Molar mass (Mr) of ethanol = 46.07 g/mol
How much energy is required to vaporize 98.6 g of ethanol (C₂H₅OH) at its boiling point?
Observe this . This means that 40.5 kJ of energy is required to vaporize every 1 mole of ethanol at its boiling point. Therefore we must first convert grams into moles.
Let us count the number of moles of 98.6 g of ethanol.
We obtain 2.14 moles of ethanol.
Let us calculate how much energy is required to vaporize 2.14 moles of ethanol at its boiling point.
The amount of energy required is
Thus, the amount of energy required to vaporize 98.6 g of ethanol at its boiling point is 86.7 kJ.
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Keywords: how much, energy required, to vaporize, ethanol, C₂H₅OH, its boiling point, ΔHvap, molar mass, moles, converts, kJ/mol