Introductory Chemistry (5th Edition)

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

Chapter 13 - Solutions - Exercises - Problems: 57

Answer

10454 mL

Work Step by Step

It is given that a Pb-contaminated water sample contains 0.0011 % Pb by mass. So, 0.0011 g of Pb is present in 100 g of water. For 115 mg of Pb → ? = (0.115 X 100)/(0.0011) = 10454 g of water Density of water = 1 g/mL 10454 g of water = 10454 mL of water
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.