Introductory Chemistry (5th Edition)

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

Chapter 13 - Solutions - Exercises - Problems - Page 480: 58

Answer

472972.97 mL

Work Step by Step

It is given that a benzene-contaminated water sample contains 0.000037% benzene by mass. So, 0.000037 g of benzene. is present in 100 g of water. For 175 mg of benzene → ? = (0.175 X 100)/0.000037 = 472972.97 g of water Density of water = 1 g/mL 472972.97 g of water = 472972.97 mL of water
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