If 50 ml of 0.1 M NaCl and 50 ml of 0.1 M `BaCl_(2)` are mixed, molarity of chloride ion in the resulting solution will be

Options

A. 0.2 M
B. 0.3 M
C. 0.15 M
D. 0.1 M

Explanation

To find the molarity of chloride ions in the resulting solution when 50 ml of 0.1 M NaCl and 50 ml of 0.1 M BaCl₂ are mixed, we can follow these steps:

### Step 1: Calculate the moles of NaCl and BaCl₂
- **NaCl**:
- Molarity (M) = 0.1 M
- Volume (V) = 50 ml = 0.050 L
- Moles of NaCl = Molarity × Volume = 0.1 mol/L × 0.050 L = 0.005 moles

- **BaCl₂**:
- Molarity (M) = 0.1 M
- Volume (V) = 50 ml = 0.050 L
- Moles of BaCl₂ = Molarity × Volume = 0.1 mol/L × 0.050 L = 0.005 moles

### Step 2: Determine the dissociation of each salt
- **NaCl** dissociates completely into:
- Na⁺ + Cl⁻
- Therefore, from 0.005 moles of NaCl, we get 0.005 moles of Cl⁻.

- **BaCl₂** dissociates completely into:
- Ba²⁺ + 2Cl⁻
- Therefore, from 0.005 moles of BaCl₂, we get 0.005 moles × 2 = 0.010 moles of Cl⁻.

### Step 3: Calculate the total moles of Cl⁻
- Total moles of Cl⁻ = moles from NaCl + moles from BaCl₂
- Total moles of Cl⁻ = 0.005 moles (from NaCl) + 0.010 moles (from BaCl₂) = 0.015 moles of Cl⁻.

### Step 4: Calculate the total volume of the solution
- Total volume = Volume of NaCl solution + Volume of BaCl₂ solution
- Total volume = 50 ml + 50 ml = 100 ml = 0.100 L.

### Step 5: Calculate the molarity of Cl⁻ in the resulting solution
- Molarity (M) = Total moles of solute / Total volume of solution in liters
- Molarity of Cl⁻ = 0.015 moles / 0.100 L = 0.15 M.

### Final Answer
The molarity of chloride ions (Cl⁻) in the resulting solution is **0.15 M**.

---