A `15mL` sample of `0.20 M" " MgCl_(2)` is added to `45ML` of `0.40 M AlCl_(3)` What is the molarity of `Cl` ions in the final solution

Options

A. `1.0M`
B. `0.60M`
C. `0.35M`
D. `0.30M`

Explanation

To find the molarity of chloride ions in the final solution after mixing the given volumes and concentrations of MgCl₂ and AlCl₃, we can follow these steps:

### Step 1: Calculate the number of moles of MgCl₂
The formula to calculate the number of moles is:
\[ \text{Number of moles} = \text{Volume (L)} \times \text{Molarity (M)} \]
Given:
- Volume of MgCl₂ = 15 mL = 0.015 L
- Molarity of MgCl₂ = 0.20 M

Calculating the moles:
\[ \text{Number of moles of MgCl₂} = 0.015 \, \text{L} \times 0.20 \, \text{M} = 0.003 \, \text{moles} \]

### Step 2: Calculate the number of moles of AlCl₃
Using the same formula:
Given:
- Volume of AlCl₃ = 45 mL = 0.045 L
- Molarity of AlCl₃ = 0.40 M

Calculating the moles:
\[ \text{Number of moles of AlCl₃} = 0.045 \, \text{L} \times 0.40 \, \text{M} = 0.018 \, \text{moles} \]

### Step 3: Determine the number of chloride ions from each compound
- Each formula unit of MgCl₂ produces 2 chloride ions (Cl⁻).
- Each formula unit of AlCl₃ produces 3 chloride ions (Cl⁻).

Calculating the total number of chloride ions:
\[ \text{Chloride ions from MgCl₂} = 2 \times 0.003 \, \text{moles} = 0.006 \, \text{moles} \]
\[ \text{Chloride ions from AlCl₃} = 3 \times 0.018 \, \text{moles} = 0.054 \, \text{moles} \]

### Step 4: Calculate the total number of moles of chloride ions
\[ \text{Total moles of Cl⁻} = 0.006 \, \text{moles} + 0.054 \, \text{moles} = 0.060 \, \text{moles} \]

### Step 5: Calculate the total volume of the solution
Total volume = Volume of MgCl₂ + Volume of AlCl₃
\[ \text{Total volume} = 15 \, \text{mL} + 45 \, \text{mL} = 60 \, \text{mL} = 0.060 \, \text{L} \]

### Step 6: Calculate the molarity of chloride ions in the final solution
Using the formula for molarity:
\[ \text{Molarity (M)} = \frac{\text{Number of moles}}{\text{Volume (L)}} \]
\[ \text{Molarity of Cl⁻} = \frac{0.060 \, \text{moles}}{0.060 \, \text{L}} = 1.0 \, \text{M} \]

### Final Answer
The molarity of chloride ions in the final solution is **1.0 M**.

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