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What is the amount of heat energy required to vaporize 50 grams of water at its boiling point of 100°C, given that the enthalpy of vaporization of water is 40.7 kJ/mol?

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To calculate the amount of heat energy required to vaporize 50 grams of water, we need to first determine the number of moles of water and then use the enthalpy of vaporization to find the energy.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 . moles of water = mass of water / molecular weight of watermoles of water = 50 g / 18.015 g/molmoles of water  2.776 moles2. Calculate the heat energy required using the enthalpy of vaporization:The enthalpy of vaporization of water is given as 40.7 kJ/mol.Heat energy required = moles of water  enthalpy of vaporizationHeat energy required = 2.776 moles  40.7 kJ/molHeat energy required  112.97 kJTherefore, the amount of heat energy required to vaporize 50 grams of water at its boiling point of 100C is approximately 112.97 kJ.
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