3.100 g of `BaCl_(2)` in 250g of water boils at `100.83^(@)C`. Calculate the value of van't Hoff factor and molality of `BaCl_(2)` in this solution. (`K_(f)` for water = 0.52 `Km^(-1)`, molar mass of `BaCl_(2) = 208.3" g mol"^(-1)`).

Explanation

To solve the problem, we need to calculate the molality of BaCl₂ and the van't Hoff factor (i) for the solution. Here are the step-by-step calculations:

### Step 1: Calculate the moles of BaCl₂
To find the moles of BaCl₂, we use the formula:
\[ \text{Moles of solute} = \frac{\text{mass of solute (g)}}{\text{molar mass of solute (g/mol)}} \]
Given:
- Mass of BaCl₂ = 3.100 g
- Molar mass of BaCl₂ = 208.3 g/mol

Calculating the moles:
\[ \text{Moles of BaCl₂} = \frac{3.100 \, \text{g}}{208.3 \, \text{g/mol}} = 0.01487 \, \text{mol} \]

### Step 2: Convert the mass of water to kilograms
Since molality is defined as moles of solute per kilogram of solvent, we need to convert the mass of water from grams to kilograms.
\[ \text{Mass of water} = 250 \, \text{g} = 0.250 \, \text{kg} \]

### Step 3: Calculate the molality of the solution
Molality (m) is given by:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \]
Substituting the values:
\[ m = \frac{0.01487 \, \text{mol}}{0.250 \, \text{kg}} = 0.05948 \, \text{mol/kg} \approx 0.0595 \, \text{mol/kg} \]

### Step 4: Calculate the boiling point elevation (ΔT_b)
The observed boiling point of the solution is given as 100.83 °C. The normal boiling point of water is 100 °C. Therefore, the boiling point elevation is:
\[ \Delta T_b = 100.83 \, °C - 100 \, °C = 0.83 \, °C \]

### Step 5: Use the boiling point elevation formula to find van't Hoff factor (i)
The boiling point elevation is related to molality and the van't Hoff factor by the formula:
\[ \Delta T_b = K_b \cdot m \cdot i \]
Where:
- \( K_b \) for water = 0.52 °C kg/mol
- \( m \) is the molality we calculated

Rearranging the formula to solve for i:
\[ i = \frac{\Delta T_b}{K_b \cdot m} \]
Substituting the known values:
\[ i = \frac{0.83 \, °C}{0.52 \, °C \cdot \text{kg/mol} \cdot 0.0595 \, \text{mol/kg}} = \frac{0.83}{0.03094} \approx 26.8 \]

### Final Results
- **Molality of BaCl₂**: 0.0595 mol/kg
- **van't Hoff factor (i)**: 26.8