Freezing-Point Depression

Salt a road in winter, and ice refuses to form until the temperature drops well below 0 °C. This is freezing-point depression, one of the four classic colligative properties. It matters because it converts an easily measured temperature shift into a direct route to a solute's molar mass. The same thermodynamic machinery that gives boiling-point elevation gives freezing-point depression, but with the sign and the constant reversed. Both trace back to the lowering of a solvent's chemical potential once solute molecules dilute it. This article derives the cryoscopic depression from first principles, defines the cryoscopic constant $K_f$ , and shows how cryoscopy is used experimentally to determine unknown molar masses.

Key takeaways

Freezing-point depression: $\Delta T_f = K_f \, b$ , where $b$ is the solute molality.
$K_f$ depends only on solvent properties: molar mass, melting point, and enthalpy of fusion.
Unlike $K_b$ , $K_f$ is usually larger because $\Delta_{fus}H$ is smaller than $\Delta_{vap}H$ .
Cryoscopy measures molar mass from $\Delta T_f$ , sample mass, and solvent mass.
Ionic solutes need the van't Hoff factor $i$ to correct for dissociation.
Key idea: dissolving solute lowers the solvent's chemical potential in the liquid, so the liquid must be cooled further below its pure freezing point before it matches the chemical potential of the solid.

The core insight

A nonvolatile solute dilutes the solvent in the liquid phase only, lowering its chemical potential there while leaving the pure solid solvent unaffected. Because equilibrium requires the liquid and solid chemical potentials to be equal, the whole freezing equilibrium must shift to a lower temperature — and how much lower depends solely on solvent properties packaged into the cryoscopic constant $K_f$ .

The physical picture

Consider a solvent A in equilibrium with its own solid at the normal freezing point $T_f^*$ . At that temperature, the chemical potentials of solid and liquid A are exactly equal:

\mu_A^*(s) = \mu_A^*(l)

Now dissolve a nonvolatile, non-volatile-in-the-solid solute B into the liquid. The solute has no appreciable solubility in the solid phase (this is the standard cryoscopy assumption — the solid that freezes out is pure solvent). The liquid's chemical potential drops because it now contains a mixture, described by

\mu_A(l) = \mu_A^*(l) + RT\ln x_A

with $x_A < 1$ , so $\ln x_A < 0$ . The solid phase, unaffected by what dissolves in the liquid, stays at $\mu_A^*(s)$ . Equilibrium between solid and liquid now occurs only when the liquid is cooled below $T_f^*$ , since cooling raises $\mu_A(l)$ relative to $\mu_A^*(s)$ less quickly than... actually more precisely, cooling changes both phases' potentials, but the solid free energy curve and the diluted-liquid free energy curve cross at a lower temperature than the pure-liquid curve does. The freezing point is depressed. This is the mirror image of boiling-point elevation, where the solute lowers the liquid's chemical potential and equilibrium with the vapor is reached only at a higher temperature. Both share the same thermodynamic root, covered in Colligative Properties: Common Thermodynamic Origin — here we specialize to the solid–liquid equilibrium.

Figure

Free energy curves: pure solvent vs. solution, solid line unaffected

The core relations

\Delta T_f = K_f \, b

where: $\Delta T_f = T_f^* - T_f$ is the freezing-point depression (positive), $b$ is the solute molality (mol solute per kg solvent), $K_f$ is the cryoscopic constant of the solvent.

K_f = \frac{R \, (T_f^*)^2 \, M_A}{\Delta_{fus}H}

where: $R$ is the gas constant, $T_f^*$ the solvent's normal freezing point (K), $M_A$ its molar mass (kg mol⁻¹), $\Delta_{fus}H$ its molar enthalpy of fusion.

M_B = \frac{K_f \, m_B}{\Delta T_f \, m_A}

where: $M_B$ is the unknown solute's molar mass, $m_B$ its mass, $m_A$ the solvent mass (kg), giving molality $b = m_B/(M_B m_A)$ .

\Delta T_f = i \, K_f \, b

where: $i$ is the van't Hoff factor, correcting for solute dissociation or association (foreshadowing The van't Hoff Factor).

Derivation AFrom chemical-potential equality to ΔTf = Kf b

Goal. Starting from solid–liquid equilibrium, derive $\Delta T_f = K_f b$ for a dilute, ideal solution of a nonvolatile solute.

Step 1 — Equilibrium condition. At the freezing point $T_f$ of the solution, solid pure A is in equilibrium with A in the liquid mixture:

\mu_A^*(s,T_f) = \mu_A^*(l,T_f) + RT_f \ln x_A

Rearranging,

\ln x_A = \frac{\mu_A^*(s,T_f) - \mu_A^*(l,T_f)}{RT_f} = \frac{-\Delta_{fus}G(T_f)}{RT_f}

since $\Delta_{fus}G = \mu_A^*(l) - \mu_A^*(s)$ is the molar Gibbs energy of fusion of pure A at $T_f$ .

