What Is Freezing Point Depression?

Freezing point depression is the decrease in the freezing point of a solvent that occurs when a solute is dissolved in it. The resulting solution freezes at a lower temperature than the pure solvent.

The most familiar example: pure water freezes at 0°C (32°F). When you dissolve salt in water, the solution freezes below 0°C — the more salt you add, the lower the freezing point drops. This is not a special property of salt — it happens with any solute dissolved in any solvent.

Key facts:

• Symbol for the change: ΔTf (delta T sub f)
• ΔTf is always a positive number representing the magnitude of the drop
• The actual new freezing point = pure solvent freezing point minus ΔTf
• The effect depends only on the number of dissolved particles, not their identity
• It is one of the four colligative properties of solutions

Why Does Freezing Point Depression Happen?

To understand why freezing point depression occurs, consider what happens at the molecular level during freezing.

When a pure solvent freezes, its molecules slow down and arrange themselves into a highly ordered crystal lattice. This requires molecules to come together in a very specific pattern at a specific temperature.

When a solute is dissolved, its particles disrupt this ordering process. Solute particles physically block solvent molecules from joining the crystal lattice, making it harder for the solid phase to form. The temperature must be lowered further — giving solvent molecules less kinetic energy — before they can successfully arrange into a crystal despite the solute interference.

In thermodynamic terms: the solute lowers the chemical potential of the liquid phase relative to the solid phase, so equilibrium between liquid and solid shifts to a lower temperature.

The more solute particles present, the greater the disruption, and the lower the temperature required to freeze the solution. This is why the effect depends on the number of particles — not what they are.

Freezing Point Depression as a Colligative Property

Freezing point depression is one of the four colligative properties of solutions. The word "colligative" comes from the Latin colligatus, meaning "bound together" — these properties all depend collectively on the number of solute particles, not their chemical identity.

The four colligative properties are:

Because colligative properties depend on particle count, a 1 molal solution of glucose (i=1) and a 1 molal solution of NaCl (i=2) produce different freezing point depressions — NaCl's effect is twice as large because each formula unit splits into two ions.

Freezing Point Depression Formula and Equation

The freezing point depression formula is:

ΔTf = i × Kf × m

And the new freezing point of the solution is found using:

Tf(solution) = Tf(pure solvent) − ΔTf

For the simplest case — a non-electrolyte solute in water (i = 1) — the freezing point depression equation reduces to:

ΔTf = Kf × m    (for non-electrolytes)

To find any one of the three variables when the other two are known, rearrange the formula:

All Variables in the Freezing Point Depression Equation Explained

Kf Values for Common Solvents — Complete Reference Table

The cryoscopic constant (Kf) is unique to each solvent. Use this table to find the correct Kf for your freezing point depression calculation:

Note: Camphor has one of the highest Kf values (37.70) — a small amount of solute causes a large, easily measurable freezing point drop, making it historically useful for determining molar masses of unknown compounds.

Van't Hoff Factor — Non-Electrolytes vs. Electrolytes

The van't Hoff factor (i) is the multiplier that accounts for how many particles a solute produces when it dissolves. It is the single most important correction when working with ionic compounds (electrolytes).

Non-Electrolytes (i = 1)

Non-electrolytes dissolve without breaking apart into ions. Each formula unit remains as one particle in solution.

• Glucose (C₆H₁₂O₆): i = 1
• Sucrose (C₁₂H₂₂O₁₁): i = 1
• Urea (CH₄N₂O): i = 1
• Ethanol (C₂H₅OH): i = 1

Electrolytes (i > 1) — Theoretical Values

Why are experimental values lower than theoretical? In concentrated solutions, oppositely charged ions attract each other and temporarily pair up (ion pairing), behaving as a single particle rather than two. This reduces the effective number of particles and lowers the observed van't Hoff factor.

For exam and homework calculations: unless told otherwise, use the theoretical integer value of i.

