Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company
ISBN 10: 0078021529
ISBN 13: 978-0-07802-152-7

Chapter 4 - Questions and Problems - Page 179: 4.69

Answer

First, I would add 3.00 mL of that $4.00$ $M$ $HNO_3$ solution to an adequate recipient, and then I would add sufficient water to reach 60.0 mL of solution.

Work Step by Step

1. Find the necessary volume: $C_1 * V_1 = C_2 * V_2$ $ 0.200* 60.0= 4.00 * V_2$ $ 12.0 = 4.00 * V_2$ $V_2 = 3.00$ mL
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