Cinnamon owes its flavor and odor to cinnamaldehyde (C9H8O). Determine the boiling point elevation of a solution of 92.7 mg of cinnamaldehyde dissolved in 1.00 g of carbon tetrachloride (Kb = 5.03°C/m).

The boiling point elevation is 3.53 °C

Explanation:

∆Tb = Kb × m

∆Tb is the boiling point elevation of the solution

Kb is the molal boiling point elevation constant of CCl4 = 5.03 °C/m

m is the molality of the solution is given by moles of solute (C9H8O) divided by mass of solvent (CCl4) in kilogram

Moles of solute = mass/MW =

mass = 92.7 mg = 92.7/1000 = 0.0927 g

MW = 132 g/mol

Moles of solute = 0.0927/132 = 7.02×10^-4 mol

Mass of solvent = 1 g = 1/1000 = 0.001 kg

m = 7.02×10^-4 mol ÷ 0.001 kg = 0.702 mol/kg

∆Tb = 5.03 × 0.702 = 3.53 °C (to 2 decimal places)

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