# Solved: The mass percentage of chloride ion in a 25.00-mL ## Problem 112 Chapter 4

Chemistry: A Molecular Approach | 3rd Edition

• 2901 Step-by-step solutions solved by professors and subject experts
• Get 24/7 help from StudySoup virtual teaching assistants Chemistry: A Molecular Approach | 3rd Edition

4 5 0 238 Reviews
30
1
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

Step 3 of 3

##### ISBN: 9780321809247

Unlock Textbook Solution

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

×
Get Full Access to Chemistry: A Molecular Approach - 3rd Edition - Chapter 4 - Problem 112

Get Full Access to Chemistry: A Molecular Approach - 3rd Edition - Chapter 4 - Problem 112

I don't want to reset my password

Need help? Contact support

Need an Account? Is not associated with an account
We're here to help