Choose the pairs having identical value of van't Hoff factor

Options

A. 0.05 M `K_(4)[Fe(CN_(6)]` (50% degree of dissociation )and 0.05 M Mohar salt (80% degree of dissociation )
B. 0.2 M NaCl (80% degree of digree of dissociation ) and `0.2 M BaCl_(2)` (40% degree of dissociation )
C. 0.05 M `NaPO_(4)` (60% degree of dissocition) and 0.05 M `K_(4)[Fe(CN)_(6)]` (45% degree of dissociation)
D. 0.01 M `NaNO_(3)` (90% degree of dissociation) and 0.01 M FeCl (30% degree of dissociation

Answer

To solve the problem of identifying pairs with identical values of van't Hoff factor (i), we will follow these steps:

### Step 1: Understand the van't Hoff factor (i)
The van't Hoff factor (i) is defined as the ratio of the number of solute particles in a solution to the number of formula units of solute dissolved. For ionic compounds that dissociate in solution, the van't Hoff factor can be calculated using the formula:
\[ i = 1 + (n - 1) \cdot \alpha \]
where:
- \( n \) = number of particles formed upon dissociation
- \( \alpha \) = degree of dissociation (expressed as a fraction)

### Step 2: Calculate the van't Hoff factor for each pair

1. **For K₄[Fe(CN)₆]**:
- Dissociation: K₄[Fe(CN)₆] → 4K⁺ + [Fe(CN)₆]⁴⁻
- Number of particles (n) = 5 (4 K⁺ + 1 [Fe(CN)₆]⁴⁻)
- Degree of dissociation (α) = 0.5
- Calculation:
\[ i = 1 + (5 - 1) \cdot 0.5 = 1 + 4 \cdot 0.5 = 1 + 2 = 3 \]

2. **For Mohr's salt (NH₂NH₄)₂[Fe(SO₄)₂]·6H₂O**:
- Dissociation: (NH₂NH₄)₂[Fe(SO₄)₂] → 2NH₄⁺ + Fe²⁺ + 2SO₄²⁻
- Number of particles (n) = 5 (2 NH₄⁺ + 1 Fe²⁺ + 2 SO₄²⁻)
- Degree of dissociation (α) = 0.8
- Calculation:
\[ i = 1 + (5 - 1) \cdot 0.8 = 1 + 4 \cdot 0.8 = 1 + 3.2 = 4.2 \]

3. **For NaCl**:
- Dissociation: NaCl → Na⁺ + Cl⁻
- Number of particles (n) = 2 (1 Na⁺ + 1 Cl⁻)
- Degree of dissociation (α) = 0.8
- Calculation:
\[ i = 1 + (2 - 1) \cdot 0.8 = 1 + 1 \cdot 0.8 = 1 + 0.8 = 1.8 \]

4. **For CaCl₂**:
- Dissociation: CaCl₂ → Ca²⁺ + 2Cl⁻
- Number of particles (n) = 3 (1 Ca²⁺ + 2 Cl⁻)
- Degree of dissociation (α) = 0.4
- Calculation:
\[ i = 1 + (3 - 1) \cdot 0.4 = 1 + 2 \cdot 0.4 = 1 + 0.8 = 1.8 \]

5. **For Na₃PO₄**:
- Dissociation: Na₃PO₄ → 3Na⁺ + PO₄³⁻
- Number of particles (n) = 4 (3 Na⁺ + 1 PO₄³⁻)
- Degree of dissociation (α) = 0.6
- Calculation:
\[ i = 1 + (4 - 1) \cdot 0.6 = 1 + 3 \cdot 0.6 = 1 + 1.8 = 2.8 \]

6. **For NaNO₃**:
- Dissociation: NaNO₃ → Na⁺ + NO₃⁻
- Number of particles (n) = 2 (1 Na⁺ + 1 NO₃⁻)
- Degree of dissociation (α) = 0.9
- Calculation:
\[ i = 1 + (2 - 1) \cdot 0.9 = 1 + 1 \cdot 0.9 = 1 + 0.9 = 1.9 \]

7. **For FeCl₃**:
- Dissociation: FeCl₃ → Fe³⁺ + 3Cl⁻
- Number of particles (n) = 4 (1 Fe³⁺ + 3 Cl⁻)
- Degree of dissociation (α) = 0.3
- Calculation:
\[ i = 1 + (4 - 1) \cdot 0.3 = 1 + 3 \cdot 0.3 = 1 + 0.9 = 1.9 \]

### Step 3: Identify pairs with identical van't Hoff factors
- NaCl (i = 1.8) and CaCl₂ (i = 1.8)
- NaNO₃ (i = 1.9) and FeCl₃ (i = 1.9)

### Final Answer
The pairs having identical values of van't Hoff factor are:
- NaCl and CaCl₂
- NaNO₃ and FeCl₃