What is the boiling point of a solution prepared by dissolving 10.00 g of naphthalene (C₁₀H₈, 128.17 g/mol) in 200.0 g of benzene (C₆H₆, 78.12 g/mol)? The normal boiling point of benzene is 80.1$\degree$C and K_b =2.53$\degree$C/m.

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Quizplus · Accepted Answer

First, calculate the molality of the solution: molality = moles of solute / kg of solvent = (10.00 g / 128.17 g/mol) / (200.0 g / 1000 g/kg) = 0.390 m. Then, calculate the boiling point elevation: ΔTb = Kb * molality = 2.53°C/m * 0.390 m = 0.987°C. Finally, add this to the normal boiling point of benzene: 80.1°C + 0.987°C = 81.087°C, which rounds to 81.1°C.

What Is the Boiling Point of a Solution Prepared by Dissolving