Determine the boiling point of a solution that contains 78.8 g of naphthalene (C₁₀H₈, molar mass = 128.16 g mol^-1) dissolved in 722 mL of benzene (d = 0.877 g mL^-1). Pure benzene has a boiling point of 80.1 °C and a boiling point elevation constant of 2.53 °C m^-1. A) 2.2 °C B) 2.5 °C C) 82.3 °C D) 80.4 °C E) 82.6 °C

Question

Quizplus · Accepted Answer

First, we need to calculate the molality of the naphthalene solution:
molality = moles of solute / mass of solvent (in kg)
mass of solvent = density x volume
mass of solvent = 0.877 g/mL x 722 mL = 640.94 g = 0.64094 kg
moles of solute = mass of solute / molar mass
moles of solute = 78.8 g / 128.16 g/mol = 0.6153 mol
molality = 0.6153 mol / 0.64094 kg = 0.9604 mol/kg
Next, we need to use the boiling point elevation equation:
ΔTb = Kb x molality
ΔTb = 2.53 °C/m x 0.9604 mol/kg = 2.43 °C
Finally, we can calculate the boiling point of the solution:
boiling point = pure solvent boiling point + ΔTb
boiling point = 80.1 °C + 2.43 °C = 82.53 °C, which rounds to 82.6 °C.

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