How to Calculate Freezing Point Depression — Step by Step

1. Identify the solute and solvent. Determine whether the solute is an electrolyte or non-electrolyte to find the van't Hoff factor (i).
2. Find the Kf value for your solvent from the reference table above. For water, Kf = 1.86 °C·kg/mol.
3. Calculate molality (m) if not given directly: m = moles of solute ÷ kg of solvent. First convert solute mass to moles using molar mass; convert solvent mass to kg.
4. Apply the freezing point depression formula: ΔTf = i × Kf × m
5. Find the new freezing point: Tf(solution) = Tf(pure solvent) − ΔTf
6. Check units — ΔTf should be in °C; m in mol/kg; Kf in °C·kg/mol.

Molality Calculation Quick Reference

m = moles of solute ÷ kg of solvent
moles of solute = mass of solute (g) ÷ molar mass (g/mol)
kg of solvent = mass of solvent (g) ÷ 1000

Worked Examples

Example 1 — Non-Electrolyte in Water (Glucose)

Problem: What is the freezing point of a solution made by dissolving 18.0 g of glucose (C₆H₁₂O₆, molar mass = 180.2 g/mol) in 500 g of water?

Given: Kf(water) = 1.86 °C·kg/mol, Tf(water) = 0°C, i = 1 (non-electrolyte)

Step 1 — Moles of glucose: 18.0 ÷ 180.2 = 0.0999 mol

Step 2 — Molality: m = 0.0999 mol ÷ 0.500 kg = 0.1998 mol/kg

Step 3 — ΔTf: ΔTf = 1 × 1.86 × 0.1998 = 0.372°C

Step 4 — New freezing point: 0 − 0.372 = −0.372°C

Example 2 — Electrolyte in Water (NaCl)

Problem: Calculate the freezing point depression when 58.44 g of NaCl (molar mass = 58.44 g/mol) is dissolved in 1.00 kg of water. Use i = 2.

Step 1 — Moles of NaCl: 58.44 ÷ 58.44 = 1.00 mol

Step 2 — Molality: m = 1.00 mol ÷ 1.00 kg = 1.00 mol/kg

Step 3 — ΔTf: ΔTf = 2 × 1.86 × 1.00 = 3.72°C

Step 4 — New freezing point: 0 − 3.72 = −3.72°C

Comparison: The same mass of glucose (non-electrolyte, i=1) would give only ΔTf = 1.86°C. NaCl produces twice the depression because it dissociates into two ions.

Example 3 — CaCl₂ with i = 3

Problem: What is the freezing point of a solution of 2.0 mol/kg CaCl₂ in water? Use i = 3, Kf = 1.86°C·kg/mol.

ΔTf = 3 × 1.86 × 2.0 = 11.16°C

New freezing point: 0 − 11.16 = −11.16°C

This is why CaCl₂ is preferred over NaCl for de-icing roads in very cold climates — its higher van't Hoff factor provides greater freezing point depression per mole.

Example 4 — Organic Solvent (Benzene)

Problem: 5.0 g of naphthalene (C₁₀H₈, molar mass = 128.2 g/mol) is dissolved in 100 g of benzene. Find the new freezing point of the solution. Kf(benzene) = 5.12°C·kg/mol, Tf(benzene) = 5.5°C, i = 1.

Step 1: moles = 5.0 ÷ 128.2 = 0.039 mol

Step 2: m = 0.039 ÷ 0.100 = 0.390 mol/kg

Step 3: ΔTf = 1 × 5.12 × 0.390 = 2.00°C

Step 4: Tf(solution) = 5.5 − 2.00 = 3.5°C

Example 5 — Finding Molality from Measured ΔTf

Problem: The freezing point of an aqueous NaCl solution is −2.79°C. What is the molality of the solution? (i = 2, Kf = 1.86)

ΔTf = 0 − (−2.79) = 2.79°C

m = ΔTf ÷ (i × Kf) = 2.79 ÷ (2 × 1.86) = 2.79 ÷ 3.72 = 0.75 mol/kg

Finding Molar Mass from Freezing Point Depression (Cryoscopy)

One of the most important historical applications of the freezing point depression equation is cryoscopy — using measured freezing point depression to calculate the unknown molar mass of a solute.

