The answer is 4.9 moles.
Solution:Â
Using the equation for boiling point elevation Δt,
   Δt = i Kb mÂ
we can rearrange the expression to solve for the molality m of the solution:
   m = Δt / i KbÂ
Since we know that pure water boils at 100 °C, and the Ebullioscopic constant Kb for water is 0.512 °C·kg/mol,Â
   m = (105°C - 100°C) / (2 * 0.512 °C·kg/mol)
     = 4.883 mol/kgÂ
From the molality m of the solution of salt added in a kilogram of water, we can now find the number of moles of salt:Â
   m = number of moles / 1.0kg
   number of moles = m*1.0kgÂ
                  = (4.883 mol/kg) * (1.0kg)
                  = 4.9 moles