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.
Answer:
Molar heat of vaporization of ethanol, 157.2 kJ/mol
Explanation:
Molar heat of vaporization is the amount heat required to vaporize 1 mole of a liquid to vapor.
The equilibrium is represented as:
C2H5OH(l) ↔ C2H5OH(g)
Given:
Mass of ethanol = 92.0 g
Energy required = 78.6 kJ
Calculation:
Molar mass of ethanol = 46 g/mol
Moles of ethanol = [tex]\frac{Mass}{Molar mass} = \frac{92 g}{46 g/mol} = 2 moles[/tex]
78.6 kJ of energy is required to vaporise 2 moles of ethanol
Therefore, the amount of energy required per mole would be:
[tex]= \frac{78.6 kJ * 2 moles}{1 mole} = 157.2 kJ[/tex]
