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The free energy of particles reads, if we assume no internal energy : $\tilde f(c)= \frac{f(c)}{k_bT}=-c \, ln(c)-(1-c)\, ln(1-c)$, with $c$ the concentration.

If we had a term $F(c)$ for the interactions between the particles : $\tilde f(c)= - c \, ln(c)-(1-c)\, ln(1-c) + F(c)$, such that $F(0)=0$, which is reasonable since it's an interaction potential.

The equilibrium between 2 phases $1$ and $2$ is determined by the relations : $\mu_1=\mu_2$ and $\pi_1=\pi_2$.

Now assume you just put your particles that interact with each other in a huge box full of water (an ocean for example). Then, a reasonnable thought is that you can have a phase with only water, and a phase with particles ($c_0$) and water ($1-c_0$). But the relation $\mu_1=\mu_2$ prevents that because that $\mu(0)=-\infty$. And I'm really not convinced by that ! If your particles attract each other, they should at some point aggregate, don't they ? In other words, entropy will never let aggregation win totally !!

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You are right, the entropy of mixing term will never allow $c\rightarrow0$ or $c\rightarrow1$.

It is possible to approach those limits very very closely. It depends on whether the interaction term is unfavourable and how unfavourable it is. An example would be $F(c)=K c(1-c)$, with some constant $K>0$, which vanishes at both extremes and has a maximum (for unfavourable interactions) in the middle. Even in the solid region of two-component systems A+B, where you see almost complete demixing into two solid phases, you can expect a tiny fraction of A atoms dissolved in B, and vice versa. The same idea applies to your solid-liquid example.

The driving force behind this is the gradient of the free energy with respect to $c$. If you calculate this from your first equation, you'll see that it diverges at both extremes $c=0$ and $c=1$, so as to make a small amount of mixing favourable, for any physically reasonable form of the interaction term. In other words, even if $K$ is very large, the gradient of the entropy of mixing term will overwhelm the $F(c)$ term near the edges of the composition range: the combination $c\ln c + (1-c)\ln (1-c) + F(c)$ will have two minima, very close to (but not exactly at) the extremes.

The compositions of the two coexisting phases will not be exactly at these minima, in general, but reasonably close. If you draw a common tangent to the curve, this corresponds to equality of the chemical potentials of each component in the two phases. You can see some relevant diagrams and a bit more explanation in this Chemistry SE answer.

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