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-

Colligative properties as elevation of boiling point part-4

Elevation of boiling point (Ebullioscopy) 

•  Boiling point – is the temperature at which the vapour pressure of the liquid becomes equal to the atmospheric pressure. 
•      Generally, we found that the B. P. of Solution is always higher than that of the pure solvent. the decrease in the boiling point of solution is called elevation of boiling point. 
• we know that vapour pressure of solvent (pure) diseases with mixing non-a volatile which occupy the surface of solvent, therefore, vapour pressure of solution is lower than that of the pure Solvent. 
• The variation of vapour pressure of pure solvent and solution is as a function of temperature. 
• hence the solution has to be heated more to 
• make the vapour pressure equal to the atmospheric 
• Elevation of boiling point may be explained on the basis of graph plotted between vapour pressure & Temperature. 

let us consider consider T°b as boiling point of pure solvent & Tb   as boiling poiling poin of 
solution.  Thus, the increase in boiling point is      
ΔTb  = Tb -   T°b          …..............(1)               

Excrementally, elevation of boing point for dilute solution is directionally proportional to molal concentration to solution   

ΔTb  α   m   …...............(2)     

 (here m= molality = w2 x 1000 /M2 x w1   , w2 = mass of solute    M2 = molar mass of solute 
w1   = mass of solvent 
ΔTb  =kbm   ------------(3)
ΔTb  =kb w2 x 1000 /M2 x w1 …. (4)                        

So   M2 =kb w2 x 1000 / ΔTb  x w1   …..(5)      

this equation is used to calculate molar mass of solute dissolved in solution. 

Kb is boiling point elevation constant or molal elevation constant or Ebullioscopic constant   which depends upon nature of solvent. The relationship between  Kb and vaporisation of enthalpy may be given as -           

kb  = R xM1 x  (T°b )2   /   1000 x  ΔvapH    …...... (6)

Problem - 1