Answer :
The boiling point elevation constant (Kb) for H2O, given the increase in boiling point from 100°C to 100.5°C, is calculated to be 0.5°C kg/mol for this problem; though more generally, it's 0.51°C kg/mol.
When 1 mole of solute is dissolved in 1 kg of H2O, the boiling point of the solution increases from the normal boiling point of water (100°C) to 100.5°C. The extent to which the boiling point increases is determined by the molal boiling point elevation constant (Kb). The constant represents the elevation in boiling point experienced when one molal (1 mol/kg) of solute is dissolved in a solvent.
The formula to determine the boiling point elevation is ΔTb = Kb ⋅ m, where ΔTb is the change in boiling point and m is the molality of the solution.
Given that m equals 1 mol/kg (by definition of the problem), we can rearrange the equation to solve for Kb: ΔTb = Kb ⋅ m0.5°C = Kb ⋅ 1 mol/kg
It is now clear that Kb = 0.5°C kg/mol for H2O within the context of this specific problem.
However, the molal boiling point elevation constant for water is typically 0.51°C kg/mol.
