The boiling point of 0.1 m K4[Fe(CN)6] is expected to be (Kb for water = 0.52 K kg mol-1)

i=1+(n−1)α  =1+(5−1)×1  =5ΔTb=i×Kb×m        =5×0.52×0.1        =0.26KTs=100+0.26       =100.26oc