3.0 molal aqueous solution of an electrolyte `A_(2)B_(3)` is 50% ionised. The boilng point of the solution at 1 atm is: `[k_(b) (H_2O) = 0.52 K kg mol^(-1)]`

Options

A. 274.76 K
B. 377.68 K
C. 374.68 K
D. 104.68 K

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
We have a 3.0 molal aqueous solution of the electrolyte \( A_2B_3 \) that is 50% ionized. We need to find the boiling point of the solution.

### Step 2: Determine the degree of ionization
The degree of ionization \( \alpha \) is given as 50%, which can be expressed as:
\[ \alpha = \frac{50}{100} = 0.5 \]

### Step 3: Write the dissociation equation
The dissociation of the electrolyte \( A_2B_3 \) can be represented as:
\[ A_2B_3 \rightarrow 2A^{3+} + 3B^{2-} \]
From this equation, we can see that one formula unit of \( A_2B_3 \) produces 5 ions (2 \( A^{3+} \) ions and 3 \( B^{2-} \) ions).

### Step 4: Calculate the van 't Hoff factor \( i \)
The van 't Hoff factor \( i \) can be calculated using the formula:
\[ \alpha = \frac{i - 1}{n - 1} \]
Where \( n \) is the total number of ions produced per formula unit. Here, \( n = 5 \).

Substituting the known values:
\[ 0.5 = \frac{i - 1}{5 - 1} \]
\[ 0.5 = \frac{i - 1}{4} \]
Multiplying both sides by 4:
\[ 2 = i - 1 \]
Thus,
\[ i = 3 \]

### Step 5: Calculate the elevation in boiling point \( \Delta T_b \)
The formula for the elevation in boiling point is:
\[ \Delta T_b = i \cdot K_b \cdot m \]
Where:
- \( K_b = 0.52 \, \text{K kg mol}^{-1} \)
- \( m = 3.0 \, \text{molal} \)

Substituting the values:
\[ \Delta T_b = 3 \cdot 0.52 \cdot 3 \]
Calculating this:
\[ \Delta T_b = 3 \cdot 0.52 = 1.56 \]
\[ \Delta T_b = 1.56 \cdot 3 = 4.68 \, \text{K} \]

### Step 6: Calculate the boiling point of the solution
The boiling point of pure water is \( 100 \, \text{°C} \) or \( 373 \, \text{K} \). Therefore, the boiling point of the solution is:
\[ \text{Boiling Point} = 373 \, \text{K} + 4.68 \, \text{K} = 377.68 \, \text{K} \]

### Final Answer
The boiling point of the solution at 1 atm is:
\[ \boxed{377.68 \, \text{K}} \]