How energy is needed to vaporize 100 g of ethanol from its boiling point?
q = mass ethanol x heat vaporization.
To determine the amount of energy needed to vaporize 100 g of ethanol from its boiling point, we need to know the heat of vaporization of ethanol. The heat of vaporization is the amount of energy required to change a substance from a liquid to a gas phase at its boiling point, while maintaining the constant temperature.
The heat of vaporization of ethanol is approximately 38.56 kJ/mol. To calculate the energy needed to vaporize a specific mass of ethanol, we can use the molar mass of ethanol, which is approximately 46.07 g/mol.
First, we need to convert the mass of ethanol into moles.
Number of moles = mass / molar mass
Number of moles = 100 g / 46.07 g/mol
Number of moles ≈ 2.17 mol
Next, we can calculate the energy required to vaporize the ethanol using the heat of vaporization.
Energy = heat of vaporization × number of moles
Energy ≈ 38.56 kJ/mol × 2.17 mol
Energy ≈ 83.85 kJ
Therefore, approximately 83.85 kJ of energy is needed to vaporize 100 g of ethanol from its boiling point.