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 Cl⁻ ions in the final solution after mixing the two solutions, we can follow these steps:

### Step 1: Calculate the moles of Cl⁻ ions from MgCl₂

1. **Identify the concentration and volume of MgCl₂**:
- Concentration (C) = 0.20 M
- Volume (V) = 15 mL = 0.015 L (convert mL to L by dividing by 1000)

2. **Calculate moles of MgCl₂**:
\[ \text{Moles of MgCl₂} = C \times V = 0.20 \, \text{mol/L} \times 0.015 \, \text{L} = 0.003 \, \text{mol} \]

3. **Determine moles of Cl⁻ ions from MgCl₂**:
- Each mole of MgCl₂ produces 2 moles of Cl⁻ ions.
\[ \text{Moles of Cl⁻ from MgCl₂} = 0.003 \, \text{mol} \times 2 = 0.006 \, \text{mol} \]

### Step 2: Calculate the moles of Cl⁻ ions from AlCl₃

1. **Identify the concentration and volume of AlCl₃**:
- Concentration (C) = 0.40 M
- Volume (V) = 45 mL = 0.045 L (convert mL to L)

2. **Calculate moles of AlCl₃**:
\[ \text{Moles of AlCl₃} = C \times V = 0.40 \, \text{mol/L} \times 0.045 \, \text{L} = 0.018 \, \text{mol} \]

3. **Determine moles of Cl⁻ ions from AlCl₃**:
- Each mole of AlCl₃ produces 3 moles of Cl⁻ ions.
\[ \text{Moles of Cl⁻ from AlCl₃} = 0.018 \, \text{mol} \times 3 = 0.054 \, \text{mol} \]

### Step 3: Calculate total moles of Cl⁻ ions

1. **Add the moles of Cl⁻ ions from both sources**:
\[ \text{Total moles of Cl⁻} = \text{Moles from MgCl₂} + \text{Moles from AlCl₃} = 0.006 \, \text{mol} + 0.054 \, \text{mol} = 0.060 \, \text{mol} \]

### Step 4: Calculate the total volume of the final solution

1. **Add the volumes of both solutions**:
\[ \text{Total volume} = 15 \, \text{mL} + 45 \, \text{mL} = 60 \, \text{mL} = 0.060 \, \text{L} \]

### Step 5: Calculate the final molarity of Cl⁻ ions

1. **Use the formula for molarity (M)**:
\[ M = \frac{\text{Total moles of Cl⁻}}{\text{Total volume in L}} = \frac{0.060 \, \text{mol}}{0.060 \, \text{L}} = 1.00 \, \text{M} \]

### Final Answer:
The molarity of Cl⁻ ions in the final solution is **1.00 M**.

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