JEE Main & Advanced Chemistry Solutions / विलयन Depression in f.pt. Of the solvent (cryoscopy)

Freezing point is the temperature at which the liquid and the solid states of a substance are in equilibrium with each other or it may be defined as the temperature at which the liquid and the solid states of a substance have the same vapour pressure. It is observed that the freezing point of a solution is always less than the freezing point of the pure solvent. Thus the freezing point of sea water is low than that of pure water. The depression in freezing point  \[(\Delta T\] or \[\Delta {{T}_{f}})\] of a solvent is the difference in the freezing point of the pure solvent \[({{T}_{s}})\] and the solution \[({{T}_{sol.}})\].

                     \[{{T}_{s}}-{{T}_{sol}}=\Delta {{T}_{f}}\] or \[\Delta T\]

\[NaCl\] or \[CaC{{l}_{2}}\] (anhydrous) are used to clear snow on roads. They depress the freezing point of water and thus reduce the temperature of the formation of ice.

Depression in freezing point is determined by Beckmann’s method and Rast’s camphor method. Study of depression in freezing point of a liquid in which a non-volatile solute is dissolved in it is called as cryoscopy.

Important relations concerning depression in freezing point.

(1) Depression in freezing point is directly proportional to the lowering of vapour pressure.  \[\Delta {{T}_{f}}\propto {{p}^{0}}-p\]

(2) \[\Delta {{T}_{f}}={{K}_{f}}\times m\]

where \[{{K}_{f}}=\] molal depression constant or cryoscopic constant; \[m=\] Molality of the solution (i.e., no. of moles of solute per \[1000g\] of the solvent);  \[\Delta {{T}_{f}}=\]Depression in freezing point

(3) \[\Delta {{T}_{f}}=\frac{1000\times {{K}_{f}}\times w}{m\times W}\] or \[m=\frac{1000\times {{K}_{f}}\times w}{\Delta {{T}_{f}}\times W}\]

where \[{{K}_{f}}\] is molal depression constant and defined as the depression in freezing point produced when 1 mole of the solute is dissolved in \[1kg\] of the solvent. \[w\] and \[W\] are the weights of solute and solvent and \[m\] is the molecular weight of the solute.

(4) \[{{K}_{f}}=\frac{R{{({{T}_{0}})}^{2}}}{{{l}_{f}}1000}=\frac{0.002{{({{T}_{0}})}^{2}}}{{{l}_{f}}}\]

where \[{{T}_{0}}=\]Normal freezing point of the solvent; \[{{l}_{f}}=\]Latent heat of fusion/g of solvent; \[{{K}_{f}}\] for water is \[1.86\ \deg -kg\ mo{{l}^{-1}}\]