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 solve the problem of calculating the degree of dissociation of BaCl₂ in an aqueous solution, we can follow these steps:

### Step 1: Calculate the elevation in boiling point (ΔT_b)
The normal boiling point of 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:

\[ \Delta T_b = \text{Boiling point of solution} - \text{Normal boiling point of water} \]
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