Dissolving a non-volatile solute lowers the vapour pressure of a solvent: some of the surface is occupied by solute particles, so fewer solvent molecules escape. The solution therefore has to be heated a little more before its vapour pressure reaches the outside pressure, and it boils at a higher temperature than the pure solvent. This rise is the boiling point elevation.
Like freezing point depression and osmotic pressure, it is a colligative property – it depends on how many solute particles are present, not on what they are. The amount is captured by the molality m (moles of solute per kilogram of solvent) and the van ‘t Hoff factor i, the number of particles each formula unit produces: 1 for sugar or urea, close to 2 for NaCl and close to 3 for CaCl₂. Each solvent has its own ebullioscopic constant Kb: 0.512 °C·kg/mol for water, 2.53 for benzene, 3.63 for chloroform.
The same equation, rearranged, is the classic ebullioscopic way to determine molar mass: dissolve a known mass of an unknown substance, measure the rise in boiling point and calculate how many moles must be present.
How to use the Boiling Point Elevation Calculator
- What do you know?
- Solvent
- Ebullioscopic constant Kb
- Boiling 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 boiling point elevation ΔTb
Results update as you type, every field has its own unit menu, and everything is calculated in your browser.
How it works
Choose the solvent (or enter your own Kb and boiling point), then give the molality directly or the masses of solute and solvent with the solute’s molar mass. The calculator multiplies i × Kb × m for ΔTb and adds it to the solvent’s boiling point. In molar-mass mode it divides the measured ΔTb by i·Kb to get the molality, then the solute mass by the moles.
Worked example
10 g of NaCl (M = 58.44 g/mol) is dissolved in 250 g of water. Moles = 10 / 58.44 = 0.1711 mol, so m = 0.1711 / 0.250 = 0.6845 mol/kg. With i = 2 and Kb = 0.512 °C·kg/mol, ΔTb = 2 × 0.512 × 0.6845 = 0.701 °C, and the solution boils at about 100.70 °C at 1 atm.
Ebullioscopic constants of common solvents
| Solvent | Boiling point (°C) | Kb (°C·kg/mol) |
|---|---|---|
| Water | 100 | 0.512 |
| Benzene | 80.1 | 2.53 |
| Chloroform | 61.26 | 3.63 |
| Acetic acid | 118.1 | 3.07 |
| Nitrobenzene | 210.9 | 5.24 |
| Ethanol | 78.4 | 1.22 |
| Diethyl ether | 34.6 | 2.02 |
| Cyclohexane | 80.7 | 2.79 |
Data: OpenStax Chemistry 2e, Table 11.2 (water, benzene, chloroform, acetic acid, nitrobenzene) and the CRC Handbook (ethanol, diethyl ether, cyclohexane).
How to interpret the result
For aqueous solutions ΔTb is small – usually well under 1 °C – which is why a sensitive thermometer is needed to use it for molar-mass work. The equation is exact only for dilute, ideal solutions (roughly below 1 mol/kg); concentrated or strongly ionic solutions deviate, and the van ‘t Hoff factor of real electrolytes is a little below the whole number.
Tips
- Use kelvin for every temperature in thermodynamic formulas – the unit menus convert for you.
- Watch the units: ΔH is usually in kJ/mol and ΔS in J/(mol·K).
- Tabulated values are for standard conditions (1 bar, usually 25 °C).
- Everything is calculated in your browser.
Common mistakes to avoid
- Mixing kJ and J between ΔH and ΔS.
- Assuming ΔH and ΔS do not change with temperature over large ranges.
- Using Celsius in ln(T₂/T₁).
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 boiling point change – find the solute’s molar mass |
| Solvent | Choose one: Water (bp 100 °C, K = 0.512), Benzene (bp 80.1 °C, K = 2.53), Chloroform (bp 61.26 °C, K = 3.63), Acetic acid (bp 118.1 °C, K = 3.07) … |
| Ebullioscopic constant Kb | In °C·kg/mol |
| Boiling 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 boiling point elevation ΔTb | 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 Kb = [K]
- Moles of solute = [Molality]×[Mass of solvent]
Formula
ΔTb = i · Kb · m · T(solution) = T(solvent) + ΔTb · m = n(solute) / m(solvent in kg) · M = m(solute) · i · Kb / (ΔTb · m(solvent in kg))Worked example
Inputs — What do you know?: Masses of solute and solvent, and the solute's molar mass; Solvent: Water (bp 100 °C, K = 0.512); Van 't Hoff factor i (1 for non-electrolytes, ≈2 for NaCl, ≈3 for CaCl₂): 2; Mass of solute: 10 g; Molar mass of the solute: 58.44 g/mol; Mass of solvent: 250 g.
Results — Boiling point elevation ΔTb = 0.7009 °C (= K); Boiling point of the solution = 100.701 °C; Molality of the solution (solute particles before dissociation) = 0.6845 mol/kg; Effective particle molality i·m = 1.3689 mol/kg; Constant used Kb = 0.512 °C·kg/mol; Moles of solute = 0.17112 mol.
Key terms
- Colligative property
- A solution property that depends on the number of dissolved particles, not their identity.
- Ebullioscopic constant (Kb)
- The boiling point rise produced by 1 mol of particles per kilogram of a given solvent.
- Molality (m)
- Moles of solute per kilogram of solvent.
- Van 't Hoff factor (i)
- The number of particles one formula unit of solute produces in solution.
Frequently asked questions
How much does salt raise the boiling point of water?
10 g of NaCl in 250 g of water (0.684 mol/kg, i ≈ 2) raises it by about 0.70 °C. A teaspoon of salt in a large pot of pasta water changes it by less than 0.1 °C.
What is the Kb of water?
0.512 °C·kg/mol (some tables give 0.513). A 1 mol/kg solution of a non-electrolyte boils at about 100.51 °C.
What is the Kb of benzene?
2.53 °C·kg/mol, about five times larger than water's, which makes benzene more sensitive for molar-mass measurements.
Why is the van 't Hoff factor of NaCl less than 2?
Oppositely charged ions attract each other and some move together as ion pairs, so the effective number of particles is lower – about 1.9 in dilute solutions.
Does boiling point elevation depend on the solute?
Only through the number of particles. 1 mol/kg of glucose and 1 mol/kg of urea give the same ΔTb in water.
Why use molality instead of molarity?
Molality is based on the mass of solvent, which does not change with temperature, whereas solution volume expands on heating.
Does this work for volatile solutes such as ethanol in water?
No – volatile solutes add their own vapour pressure. The equation applies to non-volatile solutes in dilute solutions.
How do I find molar mass from boiling point elevation?
Measure ΔTb, compute m = ΔTb/(i·Kb), multiply by the kilograms of solvent to get moles, and divide the solute mass by those moles.
What formula does the Boiling Point Elevation Calculator use?
ΔTb = i · Kb · m · T(solution) = T(solvent) + ΔTb · m = n(solute) / m(solvent in kg) · M = m(solute) · i · Kb / (ΔTb · m(solvent in kg)).