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

Options

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

Explanation

To calculate the van't Hoff factor (i) for Na₂SO₄ based on the given information, we will follow these steps:

### Step 1: Understand the given data
- Freezing point depression constant for water (Kf) = 1.86 °C kg/mol
- Mass of Na₂SO₄ (solute) = 5.00 g
- Mass of H₂O (solvent) = 45.0 g
- Change in freezing point (ΔTf) = -3.82 °C

### Step 2: Calculate the change in freezing point (ΔTf)
The change in freezing point is given as:
\[ \Delta T_f = 3.82 \, °C \]

### Step 3: Use the formula for freezing point depression
The formula for freezing point depression is:
\[ \Delta T_f = K_f \cdot m \cdot i \]
Where:
- m = molality of the solution
- i = van't Hoff factor

### Step 4: Calculate the molality (m)
First, we need to calculate the number of moles of Na₂SO₄:
- Molar mass of Na₂SO₄ = 2(23) + 32 + 4(16) = 142 g/mol
- Moles of Na₂SO₄ = \(\frac{\text{mass}}{\text{molar mass}} = \frac{5.00 \, g}{142 \, g/mol} \approx 0.0352 \, mol\)

Next, convert the mass of the solvent (water) from grams to kilograms:
- Mass of H₂O = 45.0 g = 0.045 kg

Now, calculate the molality (m):
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.0352 \, mol}{0.045 \, kg} \approx 0.7822 \, mol/kg \]

### Step 5: Substitute values into the freezing point depression formula
Now we can substitute the values into the ΔTf formula:
\[ 3.82 = 1.86 \cdot (0.7822) \cdot i \]

### Step 6: Solve for the van't Hoff factor (i)
Rearranging the equation to solve for i:
\[ i = \frac{3.82}{1.86 \cdot 0.7822} \]
Calculating the denominator:
\[ 1.86 \cdot 0.7822 \approx 1.454 \]
Now, substituting back:
\[ i = \frac{3.82}{1.454} \approx 2.63 \]

### Final Answer
The van't Hoff factor (i) for Na₂SO₄ is approximately **2.63**.

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