A solution containing 200 ml 0.5 M KCl is mixed with 50 ml 19% w/v `MgCl_(2)` and resulting solution is dilute 8 times. Molarity of chloride ion is final solution is :

Answer

To find the molarity of chloride ions in the final solution after mixing KCl and MgCl₂ and diluting the solution, we can follow these steps:

### Step 1: Calculate the moles of chloride ions from KCl
1. **Given data**:
- Volume of KCl solution = 200 ml = 0.200 L
- Molarity of KCl = 0.5 M

2. **Calculate moles of KCl**:
\[ \text{Moles of KCl} = \text{Molarity} \times \text{Volume} = 0.5 \, \text{mol/L} \times 0.200 \, \text{L} = 0.1 \, \text{mol} \]

3. **Since KCl dissociates into K⁺ and Cl⁻**, the moles of chloride ions (Cl⁻) from KCl are also 0.1 mol.

### Step 2: Calculate the moles of chloride ions from MgCl₂
1. **Given data**:
- Volume of MgCl₂ solution = 50 ml = 0.050 L
- Concentration of MgCl₂ = 19% w/v

2. **Calculate the mass of MgCl₂ in 50 ml**:
- 19% w/v means 19 g of MgCl₂ in 100 ml of solution.
- For 50 ml:
\[ \text{Mass of MgCl₂} = \frac{19 \, \text{g}}{100 \, \text{ml}} \times 50 \, \text{ml} = 9.5 \, \text{g} \]

3. **Calculate moles of MgCl₂**:
- Molar mass of MgCl₂ = 95.211 g/mol
\[ \text{Moles of MgCl₂} = \frac{9.5 \, \text{g}}{95.211 \, \text{g/mol}} \approx 0.100 \, \text{mol} \]

4. **Calculate moles of chloride ions from MgCl₂**:
- Each mole of MgCl₂ produces 2 moles of Cl⁻.
\[ \text{Moles of Cl⁻ from MgCl₂} = 0.100 \, \text{mol} \times 2 = 0.200 \, \text{mol} \]

### Step 3: Total moles of chloride ions
1. **Total moles of chloride ions**:
\[ \text{Total moles of Cl⁻} = \text{Moles from KCl} + \text{Moles from MgCl₂} = 0.1 \, \text{mol} + 0.200 \, \text{mol} = 0.300 \, \text{mol} \]

### Step 4: Calculate the final volume after dilution
1. **Total volume before dilution**:
- Volume of KCl = 200 ml
- Volume of MgCl₂ = 50 ml
\[ \text{Total volume} = 200 \, \text{ml} + 50 \, \text{ml} = 250 \, \text{ml} \]

2. **After dilution by 8 times**:
\[ \text{Final volume} = 250 \, \text{ml} \times 8 = 2000 \, \text{ml} = 2.0 \, \text{L} \]

### Step 5: Calculate the molarity of chloride ions in the final solution
1. **Molarity of Cl⁻**:
\[ \text{Molarity} = \frac{\text{Total moles of Cl⁻}}{\text{Final volume in L}} = \frac{0.300 \, \text{mol}}{2.0 \, \text{L}} = 0.150 \, \text{M} \]

### Final Answer:
The molarity of chloride ions in the final solution is **0.150 M**.

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