Answer
51.0 g
Work Step by Step
Volume in litres V= 0.300 L
Molarity M= 1.00 M
Number of moles n= MV= 1.00 M$\times$ 0.300 L=0.300 mol
Mass in grams = n$\times$ Molar mass of $AgNO_{3}$
$=0.300\,mol\times 169.87\,g/mol=51.0\,g$
Chemistry: Principles and Practice (3rd Edition)
by Reger, Daniel L.; Goode, Scott R.; Ball, David W.
Published by
Cengage Learning
ISBN 10:
0534420125
ISBN 13:
978-0-53442-012-3
Chapter 4 - Chemical Reactions in Solution - Questions and Exercises - Exercises - Page 170: 4.47
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