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