Audited 29 Jul 2026·Last updated 28 Oct 2026·6 citations·Tier 1·0 uses

Freezing Point Depression Calculator (ΔTf = i · Kf · b)

Solve ΔTf = i·Kf·b for a solution's freezing point, find an unknown solute's molar mass by cryoscopy, or derive a solvent's Kf from its enthalpy of fusion.

Freezing Point Depression Calculator

What do you want to work out?
Grams of dissolved substance. The default 62.068 g is exactly one mole of ethylene glycol.
g
Default 62.068 g/mol = ethylene glycol (C2H6O2, antifreeze), from IUPAC/CIAAW 2021 abridged atomic weights: 2(12.011) + 6(1.008) + 2(15.999). Not needed in cryoscopy mode — that mode computes it.
g/mol
Grams of solvent alone, not of the finished solution. 1000 g of water is 1.000 kg but about 1.003 L at 20 °C — weigh it, do not measure the volume.
g
Particles released per formula unit. Use 1 for a non-electrolyte such as glucose, ethylene glycol or urea. For a salt the IDEAL value is the ion count (NaCl 2, CaCl2 3), but real measured values are lower because of ion pairing — see the FAQ before using an integer for an electrolyte.
Default 1.86 K·kg/mol for water — the standard handbook value, independently confirmed as 1.8605 by deriving it from Feistel & Wagner's (2006) ice melting enthalpy. For any other solvent, use the derivation mode rather than copying a table: published values for organic solvents disagree between sources by several percent.
K·kg/mol
Normal freezing point at 101.325 kPa. Water is 0 °C by definition of the Celsius scale; benzene is 5.49 °C; cyclohexane about 6.5 °C.
°C
How far the freezing point dropped, entered as a positive magnitude. A temperature difference has the same numeric value in K and °C. Cryoscopy mode only.
K
Default 18.015 g/mol = water, IUPAC/CIAAW 2021 abridged. Used only by the Kf derivation mode. Benzene is 78.114 g/mol.
g/mol
Default 6.0067 kJ/mol for water = 333.427 kJ/kg × 0.018015 kg/mol, from Feistel & Wagner (2006), J. Phys. Chem. Ref. Data 35, 1021, melting-curve table at the normal-pressure melting point. Look your solvent up in the NIST Chemistry WebBook. Kf derivation mode only.
kJ/mol
Freezing-point depression
1.86
How far the freezing point falls, in kelvin, as a positive number; the same figure in °C because it is a difference, and reliable mainly for dilute solutions.
Freezing point of the solution
-1.86 °C
Molality
1 mol/kg
Molar mass of solute
62.068 g/mol
Cryoscopic constant used
1.86 K·kg/mol

Background.

Seawater freezes at about −1.9 °C instead of 0 °C, a glass of sweet tea in the freezer turns slushy before it turns solid, and a road sprinkled with salt stays wet on a morning that would otherwise glaze it with ice. All three are the same effect. Dissolve almost anything in a liquid and the liquid has to be cooled further before it freezes.

In dilute solutions the size of that drop depends on how many dissolved particles there are, not on what they are. The working equation is ΔTf = i · Kf · b. Here b is the molality, moles of solute per kilogram of solvent; Kf is the solvent's cryoscopic constant, 1.86 K·kg/mol for water; and i, the van 't Hoff factor, is the number of particles each dissolved formula unit releases. One mole of ethylene glycol in a kilogram of water therefore lowers the freezing point by 1.86 K, to −1.86 °C.

There are three ways to use the relation. Run it forwards to find the freezing point of a solution you have made. Run it backwards — cryoscopy — to find the molar mass of an unknown substance from how far it lowers a solvent's freezing point. Or work out Kf itself from the solvent's freezing point, molar mass and enthalpy of fusion, which matters for solvents other than water, where tabulated constants do not always agree.

The equation is a dilute-solution approximation. It holds well up to roughly 0.1 mol/kg, degrades above that, and degrades sooner for salts, whose ions do not behave as fully independent particles. Engine coolant, road brine and clinical samples are all handled with measured data in practice, and the limits further down explain why.

What is freezing point depression calculator?

