Solutions: Speed-Solving Techniques (2)

Question

A solution of glucose ($C_6H_{12}O_6$, molar mass $180$ g/mol) in water has a freezing point depression of $1.86°C$. Calculate the molality of the solution and the mass of glucose dissolved per kg of water. ($K_f$ for water $= 1.86$ K·kg/mol.) JEE Main 2023 pattern.

Solution — Step by Step

Why This Works

Freezing point depression is a colligative property — it depends only on the number of solute particles, not their identity. Glucose dissociates into zero ions in water (molecular solute), so the van ‘t Hoff factor $i = 1$. For NaCl ($i \approx 2$) or CaCl$_2$ ($i \approx 3$), we’d multiply by $i$.

The formula $\Delta T_f = K_f \cdot m$ comes from Raoult’s law applied to the freezing equilibrium. The cryoscopic constant $K_f$ is a property of the solvent only — water has $K_f = 1.86$ K·kg/mol.

Alternative Method

For a non-volatile solute, you can also use boiling point elevation: $\Delta T_b = K_b \cdot m$ with $K_b = 0.52$ K·kg/mol for water. Both give the same molality and mass — useful for cross-checking.