Freezing Point Depression Calculator
Calculate the freezing point depression of a solution using the colligative property formula ΔTf = Kf × m × i. Enter the solvent, molality of solute, and van't Hoff factor to determine how much the freezing point decreases when a solute is dissolved. Essential for antifreeze formulations, food science, and cryoscopy. See also our Boiling Point Calculator and Osmotic Pressure Calculator.
mol solute / kg solvent
NaCl=2, CaCl₂=3, glucose=1
How to Calculate Freezing Point Depression
Freezing point depression is a colligative property that describes the lowering of a solvent's freezing point when a solute is dissolved in it. Like boiling point elevation, it depends only on the number of solute particles, not their chemical identity. This phenomenon is the basis for using salt to de-ice roads, antifreeze in car radiators, and cryoscopy for determining molar masses.
The freezing point of a liquid is the temperature at which the solid and liquid phases are in equilibrium. When a solute is added, it lowers the chemical potential of the liquid phase (by increasing entropy), so a lower temperature is needed to reach equilibrium with the solid phase. The depression is proportional to the molal concentration of all solute particles in solution.
- Identify the solvent and look up its cryoscopic constant (Kf)
- Calculate the molality of the solution (moles solute / kg solvent)
- Determine the van't Hoff factor (i) based on dissociation
- Calculate ΔTf = Kf × m × i
- Subtract ΔTf from the normal freezing point to get the new freezing point
Formula
Freezing Point Depression:
ΔTf = Kf × m × i
Where:
ΔTf = freezing point depression (°C)
Kf = cryoscopic constant (°C·kg/mol)
m = molality (mol solute / kg solvent)
i = van't Hoff factor
New Freezing Point:
T_new = T_normal - ΔTf
Molar Mass Determination (Cryoscopy):
MW = (Kf × mass_solute × i) / (ΔTf × mass_solvent_kg)
Common Kf Values:
Water: 1.86 °C·kg/mol (FP = 0°C)
Benzene: 5.12 °C·kg/mol (FP = 5.5°C)
Camphor: 37.7 °C·kg/mol (FP = 176°C)
Example Calculation
Problem: Calculate the freezing point of a solution made by dissolving 58.44 g of NaCl in 1 kg of water.
Given: Kf(water) = 1.86 °C/m, MW(NaCl) = 58.44 g/mol, i = 2
Solution:
Molality = 58.44 g ÷ 58.44 g/mol ÷ 1 kg = 1.00 m
ΔTf = 1.86 × 1.00 × 2 = 3.72 °C
New FP = 0.0 - 3.72 = -3.72 °C
Problem 2: What is the freezing point of a 2.5 m glucose solution in water?
ΔTf = 1.86 × 2.5 × 1 = 4.65 °C
New FP = 0.0 - 4.65 = -4.65 °C
Cryoscopic Constants Reference Table
| Solvent | Kf (°C·kg/mol) | Normal FP (°C) | Sensitivity |
|---|---|---|---|
| Water | 1.86 | 0.0 | Low — common solvent |
| Benzene | 5.12 | 5.5 | Moderate — organic solutes |
| Cyclohexane | 20.0 | 6.6 | High — precise MW determination |
| Camphor | 37.7 | 176.0 | Very high — classic cryoscopy |
| Acetic acid | 3.90 | 16.6 | Moderate — polar solutes |
| Naphthalene | 6.94 | 80.2 | Moderate — aromatic solutes |
| Carbon tetrachloride | 29.8 | -22.9 | High — non-polar solutes |
| Ethanol | 1.99 | -114.1 | Low — miscible with water |
| Nitrobenzene | 8.1 | 5.7 | High — polar organic |
| Phenol | 7.27 | 41.0 | High — hydrogen bonding |
Practical Applications
Road de-icing is the most visible application of freezing point depression. Rock salt (NaCl) lowers the freezing point of water to about -21°C at saturation, preventing ice formation on roads. Calcium chloride (CaCl₂) is even more effective because it produces three ions (i=3) and is exothermic when dissolving, working down to about -29°C.
Automotive antifreeze (ethylene glycol) exploits freezing point depression to prevent engine coolant from freezing in winter. A 50/50 mixture of ethylene glycol and water has a freezing point of approximately -37°C. The same mixture also raises the boiling point to about 108°C, providing year-round protection.
In food science, freezing point depression determines the texture and storage properties of frozen foods. Sugar and salt in ice cream lower the freezing point, keeping it soft and scoopable at freezer temperatures. The freezing point of milk (normally -0.540°C) is used as a quality test — added water raises the freezing point, indicating adulteration.
Cryoscopy (freezing point depression measurement) was historically one of the primary methods for determining molar masses of unknown compounds. Camphor is particularly useful as a cryoscopic solvent because its very large Kf (37.7) produces easily measurable temperature changes even for small amounts of solute. This technique is still used in teaching laboratories and for compounds that cannot be analyzed by other methods.
Frequently Asked Questions
What is freezing point depression?
Freezing point depression is the decrease in the freezing point of a solvent caused by dissolving a solute. It is a colligative property that depends on the number of solute particles, not their identity. The depression is calculated as ΔTf = Kf × m × i, where Kf is the cryoscopic constant of the solvent.
Why does salt melt ice?
Salt dissolves in the thin layer of liquid water on ice surfaces, creating a solution with a lower freezing point. Since the solution's freezing point is below the ambient temperature, the ice continues to melt. NaCl can lower the freezing point to -21°C, while CaCl₂ works down to -29°C.
What is the cryoscopic constant?
The cryoscopic constant (Kf) is a property of the solvent that relates molal concentration to freezing point depression. It depends on the solvent's molar mass, freezing point, and enthalpy of fusion: Kf = RT²M / (1000 × ΔHfus). Solvents with low ΔHfus have large Kf values.
How does antifreeze work?
Antifreeze (ethylene glycol or propylene glycol) dissolves in water and lowers the freezing point through colligative effects. A 50% ethylene glycol solution freezes at about -37°C. Since glycol is a non-electrolyte (i=1), high concentrations are needed for significant depression.
Can freezing point depression determine molar mass?
Yes. By measuring ΔTf for a known mass of solute in a known mass of solvent: MW = (Kf × mass_solute × i) / (ΔTf × mass_solvent_kg). This cryoscopic method works best with solvents that have large Kf values (like camphor, Kf = 37.7) for maximum sensitivity.
Why is camphor used in cryoscopy?
Camphor has an exceptionally large cryoscopic constant (Kf = 37.7 °C·kg/mol), about 20 times that of water. This means even small amounts of solute produce large, easily measurable freezing point changes. Its high melting point (176°C) also makes it convenient to work with in the laboratory.