# Determining the depression in freezing point experimentally

I was reading about Beckmann's method for determining the depression in freezing point:

First, the solvent is placed inside the cooling chamber directly and the approximate freezing point is determined. Next, the freezing tube is again kept directly in the cooling chamber and then placed inside the air jacket.

I have the following questions:

1. Why not directly place the freezing tube into the air jacket and determine the freezing point? Why do we have to determine the approximate freezing point?

2. How is the latent heat released ?

3. Lastly, I think that the last line should say, "The temperature of the cooling bath should be 4° to 5°....." for all of this to make sense.

1) Assume you do not know the freezing point of your solvent. Even if you know it, that reference value is under different conditions such as atmospheric pressure (suppose you are in Denver of Mexico City, for instance). Thus, you need to determine the approximate freezing point of your solvent before you do careful reading. It's exactly like you are trying to determine the melting point of given compound. First time you set up MelTemp in high heating range to get an idea of melting range. Then you do careful heating with second capillary containing your solid to get melting point within $\mathrm {\pm 1 ^\circ C}$. The same approach applies here: First, direct heating to get approximate freezing point, then warm it to melt back, cool again directly to close to that temperature, wipe the outside of the tube, and this time, air jacketed it before do slow heating and get accurate reading. Whole this process is just to save your time and energy. That air jacket make long time to make your freezing tube with solvent cool (prime example: Dewar flask, with vacuum though). And, imagine if you have to stir that whole time with one stroke per second?
2) The latent heat of fusion release when a liquid undergoes liquid to solid transformation at its freezing point. If cooling is not uniform, the releasing energy may cause some solid near by to melt again (remember, at freezing or melting point, $\ce{liquid <=> solid}$). That's why frequent stirring is important, to give uniform cooling for this case.