Complete question:
Imagine two solutions with the same concentration and the same boiling point, but one has benzene as the solvent and the other has carbon tetrachloride as the solvent. Determine that molal concentration, m (or b), and boiling point, Tb.
benzene boiling point=80.1 Kb=2.53
carbon tetrachloride boiling point=76.8 Kb=5.03
Answer:
m = 1.32 mol/kg
Boiling point: 83.4°C
Explanation:
When a nonvolatile solute is added to a pure solvent, the boiling point of the solvent increases, a phenomenon called ebullioscopy. This happens because of the interactions between the solute and the solvent. The temperature variation (new boiling point - normal boiling point) can be calculated by:
ΔT = m*Kb*i
Where m is the molal concentration (moles o solute/mass of solvent in kg), Kb is the ebullioscopy constant of the solvent, and i is the van't Hoff factor, which indicates how much of the solute dissociates. Let's assume that i is equal in both solvents and equal to 1 (the solvent dissociates completely)
Calling the new boiling point as Tb, for benzene:
Tb - 80.1 = m*2.53*1
Tb = 2.53m + 80.1
For carbon tetrachloride:
Tb - 76.8 = m*5.03*1
Tb = 5.03m + 76.8
Because Tb and m are equal for both:
5.03m + 76.8 = 2.53m + 80.1
2.5m = 3.3
m = 1.32 mol/kg
So, substituting m in any of the equations (choosing the first):
Tb = 2.53 * 1.32 + 80.1
Tb = 83.4°C
