What is the expected boiling point of a brine solution containing 45.0 g of NaCl dissolved in 150.00 g of water? ?

Molar mass of NaCl = 23.0 + 35.5 = 58.5 g/mol

Each mole of NaCl contains 1 mole of Na⁺ ions and 1 mole of Cl⁻ ions, i.e. 2 moles of ions.
Hence, van't Hoff factor, i = 2
Molality of the solution, m = (45.0/58.5 mol) / (150.00/1000 kg water) = 5.128 m
Boiling point elevation constant, Kb = 0.512 °C/m

Boiling point elevation, ΔTb = i Kb m = 2 × 0.512 × 5.128 °C = 5.25°C

Normal boiling point of water = 100.00°C
Boiling point of the solution = (100.00 + 5.25) °C = 105.25 °C