Table of Contents

## Depression in Freezing Point:

It is the temperature when the solid and the liquid forms of the substance are in equilibrium with each other, i.e., the solid and the liquid form of the substance have the same vapour pressure and co-exist with each other. It is characteristic of each liquid. For Example- the freezing point of H_{2}O is 273.15 K.

It is observed that the vapour pressure of the solution is lower than the pure solvent. Thus a solution has to be cooled further to get the freezing point of the solution. The decrease in the freezing point of the solution due to the addition of non-volatile solute to solvent is called depression in freezing point. It is denoted by ΔT_{f}.

This depression in freezing point can also be explained by plotting vapour pressure versus temperature curve as shown above in the figure. T_{1} is the freezing point of the solvent and T_{2} is the freezing point of the solution (T_{2} < T_{1}). Therefore, ΔT_{f }= (T_{2} – T_{1}).

### Expression for Depression in Freezing Point:

We know that, ΔT _{f} ∝ ΔP ……….(i)Again by Raoult’s law, ΔP ∝ x _{B} ……….(ii)Therefore, ΔT _{f} = k . x_{B} [Where x_{B} is the mole fraction of solute in solution]⇒ ΔT _{f} = (k . n_{B}) / (n_{A} + n_{B}) = k (n_{B}/n_{A}) [Since solution is dilute and n_{B} << n_{A}]⇒ ΔT _{f} = k . n_{B} / (W_{A}/M_{A}) = k . M_{A}n_{B} / W_{A} = k_{f} (n_{B}/W_{A}) ……….(iii)[Where M _{A} is molecular mass of solvent and W_{A} = weight of solvent]Where k _{f} is a new constant (k_{f} = k . M_{A}) called molal depression constant or cryoscopic constant.If the weight of solvent (W _{A}) is 1 kg, then n_{B}/W_{A} = molality (m)Therefore, ΔT _{f} = k_{f} . m ……….(iv)If m = 1; ΔT _{f} = k_{f}. Thus, molal depression constant is the depression in freezing point when molality of a solution is unity. Its unit is K/m or K kg mol-1. Its value for H_{2}O is 1.86 K kg mol^{-1} at 273 K. |

### Calculation of Molecular Mass of Solute from ΔT_{f}:

From (iv), we have ΔT _{f} = k_{f} . m = k_{f} (n_{B} / W_{A}) X 1000 [Since molality is in terms of 1000 gm]Therefore, ΔT _{f} = (k_{f} . W_{B} . 1000) / (M_{B} . W_{A})⇒ M _{B} = (k_{f} . W_{B} . 1000) / (ΔT_{f} . W_{A})Hence, molecular mass of solute (M _{B}) is calculated. |