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german
3 years ago
15

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 approxim

ate molar heat of vaporization of ethanol in kj/mol?
Chemistry
2 answers:
lilavasa [31]3 years ago
8 0

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

Juli2301 [7.4K]3 years ago
7 0

The solution would be like this for this specific problem:

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

<span>39.3 </span>

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

I hope this answers your question.

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