Boiling point of 0.5 m NaCl is 373.52 K. Degree of ionization of NaCl is Kb of water=0.52 K.Kg.mol-1

Given BP of NaCl solution=373.52 K;∆Tb=373.78-373K=0.52 K; ∆Tb=i.Kb.m; i=0.520.52×0.5=2; for NaCl, n=2 i=1-α+2α=1+α; α=1;