Freezing Point Depression Calculator

Enter how much solute you dissolved, in what mass of solvent, and this calculator returns the freezing point depression ΔTf and the temperature at which the solution actually starts to freeze. It carries the cryoscopic constant Kf and normal freezing point for eight common solvents, applies the dissociation factor i for salts and acids, and will run the calculation backwards: give it a measured depression and it solves for the solute's molar mass, which is how cryoscopy has determined molecular weights since Raoult's work in the 1880s.

Calculator

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Inputs this calculator takes, with typical values
InputWhat to enterExample
What do you want to find?Pick the second option when you have measured how far the freezing point moved and want the solute's molecular weight.Freezing point of a solution
SolventChoose the pure liquid the solute is dissolved in; its constants are filled in for you.Water — Kf 1.86, freezes 0.00 °C
Cryoscopic constant KfOnly used for a custom solvent; look it up in the CRC Handbook under molal freezing-point depression constants.1.86 °C·kg/mol
Normal freezing point of the pure solventOnly used for a custom solvent; the melting point of the pure liquid at 1 atm.0 °C
Mass of solute dissolvedWeigh the solute dry, before it goes into the solvent.100 g
Molar mass of the soluteThe formula weight — 62.07 g/mol for ethylene glycol, 58.44 for NaCl. Use the molar mass calculator if you only have a formula.62.07 g/mol
Mass of solventSolvent only, not the total solution mass — molality is defined per kilogram of solvent.1000 g
Dissociation factor iParticles released per formula unit: 1 for sugar or glycol, 2 for NaCl, 3 for CaCl₂ or Na₂SO₄.1
Measured freezing point depressionHow far below the pure solvent's freezing point your solution froze, as a positive number.2.56 °C

It returns

  • Freezing point depression ΔTf — How far the freezing point moves below that of the pure solvent.
  • Freezing point of the solution
  • Molality of the solute
  • Particle (effective) molality i·b
  • Moles of solute
  • Solute molar mass — Solved from the measured depression in molar-mass mode; the value you entered otherwise.

The formula

ΔTf=iKfb
M=iKfmsoluteΔTfmsolvent

In plain text: ΔTf = i · Kf · b and Tf = Tf° − ΔTf

  • ΔTfFreezing point depression — how far the freezing point falls (°C)
  • iVan 't Hoff dissociation factor — particles released per formula unit (dimensionless)
  • KfCryoscopic (molal freezing-point depression) constant of the solvent (°C·kg/mol)
  • bMolality — moles of solute per kilogram of solvent (mol/kg)
  • Tf°Normal freezing point of the pure solvent (°C)

A limiting law: exact only as the solution becomes infinitely dilute, and valid only while the solid that separates is pure solvent.

Updated Category Colligative Properties & Phase Behavior Verified against published test cases Reading time 13 min

Why dissolving anything lowers a freezing point

A pure liquid freezes at the temperature where its liquid and solid phases have equal chemical potential. Dissolve something in it and you dilute the solvent in the liquid phase but not in the solid, because the crystal that forms excludes the solute. The liquid becomes more stable than it was, the solid does not, and the two phases only reach equilibrium again at a lower temperature. That is the entire mechanism, and it explains the property that makes this calculation so useful: the size of the shift depends on how many dissolved particles there are, not on what they are.

Properties that behave this way are called colligative. A mole of sucrose, a mole of urea and a mole of ethylene glycol all depress water's freezing point by the same 1.86 °C per kilogram of water, despite molar masses ranging from 60 to 342 g/mol. A mole of sodium chloride depresses it by twice as much, because each formula unit releases two ions. This indifference to chemical identity is exactly why cryoscopy works as a molar-mass method — and why road salt, antifreeze and ice cream all rely on the same equation.

The practical consequences run in two directions. If you want a liquid to stay liquid in the cold, dissolve something cheap and highly dissociating in it. If you want to identify an unknown compound, dissolve a weighed sample in a solvent with a large Kf, measure how far the freezing point moves, and the equation hands you the molar mass.

The formula, variable by variable

The working equation is ΔTf = i · Kf · b, and the new freezing point is the pure solvent's freezing point minus that shift. Three quantities go in and each one trips people up in a different way.

