Calculate the degree of ionisation of normal ZnSO 4 solutionat 373 k if its vapour pressure is 750mm.

							Hi, 


Recall that the total pressure of a solution is the sum of the partial pressures of the solvent and solute

psolution = psolvent + psolute = xsolvent P°solvent + xsolute P°solute 
If the solute is non-volatile (no vapour pressure: P°solute = 0) then the total vapour pressure of solution is

psolution = psolvent = xsolvent P°solvent      (where x= mol fractn)

now v.p of water at 373 k = 760 mm           



                             750= x*760


=>  x = 750/760

    1-x  = mole fractn of solute  = mol of solute/total moles(nearly = moles of water)  
                                   
                                  = mol of solute/moles of water

                        ZnSO4 -> Zn(+2) +SO4(-2)
                        0.5-y     y          y           (1N= 0.5M)

total moles of solute =  0.5+y

                 total moles of solvent(water)=55.55   (in 1 lt)

                       => 1-x = 0.5 + y /55.55


                       =>1-750/760= 1/76= 0.5+y /55.55
                   
                      => y = 0.23 
                     y = dissociation = 23%