Q.Equimolal aqueous solutions of NaCl and KCl are prepared. If the freezing point of NaCl is -2 degree C, the freezing point of KCl solution is expected to be:

Depression of Freezing Point

Imagine a cold winter morning. You see water on the road turning to ice at 0°C. But if you sprinkle salt on that ice, it melts — even though the temperature is still below zero. That’s the same phenomenon that keeps roads safe in snowy countries and makes ice cream freeze in a churn. The salt lowers the freezing point of water.

That is the core idea: when you dissolve a non-volatile solute (like salt, sugar, or urea) in a solvent (like water), the freezing point of the solution becomes lower than that of the pure solvent. This drop is called the depression of freezing point, denoted by $\Delta T_f$.

Why does this happen? The intuition

In a pure liquid, molecules at the surface escape into the solid (freeze) when the temperature is low enough — the solid and liquid are in equilibrium at the freezing point. Now add a solute. The solute particles sit between solvent molecules, getting in the way. For the solvent to freeze, its molecules must arrange themselves into an orderly crystal lattice. The solute particles disrupt this order — they make it harder for the solvent to solidify.

Think of it like trying to pack a suitcase full of neatly stacked blocks. If you throw in a few marbles, the blocks can’t settle as tightly. You’d need to cool the system further (lower the temperature) to force the blocks into place. That extra cooling is the depression.

The precise statement

For a dilute solution, the depression in freezing point is directly proportional to the molality of the solution (moles of solute per kilogram of solvent).

Where:

• $\Delta T_f = T_f^\circ - T_f$ (pure solvent freezing point minus solution freezing point)
• $K_f$ = cryoscopic constant or molal freezing point depression constant — a property of the solvent alone (units: K kg mol⁻¹)
• $m$ = molality of the solution

Each solvent has its own $K_f$. For water, $K_f = 1.86 \text{ K kg mol}^{-1}$. That means: a 1 molal aqueous solution freezes at $-1.86^\circ\text{C}$ (instead of $0^\circ\text{C}$).

How it helps find molar mass

If you dissolve a known mass $w_2$ of an unknown solute in a known mass $w_1$ of solvent, measure the freezing point depression $\Delta T_f$, you can calculate the molar mass $M_2$ of the solute.

Start from the definition of molality:

Substitute into $\Delta T_f = K_f \cdot m$:

Rearrange for $M_2$:

A quick example

You dissolve 5.00 g of a non-electrolyte in 100 g of water. The freezing point drops to $-0.93^\circ\text{C}$. Find the molar mass.

Given: $K_f = 1.86$, $\Delta T_f = 0.93$, $w_2 = 5.00$, $w_1 = 100$. …