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The boiling point elevation constant for toluene is 3.32 K kg mol^–1. The normal boiling point of toluene is 110.7°C. The enthalpy of vaporization of the toluene would be nearly:

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17.0kJ mol^–1

34.0kJ mol^–1

51.kJ mol^–1

68.0 kJ mol^–1

answer is B.

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Detailed Solution

$\begin{array}{l} {ΔH}_{vap.} = \frac{M_{solvent} \times R \times T_{b (solvent)}^{o}}{1000 \times 3 .32} \\ ∴ {ΔH}_{vap.} = \frac{92 \times 8 .314 \times (383 .7)^{2}}{1000 \times 3 .32} \\ = 33 .91 kJ / mol \approx 34 kJ / mol \end{array}$

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