Calculate molarity of all the ions produces by following aq. Solution
of electrolytes. (Assuming 100% dissociation)
(i) 0.2 M NaCl solution
(ii) 1.2 M `Al_(2)(SO_(4))_3` solution
(iii) 1.2 M `H_(2)SO_(4)` solution
(vi) 2.1 M `MgCl_(2)` solution

Answer

To calculate the molarity of all the ions produced by the given aqueous solutions of electrolytes, we will follow these steps for each solution:

### Step 1: Analyze the dissociation of each electrolyte

1. **For NaCl (0.2 M)**:
- NaCl dissociates into Na⁺ and Cl⁻.
- The dissociation can be represented as:
\[ \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- \]
- Each mole of NaCl produces 1 mole of Na⁺ and 1 mole of Cl⁻.

2. **For Al₂(SO₄)₃ (1.2 M)**:
- Al₂(SO₄)₃ dissociates into 2 Al³⁺ and 3 SO₄²⁻.
- The dissociation can be represented as:
\[ \text{Al}_2(\text{SO}_4)_3 \rightarrow 2 \text{Al}^{3+} + 3 \text{SO}_4^{2-} \]
- Each mole of Al₂(SO₄)₃ produces 2 moles of Al³⁺ and 3 moles of SO₄²⁻.

3. **For H₂SO₄ (1.2 M)**:
- H₂SO₄ dissociates into 2 H⁺ and 1 SO₄²⁻.
- The dissociation can be represented as:
\[ \text{H}_2\text{SO}_4 \rightarrow 2 \text{H}^+ + \text{SO}_4^{2-} \]
- Each mole of H₂SO₄ produces 2 moles of H⁺ and 1 mole of SO₄²⁻.

4. **For MgCl₂ (2.1 M)**:
- MgCl₂ dissociates into Mg²⁺ and 2 Cl⁻.
- The dissociation can be represented as:
\[ \text{MgCl}_2 \rightarrow \text{Mg}^{2+} + 2 \text{Cl}^- \]
- Each mole of MgCl₂ produces 1 mole of Mg²⁺ and 2 moles of Cl⁻.

### Step 2: Calculate the molarity of each ion produced

1. **For NaCl (0.2 M)**:
- Na⁺: 0.2 M
- Cl⁻: 0.2 M

2. **For Al₂(SO₄)₃ (1.2 M)**:
- Al³⁺: \(2 \times 1.2 = 2.4\) M
- SO₄²⁻: \(3 \times 1.2 = 3.6\) M

3. **For H₂SO₄ (1.2 M)**:
- H⁺: \(2 \times 1.2 = 2.4\) M
- SO₄²⁻: 1.2 M

4. **For MgCl₂ (2.1 M)**:
- Mg²⁺: 2.1 M
- Cl⁻: \(2 \times 2.1 = 4.2\) M

### Final Results

- **NaCl (0.2 M)**:
- Na⁺: 0.2 M
- Cl⁻: 0.2 M

- **Al₂(SO₄)₃ (1.2 M)**:
- Al³⁺: 2.4 M
- SO₄²⁻: 3.6 M

- **H₂SO₄ (1.2 M)**:
- H⁺: 2.4 M
- SO₄²⁻: 1.2 M

- **MgCl₂ (2.1 M)**:
- Mg²⁺: 2.1 M
- Cl⁻: 4.2 M