Chemistry: The Molecular Nature of Matter and Change 7th Edition

Published by McGraw-Hill Education
ISBN 10: 007351117X
ISBN 13: 978-0-07351-117-7

Chapter 4 - Problems: 4.55

Answer

2.206% Cl-

Work Step by Step

Cl- (aq) + AgNO3 (aq) ---> AgCl (s) + NO3- (aq) moles AgNO3 = (0.2970 M) x (0.05363 L) = 0.01593 mol AgNO3 2 Finding the mass of Cl- in the seawater: (0.01593 mol AgNO3) x (1 mol Cl- / 1 mol AgNO3) x (35.45 g Cl- / 1 mol Cl-) = 0.5647 g Cl- 3 mass % = (grams Cl- / grams seawater) x 100% grams Cl- = 0.5647 g For the mass of seawater, d = m/V, so m = dV d = 1.024 g/mL ; V = 25.00 mL mass of seawater = dV = (1.024 g/mL) x (25.00 mL) = 25.60 g seawater 4 mass % = (0.5647 g Cl- / 25.60 g seawater) x 100% = 2.206% Cl-
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