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What is the amount of heat required to completely vaporize 200 g 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 required to vaporize 200 g of water at its boiling point, we need to use the formula:q = n * Hvapwhere q is the heat required, n is the number of moles of water, and Hvap is the enthalpy of vaporization.First, we need to find the number of moles of water  n . The molar mass of water  HO  is 18.015 g/mol.n = mass / molar massn = 200 g / 18.015 g/moln  11.106 molesNow we can calculate the heat required  q  using the enthalpy of vaporization  Hvap  of water, which is 40.7 kJ/mol.q = n * Hvapq = 11.106 moles * 40.7 kJ/molq  452.1 kJTherefore, the amount of heat required to completely vaporize 200 g of water at its boiling point of 100C is approximately 452.1 kJ.
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