To find the boiling point elevation of the solution, we can use the following equation:
ΔTb = Kb x molality
where ΔTb is the boiling point elevation, Kb is the molal boiling point elevation constant (for water, Kb = 0.512°C/m), and molality is the molal concentration of the solution.
To determine the molality, we need to first calculate the number of moles of Barium Sulfate in the solution:
moles of BaSO4 = mass / molar mass
moles of BaSO4 = 600.0 g / 233.38 g/mol (molar mass of BaSO4)
moles of BaSO4 = 2.572 mol
Then, we can calculate the molality of the solution:
molality = moles of BaSO4 / mass of water (kg)
molality = 2.572 mol / 0.345 kg
molality = 7.451 m
Finally, we can use the equation above to calculate the boiling point elevation:
ΔTb = 0.512°C/m x 7.451 m
ΔTb = 3.817°C
So, the boiling point of the solution would be 3.817°C higher than the boiling point of pure water. If we assume the boiling point of pure water to be 100°C, then the boiling point of the solution would be 103.817°C.