1. Calculate the molality of the solution:
* Molar mass of NaCl: 58.44 g/mol
* Moles of NaCl: (3.40 g NaCl) / (58.44 g/mol) = 0.0582 mol NaCl
* Mass of water (in kg): 148 mL * (1 g/mL) * (1 kg/1000 g) = 0.148 kg
* Molality (m): (0.0582 mol NaCl) / (0.148 kg water) = 0.393 mol/kg
2. Calculate the boiling point elevation:
* Boiling point elevation constant (Kb) for water: 0.512 °C/m
* van't Hoff factor (i) for NaCl: 2 (NaCl dissociates into 2 ions in solution: Na+ and Cl-)
* Boiling point elevation (ΔTb): i * Kb * m = 2 * 0.512 °C/m * 0.393 mol/kg = 0.404 °C
3. Determine the new boiling point:
* Normal boiling point of water: 100 °C
* New boiling point: 100 °C + 0.404 °C = 100.404 °C
Therefore, the boiling point elevation of the salt water solution is 0.404 °C, and the new boiling point is 100.404 °C.
