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14 January, 11:41

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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  1. 14 January, 12:02
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    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
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