The most rigorous treatment comes from Edward Guggenheim's book "Thermodynamics: An Advanced Treatment for Chemists and Physicists" (ISBN 978-0444869517).
Let's assume an equilibrium between a liquid solution and the pure solid solvent 1. Let's also assume pressure is constant. At equilibrium between liquid mixture at its freezing point and the pure solid at temperatur $T$ we assume the chemical potential $\mu$ is equal:
$\mu^{\rm mixture}_1(T) = \mu^{\rm solid}_1(T)$
Let's say $T^{\rm liquid}$ is the equilibrium temperature of the pure liquid (the freezing point of the pure liquid), so we obtain:
$\mu^{\rm liquid}_1(T^{\rm liquid}) = \mu^{\rm solid}_1(T^{\rm liquid})$
If we divide equation $(2)$ by $(1)$, we get:
$\displaystyle \frac{\mu^{\rm liquid}_1(T^{\rm liquid})}{\mu^{\rm mixture}_1(T)} = \frac{\mu^{\rm solid}_1(T^{\rm liquid})}{\mu^{\rm solid}_1(T)}$
This can be rewritten to:
$\displaystyle \frac{\mu^{\rm liquid}_1(T^{\rm liquid})}{\mu^{\rm mixture}_1(T)} = \left( \large \frac{\frac{\mu^{\rm liquid}_1(T)}{\mu^{\rm liquid}_1(T^{\rm liquid})}}{\frac{\mu^{\rm solid}_1(T)}{\mu^{\rm solid}_1(T^{\rm liquid})}} \right)$
Taking the logarithms:
$\displaystyle \ln\left(\frac{\mu^{\rm liquid}_1(T^{\rm liquid})}{\mu^{\rm mixture}_1(T)}\right) = \ln\left(\frac{\mu^{\rm liquid}_1(T)}{\mu^{\rm liquid}_1(T^{\rm liquid})}\right) - \ln\left(\frac{\mu^{\rm solid}_1(T)}{\mu^{\rm solid}_1(T^{\rm liquid})}\right)$
We now apply the following equation (which we can obtain from the Guggenheim square):
$\displaystyle \frac{\partial \mu_1}{\partial T} = \frac{- H_1}{RT^2}$
We get
$\displaystyle \ln\left(\frac{\mu^{\rm liquid}_1(T)}{\mu^{\rm liquid}_1(T^{\rm liquid})}\right) = - \int_{T^{\rm liquid}}^{T} \frac{H_1^{\rm solid}}{RT^2} dT$
and
$\displaystyle \ln\left(\frac{\mu^{\rm solid}_1(T)}{\mu^{\rm solid}_1(T^{\rm liquid})}\right) = - \int_{T^{\rm liquid}}^{T} \frac{H_1^{\rm liquid}}{RT^2} dT$
Combining those equations results in:
$\displaystyle \ln\left(\frac{\mu^{\rm liquid}_1(T^{\rm liquid})}{\mu^{\rm mixture}_1(T)}\right) = - \int_{T^{\rm liquid}}^{T} = - \int_{T^{\rm liquid}}^{T} \frac{{\Delta}_f H_1^{\rm \ominus}}{RT^2} dT$
In this equation, ${\Delta}_f H_1^{\rm \ominus}$ is the proper enthalpy of fusion of the pure solvent.
We now apply the definition of the chemical potential $\mu$:
$\mu_1 = \mu_1^{\ominus} + RT \phi {\sum}_s r_s$
Here, $\phi$ denotes the osmotic coefficient of the solution at its freezing point. We can rewrite the equation to:
$\displaystyle \phi {\sum}_s r_s = \left< {\Delta}_f H_1^{\rm \ominus} \right> \left( \frac{\frac{1}{RT}}{\frac{1}{RT^{\rm liquid}}}\right)$
In this equation, $\left< {\Delta}_f H_1^{\rm \ominus} \right>$ is the average value of ${\Delta}_f H_1^{\rm \ominus}$ over the reciprocal temperature interval $1/T^{\rm liquid}$ to $1/T$. Because ${\Delta}_f H_1^{\rm \ominus}$ is always positive, it follows $T<T^{\rm liquid}$ (The freezing point of the solution is always below that of the pure solvent if the solid phase is pure solvent).
Approximation: For dilute solutions, $\Delta T = T^{\rm liquid}-T << T^{\rm liquid}$, so we get
$\displaystyle \phi {\sum}_s r_s = \frac{{\Delta}_f H_1^{\ominus}(T^{\rm liquid}-T)}{RT^2}$
We rearrange this equation to:
$T^{\rm liquid}-T = \displaystyle \phi {\sum}_s r_s \frac{RT^2}{{\Delta}_f H_1^{\ominus}}$
For calculations, we prefer the molalities $m_s$ over solute-solvent mole ratios $r_s$:
$m_s = r_s/r^{\ominus}$
We obtain:
$T^{\rm liquid}-T = \displaystyle \phi {\sum}_s m_s \left(\frac{r^{\ominus}RT^2}{{\Delta}_f H_1^{\ominus}}\right)$
The factor $\left(\frac{r^{\ominus}RT^2}{{\Delta}_f H_1^{\ominus}}\right)$ is the cryoscopic constant of the solvent.