Step 2 — Differentiate with respect to temperature. Treat $\ln x_A$ as a function of $T_f$ implicitly defined by this equilibrium. Using the Gibbs–Helmholtz relation,

\left(\frac{\partial (\Delta G/T)}{\partial T}\right)_p = -\frac{\Delta H}{T^2}

applied to $\Delta_{fus}G$ , one obtains

\frac{d(\ln x_A)}{dT} = \frac{\Delta_{fus}H}{RT^2}

This is the same maneuver used for boiling-point elevation, but with $\Delta_{fus}H$ replacing $\Delta_{vap}H$ and a sign that now points toward lower temperature as $x_A$ decreases from 1.

Step 3 — Integrate from pure solvent to the solution. As $x_A$ runs from 1 (pure solvent, $T = T_f^*$ ) to $x_A$ (solution, $T = T_f$ ),

\int_1^{x_A} d(\ln x_A') = \int_{T_f^*}^{T_f} \frac{\Delta_{fus}H}{R\,T'^2}\, dT'

\ln x_A = -\frac{\Delta_{fus}H}{R}\left(\frac{1}{T_f} - \frac{1}{T_f^*}\right)

Step 4 — Linearize for dilute solutions. Write $x_A = 1 - x_B$ with $x_B \ll 1$ , so $\ln x_A \approx -x_B$ . Also write $T_f = T_f^* - \Delta T_f$ with $\Delta T_f \ll T_f^*$ , so

\frac{1}{T_f} - \frac{1}{T_f^*} \approx \frac{\Delta T_f}{(T_f^*)^2}

Substituting both approximations,

-x_B = -\frac{\Delta_{fus}H}{R} \cdot \frac{\Delta T_f}{(T_f^*)^2}

\Delta T_f = \frac{R\,(T_f^*)^2}{\Delta_{fus}H}\, x_B

Step 5 — Convert mole fraction to molality. For a dilute solution, $x_B \approx n_B/n_A = b\,M_A$ (with $M_A$ the solvent's molar mass in kg mol⁻¹ and $b$ the molality). Substituting,

\Delta T_f = \frac{R\,(T_f^*)^2 M_A}{\Delta_{fus}H}\, b

Identifying the prefactor as the cryoscopic constant gives the boxed result:

\boxed{\Delta T_f = K_f\, b, \qquad K_f = \frac{R\,(T_f^*)^2 M_A}{\Delta_{fus}H}}

Worked examples

Example 1 — Computing $K_f$ for water from first principles, and checking against the tabulated value. Water: $T_f^* = 273.15$ K, $M_A = 0.018015$ kg mol⁻¹, $\Delta_{fus}H = 6.01$ kJ mol⁻¹ $= 6010$ J mol⁻¹, $R = 8.314$ J K⁻¹ mol⁻¹.

K_f = \frac{(8.314)(273.15)^2(0.018015)}{6010}

Compute the numerator step by step:

(273.15)^2 = 74\,611.9\ \text{K}^2

8.314 \times 74\,611.9 = 620\,222\ \text{J K mol}^{-1}

620\,222 \times 0.018015 = 11\,175\ \text{J K kg}^{-1}\text{mol}^{-1}

Dividing by $\Delta_{fus}H$ :

K_f = \frac{11\,175}{6010} = 1.86\ \text{K kg mol}^{-1}

This matches the tabulated value for water, $K_f = 1.86$ K kg mol⁻¹, exactly what is quoted in standard tables — a good internal consistency check on the derivation. Compare this to water's $K_b = 0.51$ K kg mol⁻¹: $K_f$ is over three times larger because $\Delta_{fus}H$ (6.01 kJ mol⁻¹) is much smaller than $\Delta_{vap}H$ (40.7 kJ mol⁻¹), and $K_f \propto 1/\Delta_{fus}H$ .

Example 2 — Determining the molar mass of an unknown nonelectrolyte by cryoscopy in camphor. Camphor is prized for cryoscopy because its $K_f = 40.0$ K kg mol⁻¹ is unusually large, giving big, easily measured $\Delta T_f$ even for tiny samples. Its normal melting point is $T_f^* = 452.15$ K (179.0 °C). Suppose 0.0250 g of an unknown organic compound is dissolved in 1.000 g of camphor, and the melting point of the mixture is observed at 175.3 °C, so

\Delta T_f = 179.0 - 175.3 = 3.7\ \text{K}

Step 1: find the required molality.

b = \frac{\Delta T_f}{K_f} = \frac{3.7}{40.0} = 0.0925\ \text{mol kg}^{-1}

Step 2: molality is moles of solute per kilogram of solvent, so with $m_A = 1.000\ \text{g} = 1.000\times10^{-3}$ kg,

n_B = b \times m_A = 0.0925\ \text{mol kg}^{-1} \times 1.000\times10^{-3}\ \text{kg} = 9.25\times10^{-5}\ \text{mol}

Step 3: molar mass is mass divided by moles.

