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Solved: The mass percentage of chloride ion in a 25.00-mL

Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro ISBN: 9780321809247 1

Solution for problem 112 Chapter 4

Chemistry: A Molecular Approach | 3rd Edition

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Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro

Chemistry: A Molecular Approach | 3rd Edition

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Problem 112

The mass percentage of chloride ion in a 25.00-mL sample of seawater was determined by titrating the sample with silver nitrate, precipitating silver chloride. It took 42.58 ml of 0.2997 M silver nitrate solution to reach the equivalence point in the titration. What is the mass percentage of chloride ion in seawater if its density is 1.025 g/ml?

Step-by-Step Solution:
Step 1 of 3

Solution: Here, we are going to calculate the mass percentage of chloride ion in water with the help of data given. Step1: Given, volume of silver nitrate solution used = 42.58 mL = 10008L = 0.04258 L Molarity of silver nitrate solution = 0.2997 M Therefore, number of moles of silver nitrate = volume of solution in litres x molarity = 0.04258 L x 0.2997 M = 0.013 mol Step2: Now, the reaction between silver nitrate and chloride ion takes place in the following manner: - - Cl(aq) + AgNO (aq3-------> AgCl(s) + NO (aq) 3 From the above reaction, it is clear that 1 mol of AgNO reacts with 3 mol of chloride ion for complete precipitation. Therefore, 0.013 mol of AgNO will react 3ith 0.013 mol of chloride ion for complete precipitation....

Step 2 of 3

Chapter 4, Problem 112 is Solved
Step 3 of 3

Textbook: Chemistry: A Molecular Approach
Edition: 3
Author: Nivaldo J. Tro
ISBN: 9780321809247

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Solved: The mass percentage of chloride ion in a 25.00-mL

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