Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 4 - Types of Chemical Reactions and Solution Stoichiometry - Exercises - Page 181: 38

Answer

$0.167 \text{ M}$

Work Step by Step

Moles of $HNO_3=0.05000\text{ L}*0.100\text{ M}+0.10000\text{ L}*0.200\text{ M}=0.0250\text{ moles}$ Volume of solution = 50.00 mL + 100.00 mL = 150.00 mL = 0.15000 L Molarity of solution = $\frac{0.0250\text{ moles}}{0.15000 \text{ L}}=0.167 \text{ M}$
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