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


  • $\Delta T =$ 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.

  • 2
    $\begingroup$ $K_f$ 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

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.


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 $\ce{CaCl2}$.

  • 1
    $\begingroup$ This is irrelevant. Besides, the dissolution of CaCl2 is also exothermic. $\endgroup$ – Ivan Neretin Feb 27 '18 at 18:20

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