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: 55

Answer

369 g

Work Step by Step

The amount of heat required to vaporize one mole of liquid is defined as a heat of vaporization. Heat of vaporization is represented by dH vap. The heat of vaporization of 1 mol of water is 40.7 kJ 835 kJ= 835 kJ x(1 mol/ 40.7 kJ)=20.5 mol So, with 835 kJ of dH vap we can vaporize 20.5 mol of water. Since 1mol of water = 18g. 20.5mol of water= 20.5 mol x 18 g= 369 g. So with 835 kJ of dH, we can vaporize 369 g of water.
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