Question

Cinnamon owes its flavor and odor to cinnamaldehyde (C9H8O). Determine the boiling point elevation of a solution of 97.1 mg of cinnamaldehyde dissolved in 1.00 g of carbon tetrachloride (Kb = 5.03°C/m).______ °C

Answer 1

Step 1

The equation for the boiling point elevation used here:

`\Delta Tb\text{ = Kb x m }`

m = molality, which is calculated as follows:

`m\text{ = }\frac{moles\text{ of solute}}{Mass\text{ of solvent \lparen kg\rparen}}`

Kb = boiling constant

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Step 2

Information provided:

Solute = C9H8O

Solvent = carbon tetrachloride

Mass of solute = 97.1 mg

(1 g = 1000 mg => 97.1 mg x (1 g/1000 mg) = 0.0971 g)

Mass of solvent = 1.00 g

(1 kg = 1000 g => 1.00 g x (1 kg/1000 g) = 1x10^-3 kg)

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Information needed:

The molar mass of solute = 132.1 g/mol (please, the periodic table is useful here)

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Step 3

The number of moles of solute: n

n = mass of solute/the molar mass of solute = 0.0971 g/ 132.1 g/mol = 7.35x10^-4 moles

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Molality, m:

m = moles of solute/mass of solvent (kg) = 7.35x10^-4 moles/1x10^-3 kg = 0.735 moles/kg or m

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The boiling point elevation:

ΔTb = Kb x m = 5.03 °C/m x 0.735 m = 3.69 °C

Answer: ΔTb = 3.69 °C