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 B


a) $ 0.927 \space of \space water (amount)$ b) $ 3.37 \space M \space CH_3OH$ c) $4.34 \space m \space CH_3OH$

Work Step by Step

Mass of methanol $11.3 \space mL \times \frac{0.793 \space g }{1 \space mL} = 8.96 \space g$ Amount of methanol: $8.96 \space g\times \frac{1 \space mole}{32.04 \space g} = 0.280 \space mole$ Mass of solution = $75.0 \space mL \times \frac{0.980 \space g}{1 \space mL} = 73.5 \space g$ Mass of water = 73.5 g - 8.96 g = 64.5 g of water. Amount of water = $64.5 \space g\times \frac{1 \space mole}{18.02 \space g} = 3.58 \space mole$ a) $\frac{3.58 \space mole}{0.280 \space mole + 3.58 \space mole} = 0.927$ b) $\frac{0.280 \space mole \space CH_3OH}{0.075 \space L \space solution} = 3.37 \space M$ c) $\frac{0.280 \space mole \space CH_3OH}{0.0645 \space kg \space water} = 4.34 \space m$
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