ΔTf = Kf × m. Colligative property of adding solute.

Freezing Point Depression — Why Salt Melts Ice

Question

A solution is prepared by dissolving 18 g of glucose (molecular weight = 180 g/mol) in 1 kg of water. Calculate the depression in freezing point. (Kf for water = 1.86 K kg/mol)

Also explain: why does sprinkling salt on icy roads melt the ice?

Solution — Step by Step

Moles of glucose = mass ÷ molar mass = 18 ÷ 180 = 0.1 mol.

Always start here — the formula needs molality, not grams.

Molality (m) = moles of solute ÷ mass of solvent in kg = 0.1 ÷ 1 = 0.1 mol/kg.

Molality uses kg of solvent, not solution. Don’t mix these up — it costs marks every year.

\Delta T_f = K_f \times m

\Delta T_f = 1.86 \times 0.1 = \textbf{0.186 K}

The freezing point drops by 0.186°C, so the solution now freezes at −0.186°C instead of 0°C.

Salt (NaCl) dissolves in the thin water film on ice and ionizes into Na⁺ and Cl⁻ — that’s two particles from one formula unit. More particles in solution = greater depression. The freezing point falls below 0°C, so the ice melts even in winter. Road salt works the same way, which is why it’s effective until about −9°C (beyond that, you need CaCl₂, which gives three ions).

Why This Works

Freezing point depression is a colligative property — it depends only on the number of solute particles, not what those particles are. The solute molecules get in the way of water molecules trying to arrange themselves into the ordered crystal structure of ice. More particles = more interference = lower temperature needed to freeze.

The formula $\Delta T_f = K_f \times m$ captures this perfectly. $K_f$ is a constant specific to the solvent (1.86 for water), and molality tells us how many moles of particles are dissolved per kg of solvent.

For electrolytes, we multiply by the van’t Hoff factor $i$ (number of ions per formula unit): $\Delta T_f = i \times K_f \times m$ . NaCl gives $i = 2$ , so it depresses the freezing point twice as much as glucose at the same molality.

Alternative Method — Using van’t Hoff Factor Directly

For the salt example, let’s say we dissolve 5.85 g of NaCl (M = 58.5 g/mol) in 1 kg of water.

Moles of NaCl = 5.85 ÷ 58.5 = 0.1 mol
Molality = 0.1 mol/kg
$i$ for NaCl = 2 (complete dissociation assumed)

\Delta T_f = i \times K_f \times m = 2 \times 1.86 \times 0.1 = 0.372 \text{ K}

Notice this is exactly double the glucose answer at the same molality — which confirms that the number of particles is what matters, not the identity.

NEET and JEE Main frequently give you the mass of solute and ask for ΔTf. The most reliable approach: moles → molality → plug into formula. Don’t try shortcuts until this sequence is automatic.

Common Mistake

Using mass of solution instead of mass of solvent for molality.

Molality = moles ÷ kg of solvent (the liquid you’re dissolving into). If a problem says “18 g glucose dissolved in 1 kg water”, the solvent mass is 1 kg. If it says “18 g glucose in 1 kg solution”, you need to subtract the solute mass first: solvent = 1000 − 18 = 982 g = 0.982 kg. This distinction appears regularly in CBSE board papers and costs students easy marks.

Quick Reference

\Delta T_f = K_f \times m \quad \text{(non-electrolytes)}

\Delta T_f = i \times K_f \times m \quad \text{(electrolytes)}

$K_f$ (water) = 1.86 K kg/mol
$K_f$ (benzene) = 5.12 K kg/mol
New freezing point = $T_f^{\text{pure solvent}} - \Delta T_f$