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What is the amount of energy required to vaporize 50 grams of water at its normal boiling point? The enthalpy of vaporization of water is 40.7 kJ/mol.

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To calculate the amount of energy required to vaporize 50 grams of water at its normal boiling point, we need to first determine the number of moles of water present and then use the enthalpy of vaporization to find the energy required.1. Calculate the number of moles of water:The molecular weight of water  H2O  is 18.015 g/mol  1.008 g/mol for hydrogen and 15.999 g/mol for oxygen . Number of moles = mass / molecular weightNumber of moles = 50 g / 18.015 g/mol = 2.775 moles2. Calculate the energy required using the enthalpy of vaporization:Energy required = number of moles  enthalpy of vaporizationEnergy required = 2.775 moles  40.7 kJ/mol = 112.9425 kJTherefore, the amount of energy required to vaporize 50 grams of water at its normal boiling point is approximately 112.94 kJ.
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