Addition of 1kg of non volatile solute in volatile solvent increases the boiling point from 350K to 355K and decreases the freezing point from 250K to 220K.If Kb is 15.5KKg/mol,then Kf will be?

Well, these are really the same trends, arising from some nonzero concentration of solute in a volatile solvent. It does not matter how much solute is there, if the amount of solvent is not mentioned.

Also, regardless of to what extent the solute ionizes, it will not affect the size of #K_b# relative to #K_f#. That should make sense, because those are for the solvent, not the solute.

Lastly, #K_f > K_b# makes sense. Usually it takes more solute to raise the boiling point than it does to lower the freezing point by #1^@ "C"#.

Recall that both of these arise from a given concentration:

#DeltaT_b = T_b - T_b^"*" = iK_bm#

#DeltaT_f = T_f - T_f^"*" = -iK_fm#

where:

• #T_(tr)# is the phase transition temperature, and #"*"# indicates pure solvent.
• #i# is the van't Hoff factor, i.e. the effective number of particles that dissociate from every solute particle.
• #K_b# and #K_f# are the boiling point elevation and freezing point depression constants, respectively, in units of #"K"cdot"kg/mol"#.
• #m# is the molality in #"mols solute/kg solvent"#.

You didn't really mention what kind of solute it was... so we will not identify #i#. Thus, the concentration was:

#m = (DeltaT_b)/(iK_b)#

#= ("355 K" - "350 K")/(i cdot "15.5 K"cdot"kg/mol")#

#=# #"0.3226 mol/kg"/i#

Obviously, we have been looking at the same solvent and solute the whole time. And so, the #K_f# was based on the same concentration. The #i# cancels out:

#color(blue)(K_f) = -(DeltaT_f)/(im)#

#= -("220 K" - "250 K")/(cancel(i) cdot "0.3226 mol/kg"/canceli)#

#=# #color(blue)("93 K"cdot"kg/mol")#