Kf, the cryoscopic constant, is a property of the solvent alone. It is the depression a 1 molal ideal solution of a non-dissociating solute would produce, and it derives from the solvent's own melting point and enthalpy of fusion: Kf = R·(Tf°)²·Msolvent / ΔHfus. Solvents that melt high and fuse with little enthalpy have huge constants — camphor's 37.7 °C·kg/mol is twenty times water's, which is why the Rast micro-method uses camphor to get a measurable shift from a few milligrams of sample.

b, the molality, is moles of solute per kilogram of solvent — not per litre of solution. Use molality and not molarity here for a specific reason: volume changes with temperature, and a freezing-point experiment is by definition a temperature-changing experiment. Mass does not change. If you have a molarity and need molality, you need the solution density; the molality calculator handles the conversion.

i, the van 't Hoff factor, counts the particles each formula unit actually produces. It is 1 for sugars, glycols, urea and most organics; 2 for NaCl and KCl; 3 for CaCl₂, MgCl₂ and Na₂SO₄; 4 for FeCl₃. Two caveats matter. Weak electrolytes sit between whole numbers — acetic acid in water has an i just above 1 because only a small fraction of it ionises. And in real solutions the effective i drifts away from the ideal value as concentration rises, because ions crowd each other and act as if there were fewer of them.

Rearranging for molar mass gives the cryoscopic method: dissolve a weighed mass of unknown, measure ΔTf, and M = i·Kf·msolute / (ΔTf · kgsolvent). Notice that the molar mass appears nowhere in the forward equation — it enters only through the conversion of grams to moles, which is why the method reveals it.

Worked example: antifreeze, road salt, and an unknown solid

1. Ethylene glycol in water. Dissolve 100.0 g of ethylene glycol (M = 62.07 g/mol) in 1.000 kg of water. Glycol does not dissociate, so i = 1.

  1. Moles of solute: 100.0 ÷ 62.07 = 1.6111 mol.
  2. Molality: 1.6111 mol ÷ 1.000 kg = 1.6111 mol/kg.
  3. Depression: ΔTf = 1 × 1.86 × 1.6111 = 2.997 °C.
  4. Freezing point: 0.00 − 3.00 = −3.00 °C.

Ten percent glycol buys you three degrees. That is why real coolant is sold at 50% by volume rather than 10%.

2. Road salt. Now dissolve 58.44 g of NaCl — exactly one mole — in the same kilogram of water, with i = 2.

  1. Moles: 58.44 ÷ 58.44 = 1.000 mol; molality = 1.000 mol/kg.
  2. Particle molality: i·b = 2 × 1.000 = 2.000 mol/kg.
  3. Depression: 2 × 1.86 × 1.000 = 3.72 °C, so the brine freezes at −3.72 °C.

One mole of salt beats 1.61 moles of glycol, because dissociation doubles the particle count. Per kilogram of chemical the gap is wider still: 58 g of salt outperforms 100 g of glycol.

3. Molar mass of an unknown. You dissolve 1.00 g of a white solid in 20.0 g of benzene (Kf = 5.12 °C·kg/mol, freezing at 5.50 °C) and the solution freezes at 2.94 °C.

  1. Depression: 5.50 − 2.94 = 2.56 °C.
  2. Particle molality: 2.56 ÷ 5.12 = 0.500 mol/kg. Assuming i = 1, that is the solute molality.
  3. Moles present: 0.500 mol/kg × 0.0200 kg = 0.0100 mol.
  4. Molar mass: 1.00 g ÷ 0.0100 mol = 100 g/mol.

Note how sensitive the answer is to the mass of solvent: using 20.0 g rather than 1000 g is what turns a one-gram sample into a measurable 2.56 °C shift.

How to read the result, and where it stops being true

Treat the number as an upper bound on performance in dilute solution and a rough guide beyond it. The equation is a limiting law: it is exact only in the limit of infinite dilution, and its accuracy decays smoothly as concentration climbs.

Two aqueous sodium chloride cases show the size and the direction of the error. At 10% NaCl by mass the molality is 100 ÷ 58.44 ÷ 0.900 = 1.901 mol/kg, so the ideal law predicts 2 × 1.86 × 1.901 = 7.07 °C of depression, or a freezing point of −7.07 °C; the measured value is close to −6.6 °C, so the law over-predicts. Push on to the NaCl–water eutectic at 23.3% by mass and the molality is 5.198 mol/kg, giving a predicted 19.3 °C of depression — yet the measured eutectic temperature is −21.1 °C. The law now under-predicts. The error changes sign because the osmotic coefficient of concentrated sodium chloride rises back through unity, and because a single constant Kf cannot describe a 20-degree shift. Do not assume real solutions always freeze higher than the ideal prediction.

The eutectic is the hard limit. Below 23.3% NaCl there is no benefit to adding more salt: at the eutectic the liquid, the ice and the solid salt are all in equilibrium, and additional salt simply precipitates. That is the physical reason road salt stops working near −18 °C in practice and why crews switch to calcium chloride or magnesium chloride, whose eutectics lie far lower.

