Molar Mass of an Unknown Solute from Colligative Properties
A common numerical in Class 12 solutions is: “a solute of unknown molar mass is dissolved in a solvent, and a property of the solution is measured. Find the molar mass.” The idea is simple. Colligative properties depend on the number of solute particles, so if you know the mass dissolved and the size of the effect, you can count the moles and then find the molar mass.
The formulas
Boiling point elevation: M2 = Kb × w2 × 1000 / (ΔTb × w1)
Osmotic pressure: M2 = w2 R T / (π V)
Relative lowering of vapour pressure: (p° − p) / p° = w2 M1 / (M2 w1)
M2 is the molar mass of the solute (g/mol) and M1 that of the solvent. w2 is the mass of solute (g) and w1 the mass of solvent (g). Kf and Kb are constants of the solvent in K kg/mol; the problem will give them. π is the osmotic pressure, V the solution volume in litres, T in kelvin, and R = 0.0821 L atm K−1 mol−1 when π is in atm. The 1000 in the first two formulas converts grams of solvent to kilograms.
These formulas are all the same idea written in different ways: find moles of solute, then M2 = w2 / moles.
Which property to use
- Freezing and boiling point: easy to measure, but the changes are small for dilute solutions.
- Osmotic pressure: gives a large, easy-to-read value even for very dilute solutions at room temperature, which is why it suits large and delicate molecules such as proteins. It needs a semipermeable membrane.
- Vapour pressure lowering: works only if the solute is non-volatile.
Worked example 1: freezing point depression
2.50 g of a non-volatile, non-electrolyte solute is dissolved in 100 g of benzene. The freezing point falls by 0.71 K. Kf for benzene = 5.12 K kg/mol. Find the molar mass.
M2 = (5.12 × 2.50 × 1000) / (0.71 × 100) = 12 800 / 71 = 180 g/mol (180.3 before rounding).
Check by moles: molality = 0.71 / 5.12 = 0.1387 mol/kg. Moles in 0.100 kg = 0.01387 mol. M2 = 2.50 / 0.01387 = 180 g/mol. Both routes agree.
Worked example 2: boiling point elevation
3.00 g of a solute dissolved in 100 g of water raises the boiling point by 0.26 K. Kb of water = 0.52 K kg/mol.
M2 = (0.52 × 3.00 × 1000) / (0.26 × 100) = 1560 / 26 = 60 g/mol.
Check: molality = 0.26 / 0.52 = 0.500 mol/kg, so moles in 0.100 kg = 0.0500 mol and M2 = 3.00 / 0.0500 = 60 g/mol.
Worked example 3: osmotic pressure
0.800 g of a solute is dissolved in water to make 200 mL of solution. At 300 K the osmotic pressure is 0.55 atm. Find the molar mass.
V = 0.200 L. M2 = (0.800 × 0.0821 × 300) / (0.55 × 0.200) = 19.70 / 0.110 = 179 g/mol.
Check by moles: n = πV / RT = (0.55 × 0.200) / (0.0821 × 300) = 0.110 / 24.63 = 4.466 × 10−3 mol. M2 = 0.800 / 4.466 × 10−3 = 179 g/mol.
Worked example 4: lowering of vapour pressure
6.0 g of a non-volatile solute is dissolved in 90 g of water. The vapour pressure of pure water is 23.76 mmHg and that of the solution is 23.54 mmHg at the same temperature. Take M1 = 18 g/mol.
p° − p = 23.76 − 23.54 = 0.22 mmHg. M2 = w2 M1 p° / (w1 (p° − p)) = (6.0 × 18 × 23.76) / (90 × 0.22) = 2566 / 19.8 = 129.6 g/mol, about 130 g/mol.
This short formula is valid for dilute solutions. The small pressure difference is measured to two digits, so only two significant figures can be trusted in the answer.
Worked example 5: when the molar mass comes out half
Benzoic acid (true molar mass 122 g/mol) in benzene gives an experimental molar mass of about 240 g/mol from freezing point depression.
The calculated value is about twice the true one because two molecules join through hydrogen bonds to form a dimer, so the number of particles is nearly halved. The van’t Hoff factor is i = (calculated molar mass) / (observed molar mass) = 122 / 240 = 0.51.
For association into dimers, i = 1 − α/2, so α = 2(1 − 0.51) = 0.98. About 98% of the acid molecules are associated. In water, an electrolyte like NaCl does the opposite: it dissociates, i is above 1, and the calculated molar mass is lower than the formula mass.
Common mistakes
- Using the mass of solution instead of the mass of solvent for w1.
- Forgetting the factor 1000 (grams to kilograms) in the Kf and Kb formulas.
- Using °C and K inconsistently. A change of 0.71 °C is the same as 0.71 K, but T in the osmotic pressure formula must be in kelvin.
- Using mL for V in the osmotic pressure formula. Convert to litres.
- Ignoring i for electrolytes and for associating solutes. The formulas above assume a non-electrolyte that neither dissociates nor associates.
- Using the formulas for a volatile solute in the vapour pressure method.
Exam relevance
| Given | Choose |
|---|---|
| ΔTf and Kf | M2 = Kf w2 1000 / (ΔTf w1) |
| ΔTb and Kb | M2 = Kb w2 1000 / (ΔTb w1) |
| π, V, T | M2 = w2 R T / (π V) |
| Vapour pressures | Relative lowering formula (dilute solutions) |
| Molar mass about double the expected value | Association, i about 0.5 |
| Molar mass lower than the formula mass | Dissociation, i above 1 |
The Kf and Kb values used above are the ones given in the questions. Constants in textbooks differ slightly in the last digit, so always use the value your question or book gives.
There is no separate colligative tool, but you can use the Scientific Calculator on the home page to check each multiplication and division above.
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