Step 1: Meaning in one line.
A colligative property counts particles, not identity. Whether the dissolved substance is glucose or urea, the same number of particles lowers the vapour pressure, raises the boiling point, lowers the freezing point, or creates the same osmotic pressure by the same amount.
Step 2: Start from the measured quantity.
The experimentally accessible colligative property here is the relative lowering of vapour pressure, \(\dfrac{p^{\circ}-p}{p^{\circ}}\). Raoult's law equates this to the mole fraction of the non-volatile solute \(x_2\).
Step 3: Write mole fraction in terms of masses.
\(x_2 = \dfrac{n_2}{n_1+n_2}\). In a dilute solution the solute moles are negligible in the denominator, so \(x_2 \approx \dfrac{n_2}{n_1}\). Substituting \(n_2 = \dfrac{w_2}{M_2}\) and \(n_1 = \dfrac{w_1}{M_1}\) gives \(\dfrac{p^{\circ}-p}{p^{\circ}} = \dfrac{w_2 M_1}{M_2 w_1}\).
Step 4: Solve for the unknown molar mass.
Isolating \(M_2\) gives the working formula used in the laboratory:
\[\boxed{ M_2 = \frac{w_2\, M_1\, p^{\circ}}{w_1\,(p^{\circ}-p)} }\]
Knowing the masses of solute and solvent, the molar mass of the solvent, and the two vapour pressures, the molar mass of the solute is obtained.