Question

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?

Answer 1

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 = $\frac{Mass}{Molar mass} = \frac{92 g}{46 g/mol} = 2 moles$

78.6 kJ of energy is required to vaporise 2 moles of ethanol

Therefore, the amount of energy required per mole would be:

$= \frac{78.6 kJ * 2 moles}{1 mole} = 157.2 kJ$

Answer 2

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.