When 92.0 g of ethanol (c2h5oh are vaporized at its boiling point of 78.3°c, it requires 78.6 kj of energy. what is the approximate molar heat of vaporization of ethanol in kj/mol?

The solution would be like this for this specific problem:

(78.6 kJ) / (92.0 g / (46.0684 g C2H5OH/mol)) = 39.4 kJ/mol 

39.3 

So the approximate molar heat of vaporization of ethanol in kJ/mol is 39.3.

I hope this answers your question.