Chemistry: Molecular Approach (4th Edition)

Published by Pearson
ISBN 10: 0134112830
ISBN 13: 978-0-13411-283-1

Chapter 4 - Exercises - Page 188: 57


a. $$1.3 \space mol \space KCl$$ b. $$1.5 \space mol \space KCl$$ c. $$ 0.211 \space mol \space KCl$$

Work Step by Step

1. Knowing that "M" represents "mol/L": a. $$0.556 \space L \times \frac{2.3 \space mol \space KCl}{1 \space L} = 1.3 \space mol \space KCl$$ b. $$1.8 \space L \times \frac{0.85 \space mol \space KCl}{1 \space L} = 1.5 \space mol \space KCl$$ c. $$114 \space mL \times \frac{1 \space L}{1000 \space mL} \times \frac{1.85 \space mol \space KCl}{1 \space L} = 0.211 \space mol \space KCl$$
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