#### 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 $AgNO_{3}$
$=0.300\,mol\times 169.87\,g/mol=51.0\,g$

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Published by
Cengage Learning

ISBN 10:
0534420125

ISBN 13:
978-0-53442-012-3

51.0 g

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