An aqueous solution of `3.12g` of `BaCl_(2)` in `250g` of water is found to boil at `100.0832^(@)C`. Calculate the degree of dissociation of `BaCl_(2)`. Given that the value of `K_(b)(H_(2)O)=0.52 K//m`.

Explanation

To find the degree of dissociation of BaCl₂ in the given solution, we will follow these steps:

### Step 1: Calculate the elevation in boiling point (ΔT_b)
The boiling point of pure water is 100°C. The boiling point of the solution is given as 100.0832°C. Therefore, the elevation in boiling point (ΔT_b) can be calculated as:

\[ \Delta T_b = T_{b, \text{solution}} - T_{b, \text{pure water}} = 100.0832°C - 100°C = 0.0832°C \]
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