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: 54

Answer

49.6 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 the mass of isopropyl alcohol is 65.6 g. dH vap of isopropyl alcohol at room temperature is 45.44 kJ./mol We know: 60g = 1mol 1 g= 1/60 mol So, 65.6 g contains 1.093 mol For 1 mol of isopropyl, alcohol dH is 45.4 kJ/mol For 1.093 mol dH for isopropyl alcohol is= 1.093x 45.44 kJ/mol= 49.6 kJ/mol.
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