3
$\begingroup$

I know that salt decreases the freezing point of water. We have freezing point (in Celsius), $\mathrm{T_f=-K_f\cdot m}$ (molality of salt). But, is it true that all of the solution doesn't freeze at the same temperature? (i.e. either parts of the ice form at temperature less than $0^\circ C$ but greater than $T_f$ or the process of freezing gets completed at a temperature less than $T_f$)

My question arises from the second part of the following question in my homework:

enter image description here

Two hypotheses came to my mind:

  1. the molality of salt in the frozen part at temperature $T < 0$ is always such that the freezing point of that part is $T$

  2. the molality of salt in the unfrozen part at temperature $T <T_f$ is always such that the freezing point of that part is $T$

Upon repeated trial and error, I found that the second hypothesis worked out to give the correct answer of $15g$ when I also assumed that no salt is being frozen with the ice. This was in accordance to the fact that icebergs in oceans (of salt water) are always composed of fresh water.

But, this led to another question. Like salt sugar is also a non volatile solute. How are ice lollies sweet when sugar can't be frozen into ice?

I would like to know at which part of my deduction am I going wrong.

$\endgroup$
  • $\begingroup$ iceberg in oceans come from glaciers they are not frozen ocean water . $\endgroup$ – trula Aug 6 at 18:11
  • $\begingroup$ An ice treat is not a single, solid crystal. I suspect a small treat placed in your freezer and a large container cooled slowly over days would produce very different substances. $\endgroup$ – BowlOfRed Aug 6 at 18:23
  • $\begingroup$ Freezing point depression is fundamentally linked to segregation of the impurity. If the impurity froze with the ice, then the effective concentration (technically, the chemical potential) of water would be the same in each state, and the freezing point wouldn't shift. An analogous situation arises with boiling point elevation: the impurity cannot boil along with the water. $\endgroup$ – Chemomechanics Aug 7 at 17:57
1
$\begingroup$

Your process occurs under fractional crystallization, which is a separations process. Assuming that freezing takes place relatively "slowly" (e.g., under conditions where "flash" freezing doesn't occur), ice crystals will form in the salt water solution at a temperature below 0 C, and no salt will be incorporated in those ice crystals. As freezing proceeds, this obviously leads to higher salt concentrations in the remaining water, so the freezing temperature will drop as freezing proceeds.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.