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Calculate the amount of energy required to vaporize 100 grams of liquid water at its boiling point (100°C) given its enthalpy of vaporization is 40.7 kJ/mol.

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To calculate the energy required to vaporize 100 grams of liquid water, we need to first determine the number of moles of water present and then use the enthalpy of vaporization to find the energy.1. Calculate the number of moles of water:The molar mass of water  H2O  is 18.015 g/mol.Number of moles = mass / molar massNumber of moles = 100 g / 18.015 g/mol = 5.551 moles2. Calculate the energy required using the enthalpy of vaporization:Energy = number of moles  enthalpy of vaporizationEnergy = 5.551 moles  40.7 kJ/mol = 225.93 kJTherefore, the amount of energy required to vaporize 100 grams of liquid water at its boiling point is approximately 225.93 kJ.

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