Freezing-point depression is one of the four colligative properties of solutions, alongside vapour-pressure lowering, boiling-point elevation and osmotic pressure. Colligative means the effect depends on the number of dissolved particles rather than their chemical identity: a mole of glucose and a mole of urea in the same kilogram of water lower its freezing point by the same amount. The reason lies in the liquid. Dissolved solute lowers the chemical potential of the liquid solvent, while the ice that forms is essentially pure solvent and is unaffected. Freezing happens where liquid and solid have equal chemical potential, so with the liquid's lowered, the two meet at a lower temperature. The concentration measure is molality, moles of solute per kilogram of solvent, because it contains no volume and so does not change as the solution is cooled. Clinical laboratories use the same principle in freezing-point osmometers, which measure how far a blood or urine sample's freezing point falls to report its osmolality.

How to use this calculator.

  1. Choose what to work out: the freezing point of a solution, the molar mass of a solute from a measured depression, or the cryoscopic constant of a solvent from its thermodynamic data.
  2. Enter the mass of solvent alone, in grams, not the mass or volume of the finished solution. Weigh water rather than measuring it by volume, since a litre of water is not exactly a kilogram at room temperature.
  3. Set the van 't Hoff factor: 1 for substances that dissolve as whole molecules, such as glucose, urea or ethylene glycol. For a salt the ion count — 2 for NaCl, 3 for CaCl₂ — is an upper limit; real solutions behave as if the factor were smaller.
  4. Keep Kf at 1.86 for water. For another solvent, use the derivation mode with its normal freezing point, molar mass and enthalpy of fusion, which the NIST Chemistry WebBook lists for many compounds.
  5. In cryoscopy mode, enter the measured depression as a positive number: a solvent that freezes at 0 °C and a solution that freezes at −0.62 °C give a depression of 0.62 K.
  6. Read the depression in kelvin and the solution's freezing point in °C, and check the molality output: above about 0.1 mol/kg, treat the result as an approximation.

The formula.

ΔTf = i · Kf · b Kf = R · Tf² · M_A ⁄ ΔfusH

Four relations do all the work. Molality is b = (solute mass ÷ molar mass) ÷ kilograms of solvent. The depression is ΔTf = i · Kf · b, and the solution's freezing point is the pure solvent's freezing point minus ΔTf. Run backwards, b = ΔTf ÷ (i · Kf), and the solute's molar mass is M = solute mass ÷ (b × kilograms of solvent). The cryoscopic constant is Kf = R · Tf² · M_A ÷ ΔfusH, with R = 8.314462618 J/(mol·K), Tf the solvent's freezing point in kelvin, M_A its molar mass in kg/mol and ΔfusH its molar enthalpy of fusion in J/mol.

The linear equation is a simplification of an exact condition: at the freezing point the solvent's chemical potential is the same in the liquid and in the solid. Written out for an ideal solution, that condition contains ln(1 − x), where x is the solute's mole fraction. Replacing ln(1 − x) with −x, treating the enthalpy of fusion as constant over the small temperature range, and converting mole fraction to molality give ΔTf = Kf · b. The approximations are good when x is small, which is why the equation is reliable for dilute solutions and drifts as concentration rises. It also assumes the solute does not evaporate and does not enter the solid.

The constant can be checked from first principles. Feistel and Wagner's 2006 equation of state for ice, adopted by IAPWS, gives a melting enthalpy of 333.427 kJ/kg; multiplied by water's 0.018015 kg/mol that is 6.0067 kJ/mol. Then Kf = 8.314462618 × 273.15² × 0.018015 ÷ 6,006.7 = 1.8605 K·kg/mol, within 0.03 % of the handbook 1.86. For benzene, NIST data — freezing point 278.64 K, enthalpy of fusion 9.8663 kJ/mol, molar mass 78.114 g/mol — give 5.111 K·kg/mol against the usual tabulated 5.12. A Kf derived this way is applied to the depression at full precision.

The formula shows what makes a solvent sensitive: a high freezing point, which is squared, heavy molecules and a small enthalpy of fusion.

Some entries cannot give an answer and are refused: a van 't Hoff factor, cryoscopic constant, solvent mass or solute molar mass of zero or less; in cryoscopy, a zero depression or zero solute mass, which would imply an infinite or undefined molar mass; and in the derivation, a freezing point at or below absolute zero, or a zero enthalpy of fusion.

A worked example.

