The freezing point depression constant for water is `-1.86^(@)Cm^(-1).`if `5.00g Na_(2)SO_(4)`is dissolved in `45.0gH_(2)O`,the freezing point is changed by `-3.82^(@)C`,Calculate the van't Hoff factor for `Na_(2)SO_(4)`

Options

A. `0.381`
B. `2.63`
C. `2.05`
D. `3.11`

Explanation

To calculate the van't Hoff factor (i) for Na₂SO₄, we can use the formula for freezing point depression:

\[ \Delta T_f = i \cdot K_f \cdot m \]

Where:
- \(\Delta T_f\) = change in freezing point (in °C)
- \(K_f\) = freezing point depression constant (in °C kg/mol)
- \(m\) = molality of the solution (in mol/kg)
- \(i\) = van't Hoff factor

### Step 1: Calculate the molality (m) of the solution
First, we need to calculate the molality of the Na₂SO₄ solution.

1. **Calculate the number of moles of Na₂SO₄:**
- Molar mass of Na₂SO₄:
- Na: 22.99 g/mol × 2 = 45.98 g/mol
- S: 32.07 g/mol × 1 = 32.07 g/mol
- O: 16.00 g/mol × 4 = 64.00 g/mol
- Total = 45.98 + 32.07 + 64.00 = 142.05 g/mol

- Moles of Na₂SO₄:
\[ \text{Moles of Na₂SO₄} = \frac{5.00 \, \text{g}}{142.05 \, \text{g/mol}} \approx 0.0352 \, \text{mol} \]

2. **Calculate the mass of the solvent (water) in kg:**
\[ \text{Mass of water} = 45.0 \, \text{g} = 0.0450 \, \text{kg} \]

3. **Calculate the molality (m):**
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.0352 \, \text{mol}}{0.0450 \, \text{kg}} \approx 0.7822 \, \text{mol/kg} \]

### Step 2: Substitute values into the freezing point depression formula
Now we can substitute the values into the freezing point depression formula to find the van't Hoff factor (i).

1. **Given values:**
- \(\Delta T_f = -3.82 \, °C\)
- \(K_f = -1.86 \, °C \, kg/mol\)
- \(m \approx 0.7822 \, mol/kg\)

2. **Rearranging the formula to solve for i:**
\[ i = \frac{\Delta T_f}{K_f \cdot m} \]

3. **Substituting the values:**
\[ i = \frac{-3.82}{-1.86 \cdot 0.7822} \approx \frac{3.82}{1.453} \approx 2.62 \]

### Final Answer
The van't Hoff factor (i) for Na₂SO₄ is approximately **2.62**.

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