The standard molar heat of vaporization for water is 40.79 kJ/mol. How much energy would be required to vaporize 30.0g of water? Answer in units of kJ

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Chemistry

Ethen Strickland

12 January, 18:13

The standard molar heat of vaporization for water is 40.79 kJ/mol. How much energy would be required to vaporize 30.0g of water? Answer in units of kJ

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Johan · Answer 1

The energy of vaporizing one mol is 40.79. 1 mol of water is 18g so the number of moles you have is 30/18. Once you know the number of moles you will need to multiply by the heat of vaporization per mole. Also water is not exactly 18g/mol so you might have to calculate it to a couple decimal places depending on what your teacher wants.