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The molar heat of vaporization of water is 40.7 kJ mol . How much heat must be absorbed to convert 50.0 grams of liquid water at 100oC to steam at 100oC?
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The molar heat of vaporization of water is 40.7 kJ mol . How much heat must be absorbed to convert 50.0 grams of liquid water at 100oC to steam at 100oC?

User Davina Leong
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1 Answer

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Molar mass of water 18g/mol

Number of mols = 50.0g/18g/mol =2.78 mol

Heat absorbed = 40.7 kj/mol * 2.78 mol = 113.1 kj.
User Michael Veksler
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