Calculate the degree of ionisation of normal ZnSO₄ 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%