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

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

Chapter 13 - Solutions and Their Physical Properties - Example 13-1 - Expressing a Solution Concentration in Various Units - Page 561: Practice Example A


16.2 % of ethanol by mass

Work Step by Step

Assuming 100 mL of solution: $100 \space mL \space solution \times \frac{20.0 mL \space ethanol}{100 \space mL \space solution } = 20.0 \space mL \space ethanol$ $$20.0 \space mL \space ethanol \times \frac{0.789 \space g}{1 \space mL} = 15.8 \space g$$ $$100 \space mL \space solution \times \frac{0.977 \space g}{1 \space mL} = 97.7 \space g$$ Mass percent (ethanol) = $\frac{15.8 \space g \space ethanol}{97.7 \space g \space solution} \times 100\%= 16.2 \%$
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