Freezing Point Depression Calculator
Dissolving a solute lowers the freezing point of a solvent, a colligative property that depends only on how many particles are in solution. Choose your solvent, enter the van 't Hoff factor and molality, and the calculator returns the depression and the new freezing point. Switch to reverse mode to find molality from a measured depression, or to molar mass mode to identify an unknown solute.
Formula
Worked example
For 1 mol/kg of NaCl in water (i = 2, Kf = 1.86 °C·kg/mol): ΔTf = 2 × 1.86 × 1 = 3.72 °C. The solution freezes near 0 - 3.72 = -3.72 °C instead of 0 °C. To find the molar mass of an unknown solute: dissolve 10 g in 100 g of water and measure ΔTf = 3.72 °C (i = 1), giving m = 3.72 / 1.86 = 2.0 mol/kg, moles = 2.0 × 0.1 = 0.2 mol, M = 10 / 0.2 = 50 g/mol.
What freezing point depression is
Freezing point depression is the decrease in a solvent's freezing temperature when a non-volatile solute is dissolved in it. It is a colligative property, meaning the magnitude of the effect depends only on the number of dissolved particles, not on their chemical nature. Dissolved particles disrupt the crystal lattice the solvent must adopt to freeze, so a lower temperature is needed before the solid phase can form. The relationship is governed by the equation ΔTf = i·Kf·m, where ΔTf is the temperature drop in degrees Celsius, i is the van 't Hoff factor, Kf is the cryoscopic constant of the solvent, and m is the molality of the solution.
The inputs explained: i, Kf, and molality
The van 't Hoff factor (i) counts the number of particles each formula unit releases in solution. Non-electrolytes such as glucose or sucrose give i = 1. Sodium chloride splits into Na+ and Cl- giving i close to 2, and calcium chloride gives i near 3. In real concentrated solutions the effective i is slightly below the ideal integer because of ion pairing. The cryoscopic constant Kf is a fixed property of the pure solvent: 1.86 °C·kg/mol for water, 5.12 for benzene, and 20.0 for cyclohexane. Molality (mol/kg) is used instead of molarity (mol/L) because molality is defined per kilogram of solvent, a mass that does not change with temperature, keeping the colligative relationship exact regardless of how much the liquid contracts on cooling.
Reverse-solve and molar mass modes
The reverse-molality mode lets you enter a measured depression and work backwards: m = ΔTf / (i·Kf). The molar mass mode is especially useful in chemistry labs where you dissolve a known mass of an unknown solute in a known mass of solvent, measure the freezing point drop, and calculate the molar mass. The formula is M = i·Kf·massSolute / (ΔTf·kgSolvent). This technique, called cryoscopy, was historically one of the main ways to determine molecular weights before mass spectrometry became widely available.
Real-world applications
The most familiar use of freezing point depression is road de-icing: sodium chloride (i = 2) depresses the freezing point of water by about 3.72 °C per mol/kg, while calcium chloride (i = 3) is even more effective at around 5.58 °C per mol/kg and works at lower air temperatures. Car antifreeze uses ethylene glycol (i = 1, non-electrolyte) at high concentrations to bring the freezing point below -40 °C. In biology, organisms living in cold oceans produce antifreeze proteins or accumulate solutes like glycerol to depress the freezing point of their body fluids. In food science, salt is added to ice in home ice cream making to get the brine below 0 °C so the cream mixture can freeze while churning.
Fahrenheit output and unit notes
The calculator can display the solution's new freezing point in either Celsius or Fahrenheit. The depression ΔTf is always reported in Celsius degrees (or equivalently kelvin), because the constant Kf is tabulated in those units. The Fahrenheit output is simply the final solution freezing point converted with Tf (°F) = Tf (°C) × 9/5 + 32. If you are working with a non-aqueous solvent, select the matching preset to auto-fill both Kf and the solvent's normal freezing point, or choose "Custom solvent" and enter both values manually.
Cryoscopic constants and freezing points of common solvents
| Solvent | Kf (°C·kg/mol) | Freezing point (°C) | Freezing point (°F) |
|---|---|---|---|
| Water | 1.86 | 0 | 32 |
| Benzene | 5.12 | 5.5 | 41.9 |
| Cyclohexane | 20 | 6.5 | 43.7 |
| Ethanol | 1.99 | -114.1 | -173.4 |
| Chloroform | 4.68 | -63.5 | -82.3 |
| Acetic acid | 3.9 | 16.6 | 61.9 |
| Naphthalene | 6.8 | 80.2 | 176.4 |
| Camphor | 39.7 | 179 | 354.2 |
Kf in °C·kg/mol at the normal freezing point. Use tabulated values for precise analytical work.
Frequently asked questions
What is the van 't Hoff factor and how do I choose it?
The van 't Hoff factor (i) is the number of particles a solute produces per formula unit in solution. Use 1 for molecular non-electrolytes like sugar, urea, or glucose. Use about 2 for salts that split into two ions, such as NaCl, KCl, or LiCl. Use about 3 for salts that produce three ions, such as CaCl₂ or Na₂SO₄. In real concentrated solutions the effective i is often slightly below the ideal whole number because ion pairs form, so measured depressions can be a little smaller than predicted.
Where does the cryoscopic constant Kf come from?
Kf is derived from the thermodynamic properties of the pure solvent: Kf = R·T₀²·M₀ / (1000·ΔHfus), where R is the gas constant, T₀ is the pure solvent freezing point in kelvin, M₀ is the solvent molar mass in g/mol, and ΔHfus is the molar enthalpy of fusion in J/mol. For water this gives 1.86 °C·kg/mol. Each solvent has its own value, so always use the constant that matches the solvent you are working with.
Why does salt melt ice on roads?
Spreading salt on ice dissolves into the thin film of liquid water at the surface, creating a salt solution whose freezing point is depressed below 0 °C. As long as the air temperature stays above that lowered freezing point, the ice continues melting rather than refreezing. Because sodium chloride dissociates into two ions (i approximately 2), it depresses the freezing point roughly twice as much per mole as a non-electrolyte would. Calcium chloride (i approximately 3) is even more effective and is used when temperatures drop below about -10 °C.
How do I use the molar mass mode?
Enter the mass of your unknown solute in grams, the mass of solvent in grams, and the measured freezing point depression in °C. The calculator divides the depression by i·Kf to get molality, then divides the mass of solute (in grams) by the moles of solute (molality × kg of solvent) to return the molar mass in g/mol. Leave the van 't Hoff factor at 1 (the default) if the solute is a non-electrolyte, or turn on the advanced toggle to set i for an electrolyte.
Can I use this for non-aqueous solvents?
Yes. Select the solvent from the dropdown and the calculator automatically fills in the correct Kf and pure solvent freezing point. You can also choose "Custom solvent" and type both values yourself. Solvents with high Kf values, such as camphor (39.7 °C·kg/mol) and cyclohexane (20.0 °C·kg/mol), are particularly useful for cryoscopic molar mass determinations because a small amount of solute produces a large, easily measured depression.
Why is molality used instead of molarity in these equations?
Colligative-property equations use molality (mol/kg solvent) rather than molarity (mol/L solution) because molality is a ratio of two masses and therefore does not change with temperature. Molarity, which depends on the volume of the solution, decreases slightly as the liquid cools and contracts toward its freezing point, which would introduce a systematic error. Using molality keeps the relationship exact across a range of temperatures.