Chemistry 10th Edition

by Whitten, Kenneth W.; Davis, Raymond E.; Peck, Larry; Stanley, George G.

Published by Brooks/Cole Publishing Co.

ISBN 10: 1133610668

ISBN 13: 978-1-13361-066-3

Chapter 11 - Reactions in Aqueous Solutions II: Calculations - Exercises - Molarity - Page 394: 4

Answer

.11465 mol/L

Work Step by Step

41.4 g $\div$ 120.366 g/mol = .34395 mol .34395 mol $\div$ 3 L = .11465 mol/L

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