Depression of Freezing Point

In the previous topic, you saw how a dissolved solute forces a solvent to boil at a higher temperature. Freezing tells the mirror-image story: the same solute pushes the freezing point downward. If you have ever wondered why cities spread salt on icy roads in winter, or why antifreeze is added to car radiators, you are already familiar with this effect in everyday life. Let us now understand the science behind it.

What Happens at the Freezing Point?

Before diving into solutions, recall what freezing means for a pure substance. At the freezing point (the temperature at which solidification occurs), the solid and liquid forms of a substance coexist in dynamic equilibrium (a state where the rates of melting and freezing are equal, so neither phase grows at the expense of the other).

There is a neat way to describe this equilibrium in terms of vapour pressure. At the freezing point, the vapour pressure of the liquid phase equals the vapour pressure of the solid phase. This might seem surprising, since we usually associate vapour pressure with boiling. But both solids and liquids exert a small vapour pressure, and at the freezing point, these two vapour pressures match exactly.

Why Does a Solute Lower the Freezing Point?

Now dissolve a non-volatile solute in the liquid solvent. From Raoult’s law (covered in earlier topics), you know that the solute lowers the vapour pressure of the liquid. The solid solvent, however, remains pure (the solute does not enter the crystal lattice during freezing), so its vapour pressure curve stays unchanged.

Here is the key consequence: after the solute is added, the liquid’s vapour pressure at the original freezing point ($T_\text{f}^0$) is now lower than the solid’s vapour pressure at the same temperature. The two phases are no longer in equilibrium at $T_\text{f}^0$. To re-establish equilibrium, you must cool the mixture further, moving to a lower temperature where the liquid’s (now reduced) vapour pressure curve finally meets the unchanged solid curve again. That lower temperature, $T_\text{f}$, is the new freezing point.

Fig 1.8: Depression of the freezing point of a solvent in a solution. The solution’s vapour pressure curve lies below the liquid solvent curve and intersects the frozen solvent curve at a lower temperature $T_\text{f}$, giving $\Delta T_\text{f} = T_\text{f}^0 - T_\text{f}$.

Figure 1.8 makes this clear. Three curves appear on a vapour pressure vs. temperature graph:

• Frozen solvent curve: rises steeply on the left side of the graph (solid has a rapidly increasing vapour pressure with temperature).
• Liquid solvent curve: rises more gently and crosses the frozen solvent curve at $T_\text{f}^0$, the freezing point of the pure solvent.
• Solution curve: lies below the liquid solvent curve (because the solute has lowered the vapour pressure) and meets the frozen solvent curve at $T_\text{f}$, a lower temperature.

The horizontal gap between $T_\text{f}^0$ and $T_\text{f}$ is the depression in freezing point.

Defining the Depression

Let $T_\text{f}^0$ be the freezing point of the pure solvent and $T_\text{f}$ be the freezing point of the solution. The depression in freezing point is:

$\Delta T_\text{f} = T_\text{f}^0 - T_\text{f}$

Notice the order of subtraction: pure solvent minus solution. Since the solution always freezes at a lower temperature, $\Delta T_\text{f}$ is always a positive number.

The Proportionality: Linking Depression to Molality

Just as with boiling point elevation, experiments with dilute solutions reveal a direct proportionality. The depression in freezing point grows linearly with the molality (number of moles of solute per kilogram of solvent):

$\Delta T_\text{f} \propto m$

Inserting a proportionality constant gives the working equation:

$\Delta T_\text{f} = K_\text{f} \, m \qquad \text{(1.34)}$

The constant $K_\text{f}$ depends only on the nature of the solvent. It carries several names:

• Freezing Point Depression Constant
• Molal Depression Constant
• Cryoscopic Constant (from the Greek kryos, meaning frost)

Its unit is $\text{K kg mol}^{-1}$.

Physically, $K_\text{f}$ tells you the number of kelvins by which the freezing point drops when exactly 1 mol of any non-volatile, non-electrolyte solute is dissolved in 1 kg of that solvent.

Cryoscopic and Ebullioscopic Constants for Common Solvents

Table 1.3 lists both $K_\text{b}$ (for boiling point elevation) and $K_\text{f}$ (for freezing point depression) for several common solvents. Having both values together is useful because many problems involve choosing the right constant for a given experiment.

Solvent	b.p. / K	$K_\text{b}$ / K kg mol$^{-1}$	f.p. / K	$K_\text{f}$ / K kg mol$^{-1}$
Water	373.15	0.52	273.0	1.86
Ethanol	351.5	1.20	155.7	1.99
Cyclohexane	353.74	2.79	279.55	20.00
Benzene	353.3	2.53	278.6	5.12
Chloroform	334.4	3.63	209.6	4.79
Carbon tetrachloride	350.0	5.03	250.5	31.8
Carbon disulphide	319.4	2.34	164.2	3.83
Diethyl ether	307.8	2.02	156.9	1.79
Acetic acid	391.1	2.93	290.0	3.90

Notice the wide variation across solvents. Carbon tetrachloride has the largest $K_\text{f}$ (31.8), meaning a given molal concentration produces a much bigger freezing point drop in $CCl_4$ than in water (1.86). A higher $K_\text{f}$ makes that solvent more sensitive for detecting small quantities of dissolved solute.

Deriving the Molar Mass Formula

The practical power of freezing point depression lies in determining the molar mass of an unknown solute from a simple freezing experiment. The derivation follows the same logic as for boiling point elevation.

