A aqueous solution containing 1.248 g of barium chloride (molar mass `="208.34 g mol"^(-1)`) in 100 g of water boids at `100.0832^(@)C`. Calculate the degree of dissociation of `BaCl_(2)(K_(b)" for water = 0.52 K kg mol"^(-1))`.

Explanation

To calculate the degree of dissociation of barium chloride (BaCl₂) in an aqueous solution, we will follow these steps:

### Step 1: Calculate the boiling point elevation (ΔT_b)
The boiling point elevation is given by the formula:
\[ \Delta T_b = T_b - T_{b0} \]
where:
- \(T_b\) is the boiling point of the solution (100.0832 °C)
- \(T_{b0}\) is the boiling point of pure water (100 °C)

Calculating ΔT_b:
\[ \Delta T_b = 100.0832 °C - 100 °C = 0.0832 °C \]

### Step 2: Calculate the molality (m) of the solution
The formula for molality is:
\[ m = \frac{\text{mass of solute (g)}}{\text{molar mass of solute (g/mol)} \times \text{mass of solvent (kg)}} \]
Given:
- Mass of solute (BaCl₂) = 1.248 g
- Molar mass of BaCl₂ = 208.34 g/mol
- Mass of solvent (water) = 100 g = 0.1 kg

Calculating molality:
\[ m = \frac{1.248 \, \text{g}}{208.34 \, \text{g/mol} \times 0.1 \, \text{kg}} = \frac{1.248}{20.834} \approx 0.0599 \, \text{mol/kg} \]

### Step 3: Use the boiling point elevation formula
The boiling point elevation can also be expressed as:
\[ \Delta T_b = K_b \cdot m \cdot i \]
where:
- \(K_b\) for water = 0.52 °C kg/mol
- \(i\) is the van 't Hoff factor (number of particles the solute dissociates into)

Rearranging for \(i\):
\[ i = \frac{\Delta T_b}{K_b \cdot m} \]

Substituting the values:
\[ i = \frac{0.0832}{0.52 \cdot 0.0599} \approx \frac{0.0832}{0.0311} \approx 2.67 \]

### Step 4: Determine the degree of dissociation (α)
For barium chloride (BaCl₂), it dissociates into 3 ions:
\[ \text{BaCl}_2 \rightarrow \text{Ba}^{2+} + 2\text{Cl}^- \]
Thus, \(n = 3\) (total ions formed).

The van 't Hoff factor \(i\) can also be expressed as:
\[ i = 1 + \alpha(n - 1) \]
Substituting the values:
\[ 2.67 = 1 + \alpha(3 - 1) \]
\[ 2.67 = 1 + 2\alpha \]
\[ 2\alpha = 2.67 - 1 \]
\[ 2\alpha = 1.67 \]
\[ \alpha = \frac{1.67}{2} \approx 0.835 \]

### Step 5: Convert α to percentage
To express the degree of dissociation as a percentage:
\[ \text{Degree of dissociation} = \alpha \times 100 = 0.835 \times 100 \approx 83.5\% \]

### Final Answer
The degree of dissociation of barium chloride (BaCl₂) in the solution is approximately **83.5%**.

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