Freezing Point Depression & Boiling Point Elevation
Freezing and boiling points of solutions, osmotic pressure, Raoult’s law and molar mass.
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The same solute in other solvents
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About the Freezing Point Depression & Boiling Point Elevation
Dissolve something in a liquid and four properties change by an amount that depends only on how many particles are dissolved, not on what they are: the solution freezes lower, boils higher, has an osmotic pressure, and has a lower vapour pressure. This calculator works out each of them:
- Freezing point: ΔTf = i·Kf·m, the freezing-point depression, and the solution’s freezing point;
- Boiling point: ΔTb = i·Kb·m, the boiling-point elevation;
- Osmotic pressure: π = iMRT, in atm, kPa, bar or mmHg, at any temperature;
- Vapour pressure: Raoult’s law, P = χ_solvent·P°, with water’s vapour pressure worked out from the temperature (IAPWS-IF97) or another solvent’s P° you enter;
- Molar mass: of an unknown solute, from a measured freezing point, boiling point or osmotic pressure.
The solute can be a molality, an amount in moles, or a mass with its molar mass (typed, or worked out from a formula or a name such as NaCl or glucose). The solvent constants are those of OpenStax Chemistry 2e for water, acetic acid, benzene, chloroform and nitrobenzene, or your own; the van ’t Hoff factor is the ideal number of ions or a measured value. With a Pro pass, copy the result or download it as a CSV, or unlock this one result. Without a pass the free preview shows the freezing point, boiling point, pressure or molar mass in full, with the first rows of the table and the first step of the working; every other figure hidden.
How to use it
- Choose the property: freezing point, boiling point, osmotic pressure, vapour pressure or molar mass.
- Pick the solvent; its constants fill in, and you can edit them to match your textbook.
- Give the solute as a molality, as moles and the mass of solvent, or as its mass and molar mass (or formula).
- Choose the van ’t Hoff factor: 1 for sugar or urea, 2 for NaCl, 3 for CaCl₂, a measured value, or one below 1 for a solute that associates.
- Read the result and the working. Copy it or download the CSV with a Pro pass, or after unlocking this result.
Examples
4.2 g of NaCl (58.44 g/mol) in 125 g of water, i = 2
m = 0.5749 mol/kg, ΔTf = 2.139 °C: the solution freezes at −2.14 °C
92.1 g of I₂ in 800.0 g of chloroform, i = 1
m = 0.4536 mol/kg, ΔTb = 1.647 °C: it boils at 62.91 °C
0.30 M glucose at 37 °C
π = 7.635 atm (773.6 kPa)
92.1 g of glycerin in 184.4 g of ethanol at 40 °C, P° = 0.178 atm
χ(ethanol) = 0.8001: P = 0.1424 atm
4.00 g in 55.0 g of benzene, freezing at 2.32 °C
ΔTf = 3.18 °C, 0.03416 mol: M = 117.1 g/mol
10.0 g in 0.500 L, π = 5.9 torr at 22 °C
1.603 × 10⁻⁴ mol: M = 6.24 × 10⁴ g/mol
Common uses
- Chemistry homework and exam practice on colligative properties
- Finding the molar mass of an unknown from a freezing-point experiment
- How much salt or antifreeze lowers the freezing point of water, as an estimate
- Osmotic pressure of saline, glucose or protein solutions
- Checking a lab result against the ideal value
The formulas
- Molality: m = moles of solute ÷ kilograms of solvent.
- Freezing-point depression: ΔTf = i × Kf × m, and the solution freezes at Tf° − ΔTf.
- Boiling-point elevation: ΔTb = i × Kb × m, and the solution boils at Tb° + ΔTb.
- Osmotic pressure: π = i × c × R × T, with c in mol/L, T in kelvin and R = 0.082 057 L·atm/(mol·K).
- Raoult’s law: P = χ_solvent × P°, with χ_solvent = n_solvent ÷ (n_solvent + i × n_solute).
- Molar mass: m = ΔT ÷ (i × K), then n = m × kg of solvent and M = grams ÷ n; from π, c = π ÷ (iRT), n = c × V and M = grams ÷ n.
The van ’t Hoff factor
The factor i is the number of particles each formula unit gives in solution: 1 for a nonelectrolyte such as sucrose or urea, 2 for NaCl (Na⁺ and Cl⁻), 3 for CaCl₂ or MgCl₂, 4 for FeCl₃. Real solutions fall short of these ideal numbers because ions of opposite charge pair up. OpenStax Table 11.3 gives the measured values of 0.050 m aqueous solutions: sucrose 1.0, NaCl 1.9, HCl 1.9, MgSO₄ 1.3, MgCl₂ 2.7 and FeCl₃ 3.4. A 1.0 m NaCl solution would freeze 3.7 °C below pure water with i = 2, but freezes 3.4 °C lower.
