High School

When 98.3 grams of ethanol ([tex]C_2H_5OH[/tex]) 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?

Write your answer in the box below. Round your answer to the nearest hundredths place.

Answer :

Answer:

36.84 kJ/mol

Explanation:

M(C2H5OH)=2*12.01 +6*1.01 + 16.00 = 46.07 g/mol

98.3 g* 1 mol/46.07 g = 2.1337 mol

78.6 kJ/ 2.1337 mol = 36.84 kJ/1 mol = 36.84 kJ/mol

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