2.0 molal aqueous solution of an electrolyte `X_(2) Y_(3)` is 75% ionised. The boiling point of the solution a 1 atm is `( K_(b(H_(2)O)) = 0.52K` kg `mol^(-1)`)

Options

A. 2.74 .76 K
B. 377 K
C. 376.4K
D. 377.16 K

Answer

To solve the problem of finding the boiling point of a 2.0 molal aqueous solution of the electrolyte \( X_2Y_3 \) that is 75% ionized, we will follow these steps:

### Step 1: Identify the given data
- Molality (\( m \)) = 2.0 mol/kg
- Degree of ionization (\( \alpha \)) = 75% = 0.75
- Boiling point elevation constant (\( K_b \)) = 0.52 K kg/mol

### Step 2: Determine the Van't Hoff factor (\( i \))
The Van't Hoff factor \( i \) can be calculated using the formula:
\[ i = 1 + (n - 1) \alpha \]
where \( n \) is the number of particles the solute dissociates into.

For the electrolyte \( X_2Y_3 \):
- It dissociates into 2 \( X^{2+} \) ions and 3 \( Y^{3-} \) ions.
- Therefore, \( n = 2 + 3 = 5 \).

Now substituting \( n \) and \( \alpha \) into the formula:
\[ i = 1 + (5 - 1) \times 0.75 = 1 + 4 \times 0.75 = 1 + 3 = 4 \]

### Step 3: Calculate the boiling point elevation (\( \Delta T_b \))
The boiling point elevation can be calculated using the formula:
\[ \Delta T_b = i \cdot K_b \cdot m \]
Substituting the values we have:
\[ \Delta T_b = 4 \cdot 0.52 \cdot 2.0 \]
Calculating this gives:
\[ \Delta T_b = 4 \cdot 0.52 \cdot 2 = 4.16 \text{ K} \]

### Step 4: Calculate the new boiling point of the solution
The boiling point of pure water is 373 K. Therefore, the boiling point of the solution will be:
\[ T_b = T_{b,\text{water}} + \Delta T_b = 373 \text{ K} + 4.16 \text{ K} = 377.16 \text{ K} \]

### Final Answer
The boiling point of the solution is **377.16 K**.

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