Question

It takes 38. 6 kj of energy to vaporize 1. 00 mol of ethanol (mw: 46. 07 g/mol). What will be δssys for the vaporization of 11. 0 g of ethanol at 78. 4 °c?

Answer 1

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

Step-by-step 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.