Question

What is the boiling point elevation of a 0.25 m solution of NaCl given that Kb for water = 0.512 0C kg/mol?

Answer 1

$\DeltaTb=Kb*m$

Where $\DeltaTb$ is the boiling point elevation, Kb is the boiling point elevation constant and m is the molality. Kb for water = 0.512 0C kg/mol,molality=0.25m.Therefore, $\DeltaTb$ = 0.512*0.25 =0.128.Boiling point elevation of 0.25m solution of NaCl =0.128