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My lab manual defines the freezing point as "The temperature at which the vapor pressure of the substance in its liquid phase is equal to its vapor pressure in the solid phase." It goes on to explain why adding a non-volatile solute to a solvent, which lowers the vapor pressure compared to that of the pure liquid, lowers its freezing point.

What does the freezing point, which concerns liquid and solid phase, have to do with vapor pressure? If I do an experiment in a gas-tight syringe containing liquid and solid only, there will be no equilibrium with the vapor phase, but the freezing point will still change.

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    $\begingroup$ I beg to differ - there is always an equilibrium vapor pressure over a solid or liquid. Using the vapor pressure provides a way to compare the solid to the liquid (using the vapor as an intermediary). The solid and liquid will only be in equilibrium when the vapor is in equilibrium with both of them. $\endgroup$ – Jon Custer Jan 9 at 21:10
  • $\begingroup$ Note that @JonCuster's excellent comment here is an application of the zeroth law of thermodynamics. $\endgroup$ – Zhe Jan 9 at 21:48
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A freezing point is a temperature at which solid and liquid phases are in equilibrium with each other. In equilibrium the chemical potentials of each component in all phases present are equal. It means that if there are three phases in the system, solid, liquid and gas, then the chemical potential should be equal in all the three. As a consequence, the vapor pressure of a given component over the liquid phase should equal that over the solid phase. But if there are only solid and liquid phases, I see no need to involve gas. The basic derivation, which I was taught, of the freezing point depression of liquid solvent A (when the solute is insoluble in solid A) starts with $$\mu_\mathrm{A}(\mathrm{solid})=\mu_\mathrm{A}(\mathrm{solution})=\mu_\mathrm{A}(\mathrm{pure\,liquid})+RT\ln{x_\mathrm{A}}$$ - no gases here.

I might miss something here, but my guess is that your lab manual is just trying to be more obscure than it ought to be (no offence to its authors intended). While the expression for $\mu_\mathrm{A}(\mathrm{solution})$ itself can be derived from the Raoult's law, which requires gases, that's slightly different, although related, story. Provided that the expression for $\mu_\mathrm{A}(\mathrm{solution})$ is already known, I can see no sound reason for inviting the vapor pressures to explain cryoscopy.

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