A solute that associates has fewer particles than formula units, so i is below 1: carboxylic acids such as acetic or benzoic acid form hydrogen-bonded pairs (dimers) in benzene, and if every molecule pairs up, i = 0.5. Choose “Another value” and type i from 0.1 to 10.
Solvent constants at 1 atm
OpenStax Chemistry 2e, Table 11.2:
- Water: boils at 100.0 °C, Kb 0.512 °C·kg/mol; freezes at 0.0 °C, Kf 1.86 °C·kg/mol
- Acetic acid: 118.1 °C, Kb 3.07; 16.6 °C, Kf 3.9
- Benzene: 80.1 °C, Kb 2.53; 5.5 °C, Kf 5.12
- Chloroform: 61.26 °C, Kb 3.63; −63.5 °C, Kf 4.68
- Nitrobenzene: 210.9 °C, Kb 5.24; 5.67 °C, Kf 8.1
For another solvent, choose “Another solvent” and type its constants from your textbook or data table.
Sources
- OpenStax, Chemistry 2e, section 11.4: Colligative properties (CC BY 4.0): the equations, Tables 11.2 and 11.3, and the worked examples above, which are among this calculator’s tests.
- NIST CODATA: molar gas constant, 8.314 462 618 J/(mol·K), exact.
- IAPWS, IAPWS-IF97, the industrial formulation for water and steam: the saturation-pressure equation of water, whose check values this calculator reproduces.
Limitations
- The formulas describe dilute, ideal solutions: above about 1 mol/kg of particles they are only estimates, and the calculator warns.
- The solute is taken as nonvolatile; a volatile solute such as ethanol changes the boiling point and the vapour pressure differently.
- The van ’t Hoff factor of an electrolyte depends on its concentration: the measured values of the table are for 0.050 m in water.
- Solvent constants are for the normal freezing and boiling points (1 atm).
- Water’s vapour pressure is valid from 0 °C to its critical point (373.946 °C).
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Frequently asked questions
What do I get without a pass?
Without a pass, Freezing Point Depression & Boiling Point Elevation shows the freezing point, boiling point, pressure or molar mass in full, with the first rows of the table and the first step of the working; every other figure hidden. Until you unlock it, the result can’t be downloaded or copied. A Pro, Premium or Ultimate pass, a one-time payment that never renews, unlocks the full result. The pricing page lists the passes and their prices.
How do I calculate freezing-point depression?
Multiply the van ’t Hoff factor, the solvent’s cryoscopic constant and the molality: ΔTf = i × Kf × m. For 4.2 g of NaCl in 125 g of water, m = (4.2 ÷ 58.44) ÷ 0.125 = 0.575 mol/kg and ΔTf = 2 × 1.86 × 0.575 = 2.14 °C, so the solution freezes at −2.14 °C.
What are Kf and Kb of water?
Kf = 1.86 °C·kg/mol and Kb = 0.512 °C·kg/mol (OpenStax Table 11.2): each mole of particles per kilogram of water lowers the freezing point by 1.86 °C and raises the boiling point by 0.512 °C.
What is the van ’t Hoff factor of NaCl?
Ideally 2, because NaCl gives Na⁺ and Cl⁻. Measured, it is about 1.9 at 0.050 m in water, and lower in stronger solutions, because some ions stay paired.
Can the van ’t Hoff factor be less than 1?
Yes, when the solute associates: carboxylic acids such as acetic acid form pairs (dimers) in benzene, so there are fewer particles than molecules and i is below 1, down to 0.5 when every molecule pairs up. Entered as i = 1 in the molar-mass mode, a fully paired solute seems twice as heavy: about 120 g/mol for acetic acid instead of 60.
How do I find a molar mass from a freezing point?
Work out ΔTf from the pure solvent’s freezing point and the solution’s, then m = ΔTf ÷ (i × Kf), the moles n = m × kg of solvent, and M = grams ÷ n. For 4.00 g in 55.0 g of benzene freezing at 2.32 °C: ΔTf = 3.18 °C, m = 0.621 mol/kg, n = 0.0342 mol and M = 117 g/mol.
What does Raoult’s law say?
The vapour pressure of the solvent above a solution is its mole fraction times the vapour pressure of the pure solvent: P = χ_solvent × P°. A nonvolatile solute lowers the vapour pressure in proportion to its share of the particles.
How is osmotic pressure calculated?
π = i × c × R × T, with c in mol/L, R = 0.082 057 L·atm/(mol·K) and T in kelvin. A 0.30 M glucose solution at 37 °C (310.15 K) has π = 0.30 × 0.082 057 × 310.15 = 7.64 atm.