Understanding the Freezing of Solvents

When a pure solvent freezes, its molecules arrange into an orderly lattice, strengthening intermolecular forces. In water, this results in a hexagonal hydrogen‑bond network that defines ice’s unique properties.

What Happens When a Solute Is Added?

Introducing a solute disrupts the solvent’s ordered structure. The solvent molecules can no longer align perfectly, so more energy must be removed for freezing to occur. In practice, this means the freezing point of the mixture is lower than that of the pure solvent—a phenomenon known as freezing point depression.

Defining Freezing Point Depression

The decrease in freezing temperature is directly proportional to the solute’s molality:

\(\Delta T_f = K_f \times m \times i\)

Where:

• K_f is the molal freezing‑point depression constant of the solvent.
  
• m is the molality (moles of solute per kilogram of solvent).
  
• i is the van’t Hoff factor, representing the number of particles the solute dissociates into. For NaCl, i = 2.

Freezing point depression is also expressed as the difference between the pure solvent’s freezing point (T_f⁰) and that of the solution (T_f):

\(\Delta T_f = T_f^{\circ} - T_f\)

Why Is Freezing Point Depression Useful?

Two everyday applications demonstrate its importance:

• Antifreeze: Ethylene glycol lowers water’s freezing point, protecting automotive radiators from freezing.
• Road Safety: Sprinkling salt on winter roads lowers ice’s melting point, reducing hazardous slicks.

Practical Example: Saltwater Freezing Point

Calculate the freezing point of a solution containing 100 g of NaCl in 1 kg of water.

1. Convert NaCl mass to moles:
   \(100\,\text{g} \times \dfrac{1\,\text{mol}}{58.44\,\text{g}} = 1.71\,\text{mol}\)
2. Determine molality:
   \(m = \dfrac{1.71\,\text{mol}}{1\,\text{kg}} = 1.71\,\text{m}\)
3. Apply the depression formula:
   \(\Delta T_f = 1.86\,\tfrac{\degree\text{C}}{\text{m}} \times 1.71\,\text{m} \times 2 = 6.4\,\degree\text{C}\)
4. Subtract from pure water’s 0 °C freezing point:
   \(T_f = 0\,\degree\text{C} - 6.4\,\degree\text{C} = -6.4\,\degree\text{C}\)

Thus, adding 100 g of salt to 1 kg of water depresses the freezing point to –6.4 °C.

TL;DR

Adding a solute like salt lowers a solvent’s freezing point. The more solute present, the greater the depression—explaining antifreeze and road salting.