Example

Dissolve 62.068 g of ethylene glycol, the main ingredient of most car antifreeze, in 1,000 g of water. Ethylene glycol is C₂H₆O₂, and with the 2021 IUPAC abridged atomic weights its molar mass is 2 × 12.011 + 6 × 1.008 + 2 × 15.999 = 62.068 g/mol, so the sample is exactly 1 mol. One mole in 1.000 kg of water is a molality of 1 mol/kg. Glycol dissolves as whole molecules, so i = 1, and water's Kf is 1.86 K·kg/mol. The depression is ΔTf = 1 × 1.86 × 1 = 1.86 K, and the solution freezes at 0 − 1.86 = −1.86 °C. That is a modest return for 62 g of glycol: a whole mole per kilogram buys less than two degrees of protection. A coolant that has to survive a hard winter therefore needs far more glycol than this, at concentrations where the dilute-solution equation no longer applies, which is why coolant freezing points come from measured tables rather than from this formula.

solvent Freezing Point0
solvent Mass1,000
van T Hoff Factor1
cryoscopic Constant1.86
solute Mass62.068
solve Fordepression
solute Molar Mass62.068

Frequently asked questions.

Why does dissolving something in water lower its freezing point?
Because the solute changes the liquid but not the ice. Ice that forms from a solution is almost pure water, since the dissolved particles are excluded from the crystal. In the liquid, those particles dilute the water and lower its chemical potential. Freezing happens where liquid and solid are in balance, and with the liquid's potential lowered, that balance is reached only at a lower temperature. Solutes that do enter the solid, forming a solid solution, break this picture.
How do I choose the van 't Hoff factor?
Use 1 for anything that dissolves as whole molecules: sugars, urea, alcohols, ethylene glycol. For an ionic compound the ideal value is the number of ions per formula unit — 2 for NaCl, 3 for CaCl₂ or Na₂SO₄ — but in dilute solutions the measured effect is smaller, because oppositely charged ions partly act together. The effective factor is i = ν · φ, ion count times osmotic coefficient, and Hamer and Wu's tables show φ below 1 for 1:1 salts across the dilute range.
Where does water's cryoscopic constant of 1.86 come from?
It follows from water's own properties through Kf = R · Tf² · M_A ÷ ΔfusH. Using the gas constant, 273.15 K, a molar mass of 0.018015 kg/mol and a molar enthalpy of fusion of 6,006.7 J/mol — from the 333.427 kJ/kg melting enthalpy in Feistel and Wagner's 2006 equation of state for ice — gives 1.8605 K·kg/mol. That matches the handbook 1.86 to 0.03 %, and the derivation mode reproduces it when you enter water's data.
Why is the concentration in molality rather than molarity?
Because the measurement involves cooling the solution, and molarity changes with temperature. Molarity is moles per litre of solution, and a litre shrinks or grows as the liquid cools or warms. Molality is moles per kilogram of solvent, and mass does not change with temperature, so the concentration you made up at room temperature is still the concentration at the freezing point. NIST and IUPAC both define molality per kilogram of solvent, not of solution.
How does cryoscopy find an unknown molar mass?
Dissolve a weighed amount of the unknown in a weighed amount of solvent, measure how far the freezing point falls, then work backwards. Dissolve 2.00 g in 50.0 g of water and see a 0.62 K drop: b = 0.62 ÷ 1.86 = 0.3333 mol/kg, and M = 2.00 ÷ (0.3333 × 0.0500) = 120 g/mol. If the unknown dissociates, the method returns the true molar mass divided by its van 't Hoff factor, so check that first.
Why does sugar lower the freezing point less than the same mass of antifreeze?
Because the equation counts molecules, and sucrose molecules are heavy. Sucrose, C₁₂H₂₂O₁₁, has a molar mass of 342.297 g/mol, about 5.5 times ethylene glycol's. So 62.068 g of sucrose in a kilogram of water is only 0.1813 mol/kg, and it lowers the freezing point by 1.86 × 0.1813 = 0.3373 K, against 1.86 K for the same mass of glycol. Per gram, light molecules and salts that split into ions do far more.
How accurate is the equation for concentrated solutions?
It is a dilute-solution limit, built on the ln(1 − x) ≈ −x approximation and on ideal behaviour. Up to about 0.1 mol/kg it works well for non-electrolytes; beyond that it drifts, and for salts it departs sooner. Strong brines need measured phase diagrams instead: a sodium chloride brine can stay liquid down to about −21 °C at roughly 23 % salt by mass, and no amount of extra salt takes it lower, something the linear equation cannot predict.
Can I use this to choose an antifreeze mix for my car?
No. Working coolant concentrations are far outside the dilute range, and the equation does not capture how glycol and water actually behave there — pure ethylene glycol itself freezes at about −13 °C, so more glycol is not always better. A coolant also has to protect against corrosion, carry heat and suit the engine's seals, none of which is colligative. Follow the vehicle maker's specification and the coolant manufacturer's freeze-protection chart.
Is a freezing-point depression the same in kelvin and in degrees Celsius?
Yes. The two scales differ by a fixed offset of 273.15 but have degrees of the same size, so a temperature difference has the same number in both; NIST SP 811 states this in section 8.5. A depression of 1.86 K is a drop of 1.86 °C. Absolute temperatures are another matter, which is why the derivation of Kf converts the solvent's freezing point to kelvin before squaring it.
Which solvents show the largest freezing-point depression?
Those with a large Kf: a high freezing point, heavy molecules and a small enthalpy of fusion. Benzene's Kf of about 5.11 K·kg/mol is roughly 2.7 times water's, and camphor's is roughly twenty times, which is why camphor was the classic solvent for estimating molar masses by melting-point depression. A larger constant turns a small amount of unknown into a temperature change big enough to measure with an ordinary thermometer.

