Boiling Point Elevation — Worked Problems and Formula
Dissolve anything non-volatile in a liquid and the liquid boils at a higher temperature than it did when pure. The rule is short, the arithmetic is short, and yet this is one of the easiest places in the Solutions chapter to drop marks — usually on units, occasionally on the van 't Hoff factor. This guide gives the formula, explains where the constant Kb actually comes from, and works three problems from start to finish.
Why a solute raises the boiling point
A liquid boils when its vapour pressure equals the pressure pushing down on it — 1 atm in an open vessel at sea level. By Raoult's law, a dissolved non-volatile solute lowers the vapour pressure of the solvent at every temperature. So the solution has not yet reached 1 atm of vapour pressure at the old boiling point, and you must heat it further. Boiling point elevation is simply vapour pressure lowering, viewed from the temperature axis instead of the pressure axis.
The formula
| Symbol | Meaning | Unit |
|---|---|---|
| ΔTb | elevation in boiling point (a positive number) | K or °C (same size of degree) |
| Kb | ebullioscopic constant — a property of the solvent only | K kg mol⁻¹ |
| m | molality = moles of solute ÷ kilograms of solvent | mol kg⁻¹ |
| i | van 't Hoff factor — particles per formula unit | no unit |
Rearranged for a molar mass:
with w₂ = mass of solute in grams and w₁ = mass of solvent in grams.
Ebullioscopic constants
| Solvent | Boiling point at 1 atm (°C) | Kb (K kg mol⁻¹) |
|---|---|---|
| Water | 100.0 | 0.52 |
| Ethanol | 78.4 | 1.20 |
| Benzene | 80.1 | 2.53 |
| Chloroform | 61.2 | 3.63 |
| Acetic acid | 118.1 | 3.07 |
Books quote water's Kb as 0.52 or 0.512 depending on rounding. Both are the same measurement; use whichever the question supplies.
Where Kb comes from
Kb is not an arbitrary tabulated number. It is fixed by two properties of the solvent — its boiling point and its enthalpy of vaporisation:
Check it for water. R = 8.314 J K⁻¹ mol⁻¹, Tb = 373.15 K, M₁ = 0.01802 kg/mol (the molar mass of the solvent, in kilograms), ΔvapH ≈ 40 650 J/mol.
Tb² = 373.15² = 139 241
139 241 × 8.314 = 1 157 649
1 157 649 × 0.01802 = 20 861
Kb = 20 861 ÷ 40 650 = 0.513 K kg mol⁻¹
Within rounding of the tabulated 0.52. This also explains the pattern in the table above: solvents that boil high and vaporise easily (chloroform, benzene) have large Kb values, which is why they are chosen for molar-mass work.
Worked problem 1 — glucose in water
Problem: 18.0 g of glucose is dissolved in 500 g of water. Find the boiling point of the solution at 1 atm. Kb(water) = 0.52 K kg mol⁻¹.
Step 1 — molar mass. C₆H₁₂O₆ = 6 × 12.011 + 12 × 1.008 + 6 × 15.999 = 72.066 + 12.096 + 95.994 = 180.16 g/mol
Step 2 — moles. n = 18.0 ÷ 180.16 = 0.09991 mol
Step 3 — molality. Solvent = 500 g = 0.500 kg
m = 0.09991 ÷ 0.500 = 0.19982 mol/kg
Step 4 — elevation. Glucose does not dissociate, so i = 1.
ΔTb = 0.52 × 0.19982 = 0.104 K
Step 5 — boiling point. Tb = 100.0 + 0.104 = 100.104 °C
A tenth of a degree from 18 grams of sugar. Boiling point elevation gives a much smaller signal than freezing point depression for water, because Kb (0.52) is less than a third of Kf (1.86). That is the practical reason laboratories prefer the freezing point method for aqueous work.
Worked problem 2 — molar mass of an unknown
Problem: 2.50 g of a non-volatile, non-electrolyte compound is dissolved in 100 g of chloroform. The boiling point rises by 0.907 K. Kb(chloroform) = 3.63 K kg mol⁻¹. Find the molar mass.
