Introductory Chemistry (5th Edition)

Published by Pearson
ISBN 10: 032191029X
ISBN 13: 978-0-32191-029-5

Chapter 12 - Liquids, Solids, and Intermolecular Forces - Exercises - Problems: 53

Answer

10.4 kJ/mol

Work Step by Step

The amount of heat required to vaporize one mole of liquid is defined as the heat of vaporization. Heat of vaporization is represented by dH vap. Given mass of sweat is 4.25g. dH vap of sweat at room temperature is 44 kJ./mol We know: 18g = 1mol 1 g= 1/18 mol So, 4.25g contains 0.23611 mol For 1 mol of sweat dH is 44 kJ/mol For 0.23611mol dH is= 0.23611x 44 kJ/mol= 10.4 kJ/mol
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