How much heat (in kj) is absorbed when 32.5 g h2o (l) at 100°c and 101.3 kpa is converted to steam at 100°c? (the molar heat of vaporization of water is 40.7 kj/mol.) ?

Question

Cristal

Chemistry

3 November, 15:37

How much heat (in kj) is absorbed when 32.5 g h2o (l) at 100°c and 101.3 kpa is converted to steam at 100°c? (the molar heat of vaporization of water is 40.7 kj/mol.) ?

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Justice Chung · Answer 1

The molar heat of vaporization of water is 40.7 kj/mol

The molecular mass of water is 18g/mol

Therefore; the number of moles of water will be; 32.5g/18g/mol = 1.81 mol

Hence; the amount of heat absorbed will be;

1.81 mol * 40.7 kJ/mol = 73.7 kJ

Thus; heat absorbed will be; 73.7kJ