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So I saw this equation for the freezing point depression, when you add a solute: $\Delta T= K_\mathrm fmi$

Where,

  • $\Delta T =$ The freezing point depression
  • $K_\mathrm f =$ The freezing point depression constant (Cryoscopic constant)
  • $m =$ Molality
  • $i =$ van't hoff factor.

I read online that $\ce{CaCl2}$ would be better than $\ce{MgCl2}$, but I cannot understand why in the terms of this formula since $i$ for both of them is the same and $K_\mathrm f$ seems to be independent of the solute since I saw values of $K_f$ of water online. (without any solute given)

So, could someone answer why this happens? Because it seems to me that the $K_\mathrm f$ could be the only factor that would explain this. Specifically I would like to know if the charge or the electronegativity of the dissolved ions would make a difference.

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    $\begingroup$ $K_f$ is indeed independent of the solute. It is $m$ that matters. $\ce{CaCl2}$ has better solubility, so... $\endgroup$ – Ivan Neretin Sep 10 '16 at 21:24
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As quantified in Fig. 17 of Manual of Practice for An Effective Anti-Icing Program (FHWA-RD-95-202):

By weight percent, MgCl2 depresses the freezing point of water to a greater degree than CaCl2, up to about 24%. This is consistent with the freezing point depression formula, as Mg has less mass than Ca.

Beyond 24%, CaCl2 is superior.

This is because the eutectic point of MgCl2-water is at 21.6 percent, whereas the eutectic point of CaCl2-water is at 30%.

The freezing point depression formula is only valid at reasonably low concentrations. It is certainly not valid beyond the eutectic point.

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My most recent chemistry lab experiment was determining the effectiveness of various deicers. We tested the freezing point of $\ce{NaCl}$, $\ce{KCl}$, $\ce{CaCl2}$, and $\ce{MgCl2}$. I was told by my professor that $\ce{MgCl2}$ produces an exothermic reaction. When $\ce{MgCl2}$ is dissolved in water, the temperature actually increases. This could be why the freezing point of $\ce{MgCl2}$ isn't as low as that of $\ce{CaCl2}$.

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    $\begingroup$ This is irrelevant. Besides, the dissolution of CaCl2 is also exothermic. $\endgroup$ – Ivan Neretin Feb 27 '18 at 18:20

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