Molar Mass from Colligative Properties: Worked Examples

A freezing point or osmotic pressure can tell you a solute's molar mass, but only after the units and particle-count assumptions are clear. These four original teaching examples work backwards from a property to a molar mass: a freezing-point depression, an osmotic pressure, an apparent-versus-true value, and an exact vapour-pressure lowering. The numbers are practice data, not measurements from a real experiment.

Examples 1-3 use the stated dilute ideal-property models. The van't Hoff factor i must be supplied or constrained independently: one measurement cannot determine both i and molar mass. Example 4 uses the exact Raoult law for an ideal binary solution with a nonvolatile solute that neither dissociates nor associates. This is a calculation guide for Class 12 Solutions, not a laboratory procedure.

The working forms

M

the solute's true molar mass in g/mol - the unknown in Examples 1, 2 and 4; in Example 3 the true value is known independently and the apparent molar mass (computed with i = 1) is what the measurement gives

i

the van't Hoff factor: 1 for a solute that neither dissociates nor associates, approaching the ion count for full dissociation, below 1 for association; it appears only inside the dilute property models, never multiplied into a concentrated mole fraction

Kf and Kb

the solvent's freezing-point and boiling-point constants in K kg/mol; for water Kf = 1.86 and Kb = 0.52

molality m

moles of solute per kilogram of SOLVENT, not of solution

pi, V, R, T

osmotic pressure, the final SOLUTION volume, the gas constant in matching units (0.0821 L atm K^-1 mol^-1 with atm and litres), and absolute temperature in kelvin

r

the relative vapour-pressure lowering (p0 - p)/p0, a plain ratio

Temperature-shift inversion: M = i x K x (solute mass in g) / (delta T x kg of solvent).

Osmotic inversion: M = i x (solute mass in g) x R x T / (pi x V).

Apparent versus true: compute the apparent molar mass with i = 1; then i = true M / apparent M when the true value is known independently.

Exact binary lowering: with r = (p0 - p)/p0, n_solute = r x n_solvent / (1 - r), then M = mass/n_solute. The dilute shortcut n_solute ≈ r x n_solvent drifts as r grows; use the exact form whenever r is not small.

Worked examples

Example 1: molar mass from a freezing-point depression

2.925 g of sodium chloride is dissolved in 100 g of water, and the solution freezes at -1.86 °C. Take pure water's freezing point as 0.00 °C at the same pressure. Within the ideal model, take i = 2 as the supplied value for complete dissociation of NaCl (a model assumption, not a value measured at this concentration) and Kf = 1.86 K kg/mol for water; find the molar mass.

1. Use the temperature-shift inversion: M = i x Kf x mass / (delta Tf x kg of solvent).
2. delta Tf = T_f(pure solvent) - T_f(solution) = 0.00 - (-1.86) = 1.86 K.
3. M = 2 x 1.86 x 2.925 / (1.86 x 0.100) = 58.5 g/mol.
4. Check: moles = 2.925/58.5 = 0.0500, so molality = 0.500, and the model gives delta Tf = 2 x 1.86 x 0.500 = 1.86 K, the measured depression.

The molar mass is 58.5 g/mol.

Example 2: molar mass from osmotic pressure

4.5 g of a non-electrolyte (i = 1) is dissolved to make 250 mL of solution, and the osmotic pressure at 300 K is 2.463 atm. Using R = 0.0821 L atm K^-1 mol^-1, find the molar mass.

1. Use the osmotic inversion: M = i x mass x R x T / (pi x V).
2. Convert the volume: 250 mL = 0.250 L of final solution.
3. M = 1 x 4.5 x 0.0821 x 300 / (2.463 x 0.250) = 180 g/mol.
4. Check: 4.5/180 = 0.025 mol in 0.250 L gives C = 0.100 mol/L, and the model gives pi = 1 x 0.100 x 0.0821 x 300 = 2.463 atm, the measured pressure.

The molar mass is 180 g/mol.

Example 3: apparent molar mass versus the true value

A salt's true molar mass is known independently to be 58.5 g/mol, and in this ideal model it is assumed to dissociate into two ions. A solution of 5.85 g of the salt in 100 g of water shows a freezing-point depression of 3.348 K (Kf = 1.86 K kg/mol). Find the apparent molar mass and the van't Hoff factor.

