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 - Page 441: 56

Answer

0.4766 L

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. The heat of vaporization of 1 mol of water is 40.7 kJ 1078 kJ= 1078 kJ x(1 mol/ 40.7 kJ)=26.48 mol So, with 1078 kJ of dHvap we can vaporize 26.48 mol of water. Since 1mol of water = 18g. 26.48 mol of water= 26.48 mol x 18 g= 476.64 g . We know density= mass/volume volume= mass/density density of water= 1g/mol. mass= 476.64 g. volume= (476.64 g/ 1 g/ml)= 0.4766 L
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