Chemistry: An Introduction to General, Organic, and Biological Chemistry (12th Edition)

Published by Prentice Hall
ISBN 10: 0321908449
ISBN 13: 978-0-32190-844-5

Chapter 9 - Section 9.4 - Concentrations of Solutions - Questions and Problems - Page 301: 9.37b

Answer

There are necessary 50. g of $NH_4Cl$ to prepare 1250 mL of that solution.

Work Step by Step

1. Identify the conversion factors: - Since the $NH_4Cl$ solution is 4.0% (m/v): 4.0 g (Solute) = 100 mL (Solution) So: $\frac{4.0 \space g \space (Solute)}{100 \space mL \space (Solution)}$ and $\frac{100 \space mL \space (Solution)}{4.0 \space g \space (Solute)}$ 2. Calculate the mass of solute needed to prepare 1250 mL of that solution: $1250 \space mL \space (Solution) \times \frac{4.0 \space g \space (Solute)}{100 \space mL \space (Solution)} = 50. \space g \space (Solute)$
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