Colligative properties as depression of freezing point Part-5

Depression of freezing point (cryoscopy) 

   

     freezing point - The temperature at which vapour pressure of the substance in its liquid phase is equal to its vapour pressure in the solid phase . i.e. the solid & the liquid form of substance have the same vapour pressure . 

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•      a solution will freeze when its vapour pressure becomes equal to the various pressure of the pure solid solvent. 
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•      it is observed that freezing point of solution is always lower than that of pure solvent. The decrease in freezing print is called depression of freezing point. Because lower  the temperature ,lower is the vapour pressure 
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•      Vapour pressure of solvent decreases with mixing non-volatile solute in it. Since non-volatile occupy the surface of it.  

    

      Alternatively, the depression of freezing point may be explained on the basis of plots of Vapour pressure versus temperature. Vapour pressure of solution and solvent increases with the increases of temperature while decreases on decreasing temperature. 

let us consider T°f  as freezing point of pure solvent & Tf  as freezing point of solution.

Thus depression in freezing point is  ΔTf = Tf -   T°f     …..............(1)       Experimentally, depression of freezing point for dilute solution is directionally  proportional to molal                                                                     concentration to solution  . it may be exressed as                                      

      

ΔTf  α   m   …...............(2)        

ΔTf  = Kf m  --------------(3)  

(here m= molality = w2 x 1000 /M2 x w1   , w2 = mass of solute  

ΔTf  = Kf x w2 x 1000 /M2 x w1 …. (4)               M2 = molar mass of solute                                                                              w1   = mass of solvent

So   M2 = Kf x w2 x 1000 / ΔTf x w1   …..(5)      this equation is used to calculate molar mass of solute dissolved in solution.  

Kf is molal depression constant or cryoscopy constant   which depends upon the nature of solvent. The relationship between  Kf and fusion of enthalpy may be given as -                                                                   

Kf = R X M1 X  (T°f)2   /   1000 X  ΔfusionH    …...... (5) 

some important problems-