High School

It takes 38.6 kJ of energy to vaporize 1.00 mol of ethanol (molecular weight: 46.07 g/mol).

What will be [tex]\Delta S_{sys}[/tex] for the vaporization of 11.0 g of ethanol at 78.4 °C?

Answer :

Answer:

9.21 kJ is the enrgy required to vaporize 11.0 grams of ethanol.

Explanation:

Energy of vaporization can be stated as 38.6 kj/mole for ethanol. ["It takes 38. 6 kj of energy to vaporize 1. 00 mol of ethanol"]

11.0 grams of ethanol is not 1 mole. Calculate the moles of ethanol by dividing t=it's mass by its molar mass:

(11.0 grams)/(46.07 g/mole) = 0.2388 moles ethanol

(0.2388 moles ethanol)*(38.6 kj/mole) = 9.21 kJ is the enrgy required to vaporize 11.0 grams of ethanol.

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