calculate the amount of energy required to vaporize 43.0g of water at its boiling point
q = mass H2O x heat vaporization
To calculate the amount of energy required to vaporize a given amount of water at its boiling point, you will need to use the heat of vaporization of water. The heat of vaporization is the amount of energy required to convert one gram of a substance from its liquid state to its vapor state at a constant temperature.
The heat of vaporization of water is approximately 40.7 kJ/mol. However, we need to convert grams to moles to work with this value.
Step 1: Convert grams to moles
To convert grams to moles, we need to divide the mass of water (43.0g) by its molar mass. The molar mass of water (H2O) is approximately 18.0 g/mol.
Number of moles = Mass / Molar mass
Number of moles = 43.0g / 18.0 g/mol = 2.39 mol
Step 2: Calculate the energy required
Now that we have the number of moles, we can calculate the energy required using the heat of vaporization.
Energy = Number of moles × Heat of vaporization
Energy = 2.39 mol × 40.7 kJ/mol = 97.333 kJ
Therefore, the amount of energy required to vaporize 43.0g of water at its boiling point is approximately 97.333 kJ.