General Chemistry: Principles and Modern Applications (10th Edition)

Published by Pearson Prentice Hal
ISBN 10: 0132064529
ISBN 13: 978-0-13206-452-1

Chapter 4 - Chemical Reactions - Example 4-8 - Calculating Molarity from Measured Quantities - Page 124: Practice Example B


0.524 M

Work Step by Step

Moles of acetic acid= $15.0\,mL\, acetic\,acid\times\frac{1.048\,g\,acetic\,acid}{1\,mL\, acetic\,acid}\times\frac{1\,mol\, acetic\,acid}{60.052\,g\, acetic\,acid}$ $=0.261773\,mol$ Molarity= $\frac{\text{Moles solute}}{\text{Litres solution}}=\frac{0.261773\,mol}{500.0\times10^{-3}\,L}$ $=0.524\,M$
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