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?

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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.
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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