Answer:
The answer to your question is Tb = 100.768°C
Explanation:
Data
Tb = ?
moles of KBr = 3
mass of water = 2000 g
Kb = 0.512 C/m
Process
1.- Calculate the molality of the solution
molality = moles of solute/mass of solvent (kg)
-Substitution
molality = 3/2
-Result
molality = 1.5
2.- Calculate ΔT
ΔTb = Kbm
-Substitution
ΔTb = (0.512)(1.5)
-Result
ΔTb = 0.768
3.- Calculate the boiling point, the boiliing point of pure water is 100°C
ΔTb = Tb - Tb(solvent)
-Solve for Tb
Tb = Tb(solvent) + ΔTb
-Substitution
Tb = 100 + 0.768
-Result
Tb = 100.768°C
