The heat of vaporization of 1-pentanol is 55.5 kj/mol, and its entropy of vaporization is 148 j/k. mol. what is the approximate boiling point of 1-pentanol

As you can see, the unit of heat of vaporization is in kJ/mol, while the unit for entropy of vaporization is in J/mol·K. Since it is vaporization, this occurs at the boiling point. Thus, the formula would be:

Boiling Point = ΔHvap/ΔSvap

Make sure the units are consistent.

Boiling point = 55.5 kJ/mol * 1000 J/kJ / 148 J/mol·K = 375 K or 102°C

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The correct answer is b