freezing point is a little bigger with the solute in there. VI goes down slightly at lower temperature. For example, if the freezing point is lowered to -8°C, the liquid volume is up about 0.1% compared to 0°C, but the ice volume is down about 0.2%. So it looks like the solute will slightly reduce the volume change on freezing.
To the extent that v depends on the concentration of solute, meaning that the volume change is not linear in the amount of solute, you get more complications, which I guess can have either sign.
Mike W.
(published on 12/14/2015)