For the molar-mass mode, judge your answer by how big the depression was. A shift of 0.1 °C read on a thermometer good to ±0.05 °C carries a 50% uncertainty in the molar mass. Aim for a depression of at least a degree, and check whether the compound might associate or dissociate in that solvent — an apparent molar mass exactly double the expected one is the classic signature of dimerisation.

Cryoscopic constants and freezing points of common solvents

Molal freezing-point depression constants. The last column is simply 0.100 × Kf: the depression a 0.100 molal non-dissociating solute produces.
SolventNormal freezing point (°C)Kf (°C·kg/mol)ΔTf at 0.100 m (°C)
Water0.001.860.186
Ethanol−114.61.990.199
Acetic acid16.63.900.390
Chloroform−63.54.680.468
Benzene5.505.120.512
Naphthalene80.26.940.694
Cyclohexane6.5920.02.00
Camphor179.837.73.77

Constants tabulated in the CRC Handbook of Chemistry and Physics; sources differ in the last digit for naphthalene and cyclohexane.

Molality is per kilogram of solvent, not of solution

The single most common error in this calculation is dividing by the mass of the whole solution. For a 10% brine that inflates the denominator by 11% and shrinks your predicted depression by the same fraction. Weigh the solvent, or subtract the solute mass from the solution mass before you divide.

The second most common error is entering a molarity where a molality belongs. They are numerically close only in dilute aqueous solution, where one litre weighs about one kilogram. In concentrated brine, in any organic solvent, or in anything denser than water, they diverge quickly.

Mistakes that make a freezing point prediction wrong

  • Forgetting the van 't Hoff factor. Modelling NaCl with i = 1 halves the predicted depression. Every ionic solute needs its particle count.
  • Using the ideal i in concentrated solution. The effective factor for NaCl falls below 2 in the moderately concentrated range before climbing again; a measured phase diagram beats the ideal law wherever one exists.
  • Applying the law past the eutectic. Beyond the eutectic composition the solid separating is no longer pure solvent, and the equation describes nothing physical.
  • Assuming the solute stays dissolved at temperature. Solubility falls as the solution cools; salt that crystallises out no longer counts as dissolved particles.
  • Confusing depression with the new freezing point. ΔTf is a positive shift; the freezing point is the pure solvent's value minus it. For a solvent that already freezes below zero, both numbers are negative and easily swapped.
  • Reading a cloudy point as the freezing point. Supercooling routinely carries a solution several degrees below its true freezing point before ice nucleates; take the temperature at which the ice-liquid mixture plateaus, not the minimum.
  • Ignoring solvent purity in cryoscopy. Water absorbed by benzene or camphor shifts Kf's starting point and biases every molar mass you derive from it.

Where this sits among the colligative properties

Four properties respond to particle concentration alone: freezing point depression, boiling point elevation, osmotic pressure, and the vapour-pressure lowering described by Raoult's law. They are four faces of the same thermodynamics — all of them follow from the reduction in the solvent's chemical potential when a solute dilutes it — and each has a sensitivity that suits it to a different job.

Freezing point depression is the most convenient of the four for benchtop work, because Kf is typically three to four times larger than the corresponding Kb for the same solvent (water: 1.86 against 0.512) and because a freezing point is easier to hold steady than a boiling point, which drifts with barometric pressure. Osmotic pressure is by far the most sensitive — a 0.001 molal solution generates a readable pressure but shifts a freezing point by under 0.002 °C — which is why osmometry, not cryoscopy, is used for proteins and polymers.

For clinical work the same arithmetic runs in reverse: a freezing-point osmometer measures ΔTf and reports particle concentration directly, since ΔTf ÷ 1.86 is the osmolality in osmol per kilogram. A serum depression of 0.539 °C corresponds to 0.290 osmol/kg, a normal human value. If you need the concentration side of these problems first, the mole fraction calculator and the molar mass calculator feed straight into this one.

Key terms

Colligative property
A solution property that depends on the number of dissolved particles and not on their chemical identity. Freezing point depression, boiling point elevation, osmotic pressure and vapour-pressure lowering are the four.
Cryoscopic constant (K<sub>f</sub>)
The freezing point depression produced by a 1 molal ideal solution of a non-dissociating solute in that solvent, in °C·kg/mol. A property of the solvent alone.
Molality (b)
Moles of solute per kilogram of solvent. Independent of temperature, unlike molarity, which is why colligative equations use it.
Van 't Hoff factor (i)
The number of particles produced per formula unit dissolved. Ideal values are whole numbers; effective values measured in real solutions are not.
Eutectic
The composition at which a solution freezes at the lowest possible temperature, forming solid solvent and solid solute together. For NaCl and water this is 23.3% NaCl at −21.1 °C.
Cryoscopy
Determining a solute's molar mass by measuring the freezing point depression it produces. The Rast method is the camphor-based micro-scale version.

