What is the predicted change in the boiling point of water when 4.00 g of barium chloride is dissolved in 2.00 kg of water

The predicted change in the boiling point of water is Δt = 0.0148 °C
Solution:
We will use the equation for boiling point elevation Δt
     Δt = i Kb m 
where the van't Hoff Factor i is equal to 3 since one molecule of barium chloride in aqueous solution will produce one Ba2+ ion and  two Cl- ions. The molality m of the solution of 4.00 g of barium chloride dissolved in 2.00 kg of water can be calculated using the molar mass of barium chloride: 
     m = [4.00g BaCl2 * (1 mol BaCl2 / 208.233g BaCl2)] / 2.00kg H2O
         = 0.009605 mol/kg

Therefore, the amount Δt the boiling point increases is 
     Δt = i Kb m 
          = (3) (0.512 °C·kg/mol) (0.009605 mol/kg)
          = 0.0148 °C
We can also find the new boiling point T for the solution since we know that pure water boils at 100 °C:
     Δt = T - 100°C T = Δt + 100°C = 0.0148 °C + 100°C = 100.0148°C 

What do you mean by that? like the contents? the thc level?