Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 11 - Properties of Solutions - Additional Exercises - Page 548: 96



Work Step by Step

We know that $molarity=\frac{moles \space of \space solute}{liter \space of \space solution}$ We plug in the known values to obtain: $molarity=\frac{46mL\times 0.79\frac{g}{mL}\times \frac{1mol}{46.07g}}{100.0mL\times \frac{1L}{1000mL}}=7.9M$
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