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Answer :
The molal boiling-point elevation constant (Kb) for [tex]H_2O[/tex], given the boiling point elevation of 0.5°C for a 1 molal solution, is correctly identified as 0.5°C kg/mol. Thus, the correct answer is A, 0.5.
The question deals with the concept of boiling point elevation, which is a colligative property of solutions. In this case, when 1 mole of a solute is dissolved in 1 kg of [tex]H_2O[/tex], the boiling point of water was found to be 100.5°C.
We can use the equation ΔTb = Kb times m, where ΔTb is the boiling point elevation, Kb is the molal boiling-point elevation constant (or ebullioscopic constant), and m is the molality of the solution.
To find Kb for [tex]H_2O[/tex], we rearrange the equation to Kb = ΔTb / m. Given that the boiling point elevation is 0.5°C (from the normal boiling point of water, which is 100°C) and the molality is 1 mol/kg (since we have 1 mole of solute in 1 kg of water), we get Kb = 0.5°C / 1 m, which equals 0.5°C kg/mol. Thus, the correct answer is A, 0.5.
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