`1.0` molal aqueous solution of an electrolyte `X_(3)Y_(2)` is `25%` ionized. The boiling point of the solution is `(K_(b)` for `H_(2)O = 0.52 K kg//mol)`:

Options

A. `375.5 K`
B. `374.04K`
C. `377.12K`
D. `373.25K`

Explanation

To solve the problem, we need to find the boiling point elevation of a `1.0` molal aqueous solution of the electrolyte `X₃Y₂`, which is `25%` ionized. We will use the formula for boiling point elevation:

\[ \Delta T_b = K_b \cdot m \cdot i \]

where:
- \(\Delta T_b\) = elevation in boiling point
- \(K_b\) = ebullioscopic constant of the solvent (for water, \(K_b = 0.52 \, \text{K kg/mol}\))
- \(m\) = molality of the solution (1.0 molal)
- \(i\) = van 't Hoff factor, which accounts for the number of particles the solute dissociates into.

### Step 1: Calculate the van 't Hoff factor (i)

The electrolyte `X₃Y₂` dissociates into ions as follows:
- \(X₃Y₂ \rightarrow 3X^{2+} + 2Y^{3-}\)

From this dissociation, we can see that:
- For every mole of `X₃Y₂`, we get a total of \(3 + 2 = 5\) ions.

Since the solution is `25%` ionized, we can calculate the effective number of ions (i) using the degree of ionization (\(\alpha\)):
\[ \alpha = 0.25 \]
\[ i = 1 + (n - 1) \cdot \alpha \]
where \(n\) is the total number of ions produced from one formula unit of the solute. Here, \(n = 5\):
\[ i = 1 + (5 - 1) \cdot 0.25 = 1 + 4 \cdot 0.25 = 1 + 1 = 2 \]

### Step 2: Calculate the elevation in boiling point (\(\Delta T_b\))

Now we can substitute the values into the boiling point elevation formula:
\[ \Delta T_b = K_b \cdot m \cdot i \]
Substituting the known values:
\[ \Delta T_b = 0.52 \, \text{K kg/mol} \cdot 1.0 \, \text{mol/kg} \cdot 2 = 1.04 \, \text{K} \]

### Step 3: Calculate the new boiling point

The boiling point of pure water is \(373 \, \text{K}\). Therefore, the new boiling point (\(T_b\)) is:
\[ T_b = T_{b,\text{water}} + \Delta T_b = 373 \, \text{K} + 1.04 \, \text{K} = 374.04 \, \text{K} \]

### Final Answer

The boiling point of the solution is \(374.04 \, \text{K}\).

---