The degree of dissociation of Ca(NO3)2 in a dilute aqueous solution containing 7.0 g of salt per 100 g of water at 100oC is 70%. If the vapour pressure of water at 100oC  is 760 mmHg, the vapour pressure of the solution is

Moles of Ca(NO3)2=7164 present in 100.0 g of water Since dissociation is 70%, the total number of particles 7164×0.7×3=0.0896=nAlso, moles of solvent =10018=5.55=NApplying Raoult's lawpo−psolutionpo=nn+N760−ps760=0.0890.089+5.55 760−ps = 7600.0895.637                 =760[0.0157]                          =11.99 ps=760−11.99=748.01 mmHg