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 - Exercises - Molarity - Page 593: 23


$$6.5 \times 10^{-3} \space M \space CO_2$$

Work Step by Step

$ CO_2 $ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 44.01 g/mol $$\frac{280 \space mg \space CO_2}{1 \space kg \space solution} \times \frac{1 \space g}{1000 \space mg} \times \frac{1 \space mol \space CO_2}{44.01 \space g \space CO_2} \times \frac{1027 \space kg \space solution}{1000 \space L \space solution} =6.5 \times 10^{-3} \space M \space CO_2$$
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