Step 1: Write molality in terms of masses

Suppose $w_2$ grams of solute (molar mass $M_2$) is dissolved in $w_1$ grams of solvent. The moles of solute are $w_2 / M_2$. Molality requires kilograms of solvent, so convert $w_1$ grams by dividing by 1000:

$m = \frac{w_2 / M_2}{w_1 / 1000} = \frac{w_2 \times 1000}{M_2 \times w_1} \qquad \text{(1.31)}$

Step 2: Substitute into the depression equation

Replace $m$ in Equation (1.34) with the expression from Equation (1.31):

$\Delta T_\text{f} = K_\text{f} \times \frac{w_2 \times 1000}{M_2 \times w_1} \qquad \text{(1.35)}$

This form is useful when you want to predict how much the freezing point will drop for known masses and a known solute.

Step 3: Rearrange to isolate the molar mass

If the goal is to find $M_2$, multiply both sides by $M_2$ and divide both sides by $\Delta T_\text{f}$:

$M_2 = \frac{K_\text{f} \times w_2 \times 1000}{\Delta T_\text{f} \times w_1} \qquad \text{(1.36)}$

This is the working formula for molar mass determination via freezing point depression. You need four measurable quantities:

• $w_2$: mass of the solute (in grams)
• $w_1$: mass of the solvent (in grams)
• $\Delta T_\text{f}$: measured depression of freezing point (in K)
• $K_\text{f}$: the tabulated cryoscopic constant for the solvent (in $\text{K kg mol}^{-1}$)

Where Do the Constants Come From? Thermodynamic Expressions

You might wonder how the values of $K_\text{f}$ and $K_\text{b}$ listed in Table 1.3 are calculated rather than measured. Thermodynamics provides exact formulas that connect these constants to fundamental properties of the solvent.

For the cryoscopic constant:

$K_\text{f} = \frac{R \times M_1 \times T_\text{f}^2}{1000 \times \Delta_{\text{fus}}H} \qquad \text{(1.37)}$

For the ebullioscopic constant:

$K_\text{b} = \frac{R \times M_1 \times T_\text{b}^2}{1000 \times \Delta_{\text{vap}}H} \qquad \text{(1.38)}$

Here is what each symbol represents:

• $R$: the universal gas constant (8.314 J mol$^{-1}$ K$^{-1}$)
• $M_1$: the molar mass of the solvent (in g mol$^{-1}$)
• $T_\text{f}$: the freezing point of the pure solvent (in K)
• $T_\text{b}$: the boiling point of the pure solvent (in K)
• $\Delta_{\text{fus}}H$: the enthalpy of fusion (the energy needed to melt one mole of the solid solvent at its freezing point)
• $\Delta_{\text{vap}}H$: the enthalpy of vapourisation (the energy needed to vapourise one mole of the liquid solvent at its boiling point)

Notice the structural similarity: each formula has $R$, $M_1$, and $T^2$ in the numerator and 1000 times an enthalpy change in the denominator. The only difference is which phase transition (melting or boiling) supplies the temperature and enthalpy. A solvent with a low enthalpy of fusion will have a large $K_\text{f}$, making it very sensitive to dissolved solutes.

Worked Example 1.9: Freezing Point of an Ethylene Glycol Solution

Problem: 45 g of ethylene glycol ($C_2H_6O_2$) is mixed with 600 g of water. Calculate (a) the freezing point depression and (b) the freezing point of the solution.

Solution:

Start by finding the number of moles of ethylene glycol. Its molar mass is:

$M = 2(12) + 6(1) + 2(16) = 24 + 6 + 32 = 62 \text{ g mol}^{-1}$

$\text{Moles of ethylene glycol} = \frac{45 \text{ g}}{62 \text{ g mol}^{-1}} = 0.726 \text{ mol}$

Next, calculate the molality. Convert the mass of water to kilograms:

$\text{Mass of water} = \frac{600 \text{ g}}{1000 \text{ g kg}^{-1}} = 0.6 \text{ kg}$

$m = \frac{0.726 \text{ mol}}{0.6 \text{ kg}} = 1.21 \text{ mol kg}^{-1}$

(The NCERT rounds the moles to 0.73 and the molality to 1.2 mol kg$^{-1}$.)

(a) Apply the freezing point depression formula with $K_\text{f}$ for water = 1.86 K kg mol$^{-1}$:

$\Delta T_\text{f} = K_\text{f} \times m = 1.86 \text{ K kg mol}^{-1} \times 1.2 \text{ mol kg}^{-1} = 2.2 \text{ K}$

(b) The freezing point of water is 273.15 K, so the freezing point of the solution is:

$T_\text{f} = 273.15 \text{ K} - 2.2 \text{ K} = 270.95 \text{ K}$

This is about $-2.2 \degree C$, which explains why ethylene glycol is widely used as antifreeze in car radiators: it brings the freezing point of the coolant well below the winter temperatures that would otherwise freeze plain water.

Worked Example 1.10: Finding the Molar Mass of an Unknown Solute

Problem: 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowers the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mol$^{-1}$. Find the molar mass of the solute.

Solution:

List the known values:

• $w_2 = 1.00$ g (mass of solute)
• $w_1 = 50$ g (mass of benzene, the solvent)
• $\Delta T_\text{f} = 0.40$ K
• $K_\text{f} = 5.12$ K kg mol$^{-1}$

Substitute into the molar mass formula (Equation 1.36):

$M_2 = \frac{K_\text{f} \times w_2 \times 1000}{\Delta T_\text{f} \times w_1}$

Plug in the numbers:

$M_2 = \frac{5.12 \times 1.00 \times 1000}{0.40 \times 50}$

Compute the numerator:

$5.12 \times 1.00 \times 1000 = 5120$

Compute the denominator:

$0.40 \times 50 = 20$

Divide:

$M_2 = \frac{5120}{20} = 256 \text{ g mol}^{-1}$

The molar mass of the unknown solute is 256 g mol$^{-1}$.