Numerator: 1000 × 3.63 × 2.50 = 3630 × 2.50 = 9075
Denominator: 0.907 × 100 = 90.7
M₂ = 9075 ÷ 90.7 = 100.1 g/mol
Reverse check. n = 2.50 ÷ 100.1 = 0.02498 mol in 0.100 kg, so m = 0.2498 mol/kg and ΔTb = 3.63 × 0.2498 = 0.907 K. ✓ The data are reproduced.
Note how much larger the elevation is here than in problem 1, from a smaller mass of solute. Chloroform's Kb is seven times water's, so the same experiment gives a reading you can trust.
Worked problem 3 — an ionic solute
Problem: 5.85 g of NaCl is dissolved in 250 g of water. Find the boiling point, assuming complete dissociation.
Step 1 — moles. M(NaCl) = 22.990 + 35.45 = 58.44 g/mol
n = 5.85 ÷ 58.44 = 0.10010 mol
Step 2 — molality. m = 0.10010 ÷ 0.250 = 0.40041 mol/kg
Step 3 — with i = 2 (NaCl → Na⁺ + Cl⁻):
ΔTb = 2 × 0.52 × 0.40041 = 2 × 0.20821 = 0.416 K
Step 4 — boiling point. 100.0 + 0.416 = 100.416 °C
If you had forgotten the factor of 2, you would have reported 100.208 °C — exactly half the elevation, and a full mark gone.
Does salting the pasta water really matter?
Worth doing as a quick reality check, because students often hear that salt makes water boil hotter. Suppose you add a generous 10 g of salt to 2.0 litres of water (2.0 kg):
n(NaCl) = 10 ÷ 58.44 = 0.1711 mol
particles = 2 × 0.1711 = 0.3422 mol
molality of particles = 0.3422 ÷ 2.0 = 0.1711 mol/kg
ΔTb = 0.52 × 0.1711 = 0.089 K
Under a tenth of a degree. Salt is added to pasta water for flavour, not for temperature. The chemistry is real but the size of the effect is not — and being honest about magnitudes is a habit that pays off in physical chemistry generally.
Common mistakes that cost marks
- Using molarity instead of molality. The formula is defined per kilogram of solvent, and unlike molarity it does not change when the solution is heated.
- Dividing by grams of solvent rather than kilograms. An answer 1000 times too small.
- Swapping Kb and Kf. For water they are 0.52 and 1.86. Using the wrong one is an instant loss of the whole numerical.
- Forgetting i. Ionic solutes give more than one particle; molecular solutes give one.
- Reporting ΔTb as the boiling point. Add it to the solvent's boiling point — and remember that number is not 100 °C for anything except water.
- Assuming the boiling point of water is always 100 °C. It is 100 °C at 1 atm only. At a hill station where the pressure is lower, water boils below 100 °C, and the elevation is measured from that lower value.
- Applying the formula to a volatile solute. If the solute evaporates it raises the total vapour pressure, and the boiling point can fall instead.
Where this appears in exams
| Exam | Typical question |
|---|---|
| CBSE/ICSE Class 12 | Boiling point of a given solution; molar mass by the ebullioscopic method |
| JEE/NEET | Ranking solutions by boiling point; combined ΔTb and ΔTf problems |
| IIT-JAM / CUET-PG | Deriving Kb from ΔvapH; abnormal molar masses |
| GATE / CSIR-NET | Solvent activity and non-ideal corrections to colligative formulas |
Every one of these problems begins or ends with a molar mass. Problem 1 needed M(C₆H₁₂O₆) before anything else could happen; problem 2 produced a molar mass as the answer and had to be checked against a candidate formula. The Molar Mass & Composition tool gives you that number for any formula, with the element-wise breakdown shown.
Open the Molar Mass & Composition Calculator →ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre and online classes across India, with the Solutions chapter taught through worked numericals rather than definitions alone: abcchemistry.in.