Here is an estimate of the effect described in Rob Hooft's answer to Why does salt lower the freezing point of water?

We imagine some ice sitting in some salt water. The water has volume V V V and there are n n n salt ions in it. The entropy of the salt ions is S∝nlogV S∝nlogV S \propto n \log V

Suppose a small volume of water dV dV \mathrm{d}V freezes. Then the entropy of the ions decreases by

dS=−nV dV dS=−nVdV \mathrm{d}S = -\frac{n}{V} \mathrm{d}V

The second law says we had better pay for that entropy decrease. When the water freezes, it gives off some heat dQ dQ \mathrm{d}Q . If this is to pay for the entropy cost we need dQT+ΔT −dQT ≈dS dQT+ΔT−dQT≈dS \frac{\mathrm{d}Q}{T + \Delta T} – \frac{\mathrm{d}Q}{T} \approx \mathrm{d}S where ΔT ΔT \Delta T is the increase in the freezing point from when there are no ions.

We will approximate this as

dQT 2 ΔT≈−nV dV dQT2ΔT≈−nVdV \frac{\mathrm{d}Q}{T^2}\Delta T \approx -\frac{n}{V}\mathrm{d}V

dQdV dQdV \frac{\mathrm{d}Q}{\mathrm{d}V} is nearly the heat of fusion, ΔH≡dQρdV ΔH≡dQρdV \Delta H \equiv \frac{\mathrm{d}Q}{\rho \mathrm{d}V} with ρ ρ \rho the density.

Thus

ΔT≈−nT 2 VρΔH ΔT≈−nT2VρΔH \Delta T \approx -\frac{n T^2}{V \rho \Delta H}

For water, we have T=273K T=273K T = 273K , ρ=1000kgm 3 ρ=1000kgm3 \rho = 1000 \frac{kg}{m^3} , and ΔH=333000Jkg ΔH=333000Jkg \Delta H = 333000 \frac{J}{kg} . This gives 1.86K⋅kg/mol 1.86K⋅kg/mol 1.86 K \cdot kg / mol , in good agreement with the figure of 1.853K⋅kg/mol 1.853K⋅kg/mol 1.853 K \cdot kg/mol quoted in

wikipedia.org

Freezing-point depression on Wikipedia.

It's important to note that it doesn't matter what the solute is; individual atoms or large molecules behave the same way for freezing point depression because the effect is entropic.

Written Feb 3 • View Upvotes

Answer by Mark Eichenlaub:

Here is an estimate of the effect described in Rob Hooft's answer to Why does salt lower the freezing point of water?We imagine some ice sitting in some salt water. The water has volume [math]V[/math] and there are [math]n[/math] salt ions in it. The entropy of the salt ions is [math]S \propto n \log V[/math]Suppose a small volume of water [math]\mathrm{d}V[/math] freezes. Then the entropy of the ions decreases by[math]\mathrm{d}S = -\frac{n}{V} \mathrm{d}V[/math]The second law says we had better pay for that entropy decrease. When the water freezes, it gives off some heat [math]\mathrm{d}Q[/math]. If this is to pay for the entropy cost we need [math]\frac{\mathrm{d}Q}{T + \Delta T} – \frac{\mathrm{d}Q}{T} \approx \mathrm{d}S[/math] where [math]\Delta T[/math] is the increase in the freezing point from when there are no ions.We will approximate this as[math]\frac{\mathrm{d}Q}{T^2}\Delta T \approx -\frac{n}{V}\mathrm{d}V[/math][math]\frac{\mathrm{d}Q}{\mathrm{d}V}[/math] is nearly the heat of fusion, [math]\Delta H \equiv \frac{\mathrm{d}Q}{\rho \mathrm{d}V}[/math] with [math]\rho[/math] the density.Thus[math]\Delta T \approx -\frac{n T^2}{V \rho \Delta H}[/math]For water, we have [math]T = 273K[/math], [math]\rho = 1000 \frac{kg}{m^3}[/math], and [math]\Delta H = 333000 \frac{J}{kg}[/math]. This gives [math]1.86 K \cdot kg / mol[/math], in good agreement with the figure of [math]1.853 K \cdot kg/mol[/math] quoted in Freezing-point depression on Wikipedia.It's important to note that it doesn't matter what the solute is; individual atoms or large molecules behave the same way for freezing point depression because the effect is entropic.

Why does salt lower the freezing point of water?