What is the boiling point of a solution prepared by dissolving 10.00 grams of naphthalene (C₁₀H₈,128.2 g/mol)in 200.0 grams of benzene (C₆H₆,78.12 g/mol)? The normal boiling point of benzene is 80.1 $\circ$ C and K_b = 2.53 $\circ$ C/m.

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Quizplus · Accepted Answer

The boiling point elevation can be calculated using ΔT = i * Kb * m, where i = 1 for naphthalene (non-electrolyte), Kb = 2.53 °C/m, and m is the molality. Molality (m) = moles of solute/kg of solvent = (10.00 g / 128.2 g/mol) / (0.200 kg) = 0.3906 m. ΔT = 1 * 2.53 °C/m * 0.3906 m = 0.988 °C. The new boiling point = 80.1 °C + 0.988 °C = 81.1 °C.

What Is the Boiling Point of a Solution Prepared by Dissolving