1. Compute the apparent molar mass with i = 1: M_app = 1 x Kf x mass / (delta Tf x kg of solvent).
2. M_app = 1.86 x 5.85 / (3.348 x 0.100) = 32.5 g/mol.
3. Split with the known true value: i = M_true / M_app = 58.5 / 32.5 = 1.8.
4. Interpret under the stated two-ion assumption: i between 1 and 2 corresponds to partial dissociation, with degree of dissociation (1.8 - 1)/(2 - 1) = 0.8 in this ideal model.
5. Check: the model gives delta Tf = 1.8 x 1.86 x (5.85/58.5) / 0.100 = 1.8 x 1.86 x 1.0 = 3.348 K, the measured depression.

The apparent molar mass is 32.5 g/mol and i = 1.8. Without the independently known true molar mass, the single measurement would have fixed only the apparent value - one measurement cannot give both M and i.

Example 4: molar mass from exact binary vapour-pressure lowering

180 g of a non-volatile solute that neither dissociates nor associates is dissolved in 432 g of water, and the vapour pressure falls by 4.0 percent at a fixed temperature. Using the exact binary relation in this ideal model, find the molar mass.

1. Write the relative lowering: r = 0.040, and moles of water = 432/18 = 24.
2. Exact binary inversion: n_solute = r x n_water / (1 - r) = 0.040 x 24 / 0.96 = 1.00 mol.
3. Finish with the mass: M = 180/1.00 = 180 g/mol.
4. Using the dilute approximation would give n_solute ≈ 0.040 x 24 = 0.96 mol and M ≈ 187.5 g/mol - visibly different here, which is why the exact form is used when r is not small.
5. Check: x_solute = 1.00 / (1.00 + 24) = 0.0400, exactly the measured relative lowering in this ideal binary model.

The molar mass is 180 g/mol.

Summary

From a temperature shift

M = i x K x mass / (delta T x kg of solvent) - molality counts solvent kilograms only

From osmotic pressure

M = i x mass x R x T / (pi x V) - V is the final solution volume, T in kelvin

Apparent versus true

apparent M assumes i = 1; with the true M known, i = true/apparent; one measurement alone cannot fix both

Exact lowering

n_solute = r x n_solvent / (1 - r), then M = mass/n_solute; the dilute ratio is only the small-r shortcut

Where i belongs

inside the dilute property models only - never multiplied into a concentrated mole fraction

Frequently asked questions

Can one measurement give both the molar mass and i?

No. One measurement fixes the effective particle count i x n. Separating M from i needs data that constrain one independently, such as a non-colligative molar mass; another colligative reading under the same model generally gives the same i/M ratio.

Why can dissociation make the apparent molar mass too small?

Dissociation raises the particle count, so the measured effect is larger than one particle per formula unit. Dividing it out with i = 1 gives too small a molar mass; i = true/apparent corrects it.

Do I divide by kg of solution or kg of solvent?

Solvent for the temperature shifts: delta Tb and delta Tf use molality, moles per kilogram of solvent. Osmotic pressure is the exception - it uses the final solution volume.

When is the exact mole-fraction relation needed?

Whenever the relative lowering r is not small. The exact ideal binary form is n_solute = r x n_solvent/(1 - r); the dilute shortcut n_solute ≈ r x n_solvent drifts further off as r grows.

What does this article add to the Solutions page?

The linked class 12 Solutions page drills the forward direction, predicting a property from a known solute. This article works the inverse direction only: molar mass from a temperature shift or osmotic pressure, the apparent/true split, and the exact binary lowering.

Sources and scope

• CBSE Chemistry (043) Classes XI-XII curriculum 2026-27 (retrieved 15 November 2026): Class XII Theory Unit 1 "Solutions", 7 of 70 marks, lists "Colligative Properties and Determination of Molar Mass, Abnormal Molecular Masses".
• CBSE Class XII Chemistry Sample Question Paper 2026-27 (retrieved 15 November 2026): Section C Question 28 includes a molar-mass calculation from vapour-pressure lowering, confirmed against the official marking scheme.
• Colligative properties (retrieved 15 November 2026; formula reference).
• Scope: only the inverse calculations above. Forward drills, Henry's law, azeotropes and the van't Hoff background stay with the linked class 12 Solutions practice page, whose FAQ already explains i for dissociation and association.