M_B = \frac{m_B}{n_B} = \frac{0.0250\ \text{g}}{9.25\times10^{-5}\ \text{mol}} = 270\ \text{g mol}^{-1}

Sanity check: this is a plausible molar mass for a small organic molecule (comparable to naphthalene derivatives or steroids), and the depression of 3.7 K is easily read on an ordinary thermometer — confirming why camphor was historically favored (the Rast method) over water, whose $K_f = 1.86$ K kg mol⁻¹ would give a depression of only about 0.17 K for the same sample, at the edge of measurement precision.

Limiting cases & common traps

Situation	What happens	Why
Ionic solute (e.g. NaCl)	$\Delta T_f$ roughly doubles per mole dissolved	Must use $i \approx 2$ (van't Hoff factor) for full dissociation
Very dilute solution	$\Delta T_f = K_f b$ holds well	Linearization in Step 4 requires $x_B \ll 1$
Concentrated solution	Deviations grow, $\Delta T_f$ vs. $b$ curves	Activity ≠ concentration; ideal-dilute assumption fails
Solute soluble in solid phase	Formula overestimates $\Delta T_f$	Derivation assumes solute excluded from the freezing solid
Choosing solvent for cryoscopy	Camphor, cyclohexane favored over water	Large $K_f$ (small $\Delta_{fus}H$ , or favorable $T_f^{*2}M_A$ ) gives measurable $\Delta T_f$

The chemist behind it

François-Marie Raoult built the quantitative foundation for colligative properties in the 1880s, measuring freezing-point depressions across dozens of solvents and solutes to establish that $\Delta T_f$ was proportional to solute concentration, independent of solute identity, for dilute solutions. This empirical law let chemists determine molar masses of unknown organic compounds — including many natural products — years before modern spectroscopic methods existed. Jacobus van't Hoff later placed Raoult's empirical rule on rigorous thermodynamic footing, connecting it to chemical potential and osmotic pressure, and introduced the correction factor for electrolytes that bears his name.

Origins · 1 minFrançois-Marie Raoult and how this was discovered

François-Marie Raoult

1830–1901 · French chemist who established the empirical laws of freezing-point depression and vapor-pressure lowering that underpin cryoscopy

Ask the tutor: How does the van't Hoff factor change the molar-mass calculation when the solute is a weak electrolyte that only partially dissociates?

Why does the solute have to be nonvolatile for the simple formula to work?

The derivation assumes the solid phase in equilibrium is pure solvent, and it treats the liquid's chemical potential using Raoult's law ideal-solution form. If the solute itself could crystallize into the solid, or if it altered the solid's composition, the simple $\mu_A^*(s)$ term would need correction. Volatility of the solute matters more directly for vapor-pressure lowering and boiling-point elevation; for freezing-point depression, what truly matters is that the solute stays entirely in the liquid and is excluded from the growing solid, which is true for most molecular solutes in water or camphor.

How is Kf different for every solvent, and can I look it up?

Yes — $K_f$ is tabulated for common solvents because it depends only on solvent properties: melting point, molar mass, and enthalpy of fusion. Water has $K_f = 1.86$ K kg mol⁻¹, benzene about 5.12 K kg mol⁻¹, and camphor a strikingly large 40.0 K kg mol⁻¹. Solvents with large $K_f$ values are preferred for cryoscopic molar-mass determinations because they produce a bigger, more easily measured temperature shift for the same small amount of dissolved solute.

Does freezing-point depression work for gases dissolved in a liquid solvent?

In principle yes, if the gas dissolves and stays in solution as the liquid is cooled, but in practice this is rarely used for molar mass determination because gas solubility is small and temperature-dependent, and gases often escape before or during freezing. Cryoscopy is almost always applied to solid or liquid solutes dissolved in a liquid solvent, where the amount of dissolved material is fixed and easily weighed.

Why does adding salt to icy roads work, and how much depression can you realistically get?

Salt (NaCl) dissociates into Na⁺ and Cl⁻, so its van't Hoff factor is close to 2, doubling the effective molality. Even so, there's a practical limit: as more salt dissolves, the solution becomes concentrated and deviates from the ideal dilute law, so $\Delta T_f$ does not grow indefinitely with more salt. In practice, saturated brine can depress the freezing point of water by around 20 K, useful for typical winter temperatures but insufficient in extreme cold.

How does freezing-point depression relate to osmotic pressure and boiling-point elevation?

All colligative properties share one origin: a nonvolatile solute lowers the solvent's chemical potential in the liquid phase. Boiling-point elevation compares liquid to vapor; freezing-point depression compares liquid to solid; osmotic pressure compares solution to pure solvent across a semipermeable membrane. Each derivation uses the same Gibbs–Helmholtz technique, just applied to a different pair of phases, which is why the resulting formulas all take the compact form (constant) × (concentration).