Depression in freezing point

Why freezing point drops at all. A pure solvent freezes at the one temperature where its solid and liquid forms have equal vapour pressure. Dissolving a non-volatile solute lowers the liquid's vapour pressure at every temperature (Raoult's law), while the solid solvent's vapour-pressure curve is untouched. The two curves now cross at a lower temperature, so the solution must be cooled further before freezing sets in. This shortfall is the depression in freezing point:

where: $T_f^{0}$ is the freezing point of the pure solvent and $T_f$ is the freezing point of the solution (always $T_f < T_f^{0}$).

Why molality, not molarity. For dilute solutions, experiment shows $\Delta T_f$ is directly proportional to the solution's molality $m$:

where: $K_f$ is the molal depression constant (also called the cryoscopic constant), with units $\text{K kg mol}^{-1}$. Molality is used rather than molarity because molality is defined per kilogram of solvent and does not change with temperature, whereas molarity (per litre of solution) drifts as the solution expands or contracts on cooling — a serious problem right at the freezing point.

What $K_f$ actually is. $K_f$ depends only on the solvent, never on the solute (as long as the solute is a non-volatile non-electrolyte). It equals the freezing-point depression a hypothetical 1 molal solution would show. Water has a modest $K_f = 1.86\ \text{K kg mol}^{-1}$; benzene's is larger at $5.12\ \text{K kg mol}^{-1}$; camphor's is unusually large, near $40\ \text{K kg mol}^{-1}$, which is why camphor historically made a convenient solvent for molar-mass work — a small sample gives an easily measurable $\Delta T_f$.

Turning it into a molar-mass tool. Writing molality in terms of masses, with $w_2$ g of solute of molar mass $M_2$ dissolved in $w_1$ g of solvent:

Substituting into $\Delta T_f = K_f m$ and rearranging for $M_2$ gives the standard working formula:

where: all masses are in grams, $\Delta T_f$ in kelvin, and $K_f$ in $\text{K kg mol}^{-1}$. This single equation is how an unlabelled non-electrolyte's molar mass gets found from nothing more than a thermometer and a scale.

A solution of 2.5 g of an unknown non-electrolyte in 100 g of water freezes at $-0.372\ ^{\circ}\text{C}$. Pure water freezes at $0\ ^{\circ}\text{C}$ and $K_f$ for water is $1.86\ \text{K kg mol}^{-1}$. Find the molar mass of the solute, and then find its freezing point if the same solute-solvent ratio were dissolved in benzene instead.

Step 1. Find the depression: $\Delta T_f = T_f^{0} - T_f = 0 - (-0.372) = 0.372\ \text{K}$.

Step 2. Apply the molar-mass formula with $w_2 = 2.5\ \text{g}$, $w_1 = 100\ \text{g}$, $K_f = 1.86\ \text{K kg mol}^{-1}$: $M_2 = \dfrac{1000 \times 1.86 \times 2.5}{0.372 \times 100} = \dfrac{4650}{37.2}$.

Step 3. Evaluate: $M_2 = 125\ \text{g mol}^{-1}$. The unknown solute's molar mass is 125 g/mol.

Competitive stretch

Step 4. Now dissolve the same 2.5 g of this 125 g/mol solute in 100 g of benzene ($K_f = 5.12\ \text{K kg mol}^{-1}$, $T_f^{0} = 5.5\ ^{\circ}\text{C}$) instead of water. The molality is unchanged in form: $m = \dfrac{2.5/125}{100/1000} = \dfrac{0.020}{0.100} = 0.200\ \text{mol kg}^{-1}$.

Step 5. Compute the new depression and freezing point: $\Delta T_f = 5.12 \times 0.200 = 1.024\ \text{K}$, so $T_f = 5.5 - 1.024 = 4.48\ ^{\circ}\text{C}$. The same solute in benzene depresses the freezing point almost three times as much as in water — entirely because benzene's $K_f$ is larger, not because the solute changed.

• Using molarity instead of molality in $\Delta T_f = K_f m$ — the law needs a temperature-independent concentration measure, and only molality (mass-based) qualifies.
• Dropping the factor of 1000 when converting the solvent mass from grams to kilograms inside the molar-mass formula, which throws the final $M_2$ off by three orders of magnitude.
• Applying $\Delta T_f = K_f m$ unmodified to an electrolyte such as NaCl or $CaCl_2$ — dissociating solutes need the van't Hoff factor $i$, giving $\Delta T_f = i\, K_f\, m$, since each formula unit produces more than one particle in solution.
• Treating $K_f$ as if it belonged to the solute — it is fixed by the solvent alone; switching solutes at the same molality never changes $K_f$, only switching solvents does.

Ask the tutor: want to work through why camphor's unusually large $K_f$ made it historically useful for molar-mass determination, or try a problem that mixes freezing-point depression with the van't Hoff factor for an electrolyte?