Molar Mass Formula

M = (i × Kf × w × 1000) ÷ (ΔTf × W)

Where:

• M = molar mass of solute (g/mol)
• w = mass of solute (g)
• W = mass of solvent (g)
• Kf = cryoscopic constant of solvent
• ΔTf = measured freezing point depression (°C)
• i = van't Hoff factor (= 1 for unknown non-electrolytes)

Cryoscopy Example

Problem: 6.0 g of an unknown organic compound is dissolved in 200 g of benzene. The freezing point drops from 5.5°C to 3.68°C. Find the molar mass. (Kf = 5.12°C·kg/mol, i = 1)

ΔTf = 5.5 − 3.68 = 1.82°C

M = (1 × 5.12 × 6.0 × 1000) ÷ (1.82 × 200) = 30,720 ÷ 364 = 84.4 g/mol

This method was widely used before modern spectroscopy was available and is still taught as a fundamental experimental technique in physical chemistry.

Real-Life Applications of Freezing Point Depression

1. Road De-icing (Salt on Ice)

Spreading salt (NaCl or CaCl₂) on icy roads is the most widely recognized application. NaCl lowers water's freezing point to about −9°C; CaCl₂ (i=3) is used in extreme cold because it lowers it further to about −29°C. The de-icer works by dissolving in the thin layer of liquid water on ice, creating a salt solution with a much lower freezing point — so the ice melts even when air temperature is below 0°C.

2. Antifreeze in Car Radiators

Ethylene glycol mixed with water is the standard automotive antifreeze. A 50/50 mixture by volume lowers the freezing point to approximately −37°C and raises the boiling point to about 106°C, protecting the engine year-round. The concentration is calculated directly using the freezing point depression equation.

3. Ice Cream and Food Science

Ice cream is kept soft and scoopable at freezer temperatures because sugar and other solutes depress the freezing point of the water in the mixture. At −18°C (typical freezer), only a portion of the water is frozen — the unfrozen liquid phase gives ice cream its creamy texture. If ice cream were made from pure water, it would be rock-solid.

4. Pharmaceutical Cryopreservation

Biological samples, vaccines, and donor organs must be stored at sub-zero temperatures without forming damaging ice crystals. Cryoprotectant agents (glycerol, DMSO) work by depressing the freezing point of biological fluids, preventing the sudden crystallization that would rupture cell membranes.

5. Blood and Body Fluids

Human blood plasma freezes at approximately −0.56°C rather than 0°C because of dissolved salts, proteins, and glucose. This slight depression is used clinically as a measure of plasma osmolality — an important indicator of dehydration and kidney function.

6. Oceanography and Sea Ice

Seawater (approximately 3.5% NaCl by mass) freezes at about −1.8°C rather than 0°C. This difference has major implications for ocean circulation, marine ecology, and polar climate systems. Oceanographers use freezing point depression equations to model sea ice formation.

7. Determining Molar Mass (Cryoscopy)

Chemists historically used cryoscopy — dissolving unknown compounds in solvents with high Kf values like camphor — to determine molar masses before mass spectrometry. The technique is still used in teaching labs today.

Freezing Point Depression vs. Boiling Point Elevation

These two colligative properties are closely related and are often taught together. Both depend on solute concentration and the van't Hoff factor, but they act in opposite directions.

Key insight: For water, Kf (1.86) is about 3.6× larger than Kb (0.512). This means dissolving a solute has a bigger impact on the freezing point than on the boiling point. Adding 1 mol/kg of NaCl (i=2) to water lowers the freezing point by 3.72°C but only raises the boiling point by 1.024°C.

Frequently Asked Questions