General, Organic, and Biological Chemistry: Structures of Life (5th Edition)

Published by Pearson
ISBN 10: 0321967461
ISBN 13: 978-0-32196-746-6

Chapter 9 - Solutions - 9.4 Solution Concentrations and Reactions - Questions and Problems - Page 348: 9.47

Answer

a. 1.50 L of solution b. 10.0 L of solution c. 62.5 mL of solution

Work Step by Step

a. 1. Find the conversion factor: - 2.00 M $KBr$ solution: $$2.00 \space moles \space KBr = 1 \space L \space solution$$ 2. Calculate the volume in L: $$3.00 \space moles \space KBr \times \frac{1 \space L \space solution}{2.00 \space moles \space KBr} = 1.50 \space L \space of \space solution$$ b. 1. Find the conversion factors - 1.50 M $NaCl$ solution: $$1.50 \space moles \space NaCl = 1 \space L \space solution$$ 2. Calculate the volume in L: $$15.0 \space moles \space NaCl \times \frac{1 \space L \space solution}{1.50 \space moles \space NaCl} = 10.0 \space L \space of \space solution$$ c. 1. Find the conversion factors - 0.800 M $Ca(NO_3)_2$ solution: $$0.800 \space mole \space Ca(NO_3)_2 = 1 \space L \space solution$$ - 1 L = 1000 mL 2. Calculate the volume in mL: $$0.0500 \space mole \space Ca(NO_3)_2 \times \frac{1 \space L \space solution}{0.800 \space mole \space Ca(NO_3)_2} \times \frac{1000 \space mL}{1 \space L}= 62.5 \space mL \space of \space solution$$
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