Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question
What concentration of Cl−Cl− results when 985 mL985 mL of 0.507 M LiCl0.507 M LiCl is mixed with 801 mL801 mL of 0.365 M MgCl2?
![**Question:**
What concentration of Cl⁻ results when 985 mL of 0.507 M LiCl is mixed with 801 mL of 0.365 M MgCl₂?
**Answer:**
\[ [\text{Cl}^-] = \text{__________ M} \]
**Explanation:**
To find the concentration of Cl⁻ ions in the final solution:
1. **Calculate the moles of Cl⁻ from LiCl:**
\[ \text{Moles of Cl⁻ from LiCl} = \text{Volume (L)} \times \text{Molarity (M)} = 0.985 \times 0.507 \]
2. **Calculate the moles of Cl⁻ from MgCl₂:**
MgCl₂ provides two moles of Cl⁻ per mole of MgCl₂.
\[ \text{Moles of Cl⁻ from MgCl₂} = \text{Volume (L)} \times \text{Molarity (M)} \times 2 = 0.801 \times 0.365 \times 2 \]
3. **Total moles of Cl⁻:**
Add the moles of Cl⁻ from LiCl and MgCl₂ calculated above.
4. **Calculate the final concentration of Cl⁻:**
Total volume of the solution = 985 mL + 801 mL = 1786 mL = 1.786 L
\[ [\text{Cl}^-] = \frac{\text{Total moles of Cl}^-}{\text{Total volume (L)}} \]
This formula will help determine the molarity of Cl⁻ ions in the final solution after mixing the two solutions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F840f535b-f075-49f5-9fe8-c5a30f6f8dd7%2F84c44b75-b32b-4e14-ada9-829fdf4a003f%2Fy8w3irc_processed.png&w=3840&q=75)
Transcribed Image Text:**Question:**
What concentration of Cl⁻ results when 985 mL of 0.507 M LiCl is mixed with 801 mL of 0.365 M MgCl₂?
**Answer:**
\[ [\text{Cl}^-] = \text{__________ M} \]
**Explanation:**
To find the concentration of Cl⁻ ions in the final solution:
1. **Calculate the moles of Cl⁻ from LiCl:**
\[ \text{Moles of Cl⁻ from LiCl} = \text{Volume (L)} \times \text{Molarity (M)} = 0.985 \times 0.507 \]
2. **Calculate the moles of Cl⁻ from MgCl₂:**
MgCl₂ provides two moles of Cl⁻ per mole of MgCl₂.
\[ \text{Moles of Cl⁻ from MgCl₂} = \text{Volume (L)} \times \text{Molarity (M)} \times 2 = 0.801 \times 0.365 \times 2 \]
3. **Total moles of Cl⁻:**
Add the moles of Cl⁻ from LiCl and MgCl₂ calculated above.
4. **Calculate the final concentration of Cl⁻:**
Total volume of the solution = 985 mL + 801 mL = 1786 mL = 1.786 L
\[ [\text{Cl}^-] = \frac{\text{Total moles of Cl}^-}{\text{Total volume (L)}} \]
This formula will help determine the molarity of Cl⁻ ions in the final solution after mixing the two solutions.
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