A solution freezes at a lower temperature than the pure solvent. Dissolved particles get in the way of solvent molecules locking into a crystal and they lower the solvent’s vapour pressure, so the liquid stays stable below the usual freezing point. The size of the effect depends only on the concentration of dissolved particles – it is a colligative property, like boiling point elevation and osmotic pressure.
Everyday examples are everywhere: salt spread on icy roads, ethylene glycol antifreeze in car radiators, sugar keeping ice cream scoopable, and fish and insects that survive sub-zero temperatures thanks to dissolved sugars and proteins. In the laboratory the same effect – cryoscopy – is a simple way to find the molar mass of a new compound, especially with solvents that have a large cryoscopic constant such as cyclohexane (Kf = 20.0) or camphor.
The equation needs three numbers: the solvent’s cryoscopic constant Kf (1.86 °C·kg/mol for water), the molality m of the solution and the van ‘t Hoff factor i, which counts the particles each formula unit releases (1 for glucose, about 2 for NaCl, about 3 for CaCl₂).
How to use the Freezing Point Depression Calculator
- What do you know?
- Solvent
- Cryoscopic constant Kf
- Freezing point of the pure solvent
- Van ‘t Hoff factor i (1 for non-electrolytes, ≈2 for NaCl, ≈3 for CaCl₂)
- Molality of the solution
- Mass of solute
- Molar mass of the solute
- Mass of solvent
- Measured freezing point depression ΔTf
Results update as you type, every field has its own unit menu, and everything is calculated in your browser.
How it works
Select the solvent or enter your own Kf and freezing point. Give the molality, or the masses of solute and solvent with the molar mass, and the calculator multiplies i × Kf × m and subtracts the result from the pure solvent’s freezing point. In molar-mass mode the measured ΔTf gives the molality, the molality gives the moles of solute, and the solute mass divided by the moles is the molar mass.
Worked example
A 0.5 mol/kg glucose solution in water (i = 1, Kf = 1.86 °C·kg/mol): ΔTf = 1 × 1.86 × 0.5 = 0.93 °C, so it freezes at −0.93 °C.
Molar mass mode: 5.00 g of an unknown non-electrolyte in 100 g of benzene (Kf = 5.12) lowers the freezing point by 1.28 °C. m = 1.28 / 5.12 = 0.250 mol/kg; moles = 0.250 × 0.100 = 0.0250 mol; M = 5.00 / 0.0250 = 200 g/mol.
Cryoscopic constants of common solvents
| Solvent | Freezing point (°C) | Kf (°C·kg/mol) |
|---|---|---|
| Water | 0 | 1.86 |
| Benzene | 5.5 | 5.12 |
| Chloroform | −63.5 | 4.68 |
| Acetic acid | 16.6 | 3.9 |
| Nitrobenzene | 5.67 | 8.1 |
| Ethanol | −114.1 | 1.99 |
| Diethyl ether | −116.2 | 1.79 |
| Cyclohexane | 6.6 | 20 |
| Naphthalene | 80.2 | 6.94 |
Data: OpenStax Chemistry 2e, Table 11.2 (water, benzene, chloroform, acetic acid, nitrobenzene) and the CRC Handbook (ethanol, diethyl ether, cyclohexane, naphthalene).
How to interpret the result
The linear equation is reliable for dilute solutions (below about 1 mol/kg). At higher concentrations, particularly for salts and antifreeze mixtures, real freezing points fall further than predicted, and the van ‘t Hoff factor of electrolytes is slightly below the ideal whole number because of ion pairing.
Inputs explained
| Field | What to enter |
|---|---|
| What do you know? | Choose one: The molality of the solution, Masses of solute and solvent, and the solute’s molar mass, The measured freezing point change – find the solute’s molar mass |
| Solvent | Choose one: Water (fp 0 °C, K = 1.86), Benzene (fp 5.5 °C, K = 5.12), Chloroform (fp −63.5 °C, K = 4.68), Acetic acid (fp 16.6 °C, K = 3.9) … |
| Cryoscopic constant Kf | In °C·kg/mol |
| Freezing point of the pure solvent | Units: °C, K, °F |
| Van ‘t Hoff factor i (1 for non-electrolytes, ≈2 for NaCl, ≈3 for CaCl₂) | A number |
| Molality of the solution | Units: mol/kg, mmol/kg |
| Mass of solute | Units: g, mg, µg, kg, oz, lb |
| Molar mass of the solute | Units: g/mol, kg/mol, Da (u), kDa |
| Mass of solvent | Units: g, kg, mg, lb |
| Measured freezing point depression ΔTf | Units: °C (= K), °F |
What each result means
- Molar mass of the solute = [Mass of solute]÷([Molality]×[Mass of solvent])
- Molality of the solution (solute particles before dissociation) = [Molality]
- Effective particle molality i·m = [Van ‘t Hoff factor i (1 for non-electrolytes, ≈2 for NaCl, ≈3 for CaCl₂)]×[Molality]
- Constant used Kf = [K]
- Moles of solute = [Molality]×[Mass of solvent]
Formula
ΔTf = i · Kf · m · T(solution) = T(solvent) − ΔTf · m = n(solute) / m(solvent in kg) · M = m(solute) · i · Kf / (ΔTf · m(solvent in kg))
Worked example
Inputs — What do you know?: The molality of the solution; Solvent: Water (fp 0 °C, K = 1.86); Van 't Hoff factor i (1 for non-electrolytes, ≈2 for NaCl, ≈3 for CaCl₂): 1; Molality of the solution: 0.5 mol/kg.
Results — Freezing point depression ΔTf = 0.93 °C (= K); Freezing point of the solution = -0.93 °C; Molality of the solution (solute particles before dissociation) = 0.5 mol/kg; Effective particle molality i·m = 0.5 mol/kg; Constant used Kf = 1.86 °C·kg/mol.
Key terms
- Freezing point depression
- The lowering of a solvent's freezing point caused by dissolved solute particles.
- Cryoscopic constant (Kf)
- The freezing point drop produced by 1 mol of particles per kilogram of solvent.
- Cryoscopy
- Determining molar mass by measuring freezing point depression.
- Eutectic point
- The lowest temperature at which a mixture of fixed composition can stay liquid.
Frequently asked questions
What is the Kf of water?
1.86 °C·kg/mol (1.853 in precise tables). A 1 mol/kg sugar solution freezes at about −1.86 °C.
How much does salt lower the freezing point of water?
A 1 mol/kg NaCl solution (58.4 g per kg of water, i ≈ 2) freezes near −3.7 °C. Saturated brine can stay liquid down to about −21 °C, the NaCl–water eutectic.
Why is CaCl₂ better than NaCl for de-icing?
Each formula unit releases three ions instead of two, and calcium chloride stays effective down to about −30 °C.
How does antifreeze work?
Ethylene glycol dissolved in the coolant lowers its freezing point; a 50:50 mixture by volume protects to about −37 °C, though at that concentration the simple equation underestimates the effect.
How do I find molar mass from freezing point depression?
Divide the measured ΔTf by i·Kf to get the molality, multiply by the kilograms of solvent to get moles, then divide the grams of solute by those moles.
Why is cyclohexane or camphor used for cryoscopy?
Their Kf values (20.0 and about 40 °C·kg/mol) are 10–20 times larger than water's, so the temperature change is easy to measure.
Is the freezing point depression positive or negative?
ΔTf is usually reported as a positive number; the freezing point of the solution is the solvent's freezing point minus ΔTf.