Frequently asked questions

What is the freezing point of salt water?

It depends entirely on how much salt is dissolved. A 1 molal solution — 58.4 g of NaCl per kilogram of water — freezes at about −3.7 °C by the ideal law. Seawater, at roughly 0.6 molal in total salts, freezes near −1.9 °C. Fully saturated brine reaches the eutectic at −21.1 °C and 23.3% salt by mass, which is the lowest temperature any sodium chloride solution can reach before ice and salt crystallise together.

Why do we use molality instead of molarity in this equation?

Because molarity changes with temperature and molality does not. Molarity is moles per litre of solution, and a litre expands or contracts as the solution cools toward its freezing point, so the concentration would drift during the very measurement you are making. Molality is moles per kilogram of solvent, and mass is temperature-independent. In dilute aqueous solution the two are numerically close; in anything concentrated or non-aqueous they are not.

What van 't Hoff factor should I use for my solute?

Use the number of particles one formula unit releases: 1 for sugars, glycols, urea and most covalent organics; 2 for NaCl, KCl and NaNO₃; 3 for CaCl₂, MgCl₂ and Na₂SO₄; 4 for FeCl₃ and K₃PO₄. Weak acids and bases fall between 1 and 2 depending on how far they ionise. In concentrated solution the measured factor differs from these ideal values, so treat them as the dilute-solution limit.

How does this calculator find a molar mass?

Switch the mode to Solute molar mass from a measured depression and enter the mass of solute you weighed, the mass of solvent, and how far below the pure solvent's freezing point your solution froze. The calculator converts the depression to a molality by dividing by i·Kf, multiplies by the kilograms of solvent to get moles, then divides your weighed mass by those moles. Accuracy is limited by the thermometer, so aim for a depression of at least one degree.

Why does calcium chloride outperform sodium chloride for de-icing?

Two reasons, one of which this equation captures. Each CaCl₂ unit releases three ions rather than two, so at equal molality it depresses the freezing point 50% more. It also has a much lower eutectic temperature, so it keeps working far below the point where sodium chloride brine simply freezes. Calcium chloride additionally dissolves exothermically, which helps it melt through existing ice rather than merely preventing new ice.

Can I use this for antifreeze mixtures?

Only for dilute mixtures. Automotive coolant is typically 50% ethylene glycol by volume, which is far outside the range where a linear colligative law holds — at that concentration the mixture is better described as a binary phase diagram than as a dilute solution, and glycol blends show a freezing-point minimum near 60–70% glycol beyond which the freezing point rises again. Use the manufacturer's chart or a refractometer for real coolant; use this calculator to understand why the effect exists.

What is a normal depression for a laboratory measurement?

For a molar-mass determination, aim for somewhere between 1 and 5 °C. Below one degree the thermometer error dominates; above five degrees the ideal law itself starts contributing error. Choose the solvent to hit that window: with a 100 g/mol unknown and one gram of sample, benzene needs about 20 g of solvent and camphor needs about 150 g to land in range.

Does the solute's own freezing point matter?

Not for the calculation, which is why the property is colligative — but it does set a practical limit. The equation assumes the solute stays dissolved and only pure solvent crystallises. Once you cool past the point where the solute's solubility is exceeded, it precipitates and stops contributing particles, and the freezing point stops falling. That is exactly what the eutectic is.

Why did my solution cool below the calculated freezing point without freezing?

Supercooling. Ice needs a nucleation site, and a clean, still solution can sit several degrees below its equilibrium freezing point before crystals appear. When they do, the latent heat released warms the mixture back up to the true freezing point, where it plateaus. Read the plateau, not the minimum — and seed with a crystal or a scratch on the glass if you want the plateau to arrive quickly.

References

  • CRC Handbook of Chemistry and Physics — cryoscopic constants and freezing-point depressions of aqueous solutions — CRC Press / Taylor & Francis
  • Atkins' Physical Chemistry, 11th ed. — colligative properties and the origin of the cryoscopic constant — Oxford University Press
  • Chemistry: The Central Science, 14th ed. — colligative properties and the van 't Hoff factor — Pearson (Brown, LeMay, Bursten)
  • NIST Chemistry WebBook — phase-change data for pure solventsNational Institute of Standards and Technology