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When 98.3 grams 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? Write your answer in the box below. Round your answer to the
nearest hundredths place. *grams to moles

Respuesta :

SvetkaChem

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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