Chemistry · Solutions · NEET
The formula is ΔTf = Kf × m. Here ΔTf is the drop in freezing point (in kelvin), Kf is the molal depression constant of the solvent, and m is the molality of the solution. This is true only for dilute solutions of a non-volatile, non-electrolyte solute. For NEET, always start every freezing-point numerical from this one equation.
No. Kf depends ONLY on the solvent, never on the solute or its amount. Kf is a fixed property of the solvent (like water = 1.86, benzene = 5.12). If you double the molality, ΔTf doubles, but Kf stays exactly the same. This exact point was asked in NEET 2017, so remember it.
Adding a non-volatile solute lowers the vapour pressure of the solution. A liquid freezes when its vapour pressure equals the vapour pressure of the solid solvent. Because the solution now has lower vapour pressure, it must be cooled to a lower temperature to reach that point. So the solution freezes at a lower temperature than the pure solvent.
Kf has units of K kg mol⁻¹ (kelvin kilogram per mole). This matches the formula ΔTf = Kf × m, because molality m has units of mol kg⁻¹, and multiplying gives kelvin. Kf is also called the cryoscopic constant. NEET options sometimes hide the wrong units to trap you, so check them.
Kf is the molal depression constant used for freezing point (ΔTf = Kf·m), and Kb is the molal elevation constant used for boiling point (ΔTb = Kb·m). Both depend only on the solvent. For water, Kf = 1.86 and Kb = 0.52, so freezing point is depressed MORE than boiling point is elevated for the same molality.
Yes. It is a colligative property, so it depends on the NUMBER of solute particles, not their type. For electrolytes like NaCl that split into ions, you must multiply by the van't Hoff factor i: ΔTf = i × Kf × m. For a non-electrolyte (like glucose), i = 1, so you use the plain formula.
If molality of the dilute solution is doubled, the value of molal depression constant (Kf) will be
The freezing point depression constant (Kf) of benzene is 5.12 K kg mol⁻¹. The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is (rounded off upto two decimal places):
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes. It depends only on the number of solute particles in the solution, not on what the solute is. That is the definition of a colligative property.
For water, Kf = 1.86 K kg mol⁻¹. This means a 1 molal solution of a non-electrolyte lowers water's freezing point by 1.86 K (from 0 °C to about −1.86 °C).
These are electrolytes that break into ions, so use ΔTf = i × Kf × m. For NaCl, i = 2 (Na⁺ + Cl⁻); for CaCl2, i = 3 (Ca²⁺ + 2Cl⁻). More particles means a bigger freezing point drop.
Salt dissolves in the thin water layer and lowers its freezing point. So the water needs a much colder temperature to freeze, and the roads stay ice-free. This is depression of freezing point in real life.
No. The formula uses molality (mol per kg of solvent), because molality does not change with temperature. Using molarity gives a wrong answer in NEET numericals.