References& sources.

  1. [1]Feistel, R. & Wagner, W. (2006). 'A New Equation of State for H2O Ice Ih', Journal of Physical and Chemical Reference Data 35(2), 1021–1047 — the equation of state adopted by IAPWS in 2006. Melting-curve table at the normal-pressure melting point (T = 273.152519 K, p = 0.1013 MPa) gives Δh_melt = 333.427 kJ/kg; §3.5 compares this with the experimental determinations of Giauque & Stout (1936: 333.49 and 333.42), Osborne (1939: 333.54) and Haida et al. (1974: 333.41). This is the value from which this page derives Kf(water) = 1.8605 K·kg/mol. Peer-reviewed, free PDF; retrieved 2026-07-29.
  2. [2]NIST Chemistry WebBook, SRD 69 — phase-change data. For benzene: T_fus = 278.64 ± 0.08 K (average of 57 values) and Δ_fus H = 9.8663 kJ/mol at 278.69 K from Oliver, Eaton & Huffman (1948), compilation DH. Used for the worked non-water derivation on this page, which returns Kf = 5.111 K·kg/mol against the commonly tabulated 5.12. Retrieved 2026-07-29.
  3. [3]NIST Special Publication 811, 2008 edition, section 8.6.8 'Molality of solute B' — b_B = n_B/m_A, SI unit mol/kg, defined per mass of SOLVENT. Section 8.5 'Temperature interval and temperature difference' establishes that {Δt}°C = {ΔT}K, which is why the depression on this page needs no unit conversion. Retrieved 2026-07-29.
  4. [4]IUPAC, 'Quantities, Units and Symbols in Physical Chemistry' (Green Book), 3rd edition, 2nd printing 2012, section 2.10 composition table: molality m_B, b_B = n_B/m_A with SI unit mol kg⁻¹, and note 14 on the m/b symbol collision. Confirms independently of NIST that the concentration variable in the colligative equations is per kilogram of solvent. Retrieved 2026-07-29.
  5. [5]Hamer, W. J. & Wu, Y.-C. (1972). 'Osmotic Coefficients and Mean Activity Coefficients of Uni-univalent Electrolytes in Water at 25 °C', Journal of Physical and Chemical Reference Data 1(4), 1047–1100. Tabulates practical osmotic coefficients φ for 79 uni-univalent electrolytes; the effective van 't Hoff factor is i = ν·φ, and φ is below 1 throughout the dilute range, which is why a real salt depresses less than its ion count predicts. Cited for the direction and origin of the deviation; specific tabulated coefficients were not extracted from the scanned tables and no numeric φ is asserted on this page. Free NIST reprint; retrieved 2026-07-29.
  6. [6]IUPAC Commission on Isotopic Abundances and Atomic Weights (CIAAW), Standard Atomic Weights 2021, abridged values — the source of every molar mass on this page: H 1.008, C 12.011, O 15.999, giving M(H2O) = 18.015, M(C2H6O2) = 62.068, M(C6H6) = 78.114 g/mol. Retrieved 2026-07-29.

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ΔTf = i · Kf · b Kf = R · Tf² · M_A ⁄ ΔfusH
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