Answer: 106.58°C
1) Data
- solute: NaCl
- n solute = 1.00 moles
- solvent: water
- grams of solvent = 155 g
- kb = 0.51 °C / m
- Tb ?
2) Formulas
- ΔTb = Tb - normal Tb
- ΔTb = i * kb * m
- m = moles of solute / kg of solvent
3) Solution:
3.1) m = n solute / kg solvent = 1.00 moles / 0.155 kg = 6.4516 m
3.2) i for NaCl is 2, because each molecule of NaCl dissolves into 2 ions.
3.3) ΔTb = 2 * 0.51 °C/m * 6.4516m = 6.58°C
3.3) ΔTb = Tb - normal Tb => Tb = ΔTb + normal Tb
normal Tb of water = 100°C
Tb = 6.58 °C + 100 °C = 106.58°C
Answer: 106.58°C
