1.0 molal aqueous solution of an electrolyte `"A"_(2)"B"_(3)` is 60% ionised. The boiling point of the solution at 1 atm is `("K"_("b"("H"_(2)"O"))=0.52 "K kg mol"^(-1)")`

Options

A. 274.76K
B. 377K
C. 376.4K
D. 374.76K

Explanation

To solve the problem, we need to determine the boiling point elevation of a 1.0 molal aqueous solution of the electrolyte \( A_2B_3 \) that is 60% ionized. We will use the formula for boiling point elevation and the provided data.

### Step-by-Step Solution:

**Step 1: Determine the van 't Hoff factor (i)**
The electrolyte \( A_2B_3 \) dissociates into ions as follows:
\[ A_2B_3 \rightarrow 2A^+ + 3B^{3-} \]
This means that one formula unit of \( A_2B_3 \) produces a total of 5 ions (2 from \( A \) and 3 from \( B \)). Therefore, the theoretical van 't Hoff factor \( i \) is 5.

Since the solution is 60% ionized, we can calculate the effective van 't Hoff factor \( i_{\text{effective}} \):
\[ i_{\text{effective}} = 1 + (i - 1) \cdot \alpha \]
where \( \alpha \) is the degree of ionization. Here, \( \alpha = 0.6 \) (60% ionization).

Substituting the values:
\[ i_{\text{effective}} = 1 + (5 - 1) \cdot 0.6 = 1 + 4 \cdot 0.6 = 1 + 2.4 = 3.4 \]

**Step 2: Calculate the boiling point elevation (\( \Delta T_b \))**
The formula for boiling point elevation is given by:
\[ \Delta T_b = i \cdot K_b \cdot m \]
where:
- \( K_b \) is the ebullioscopic constant of water, given as \( 0.52 \, \text{K kg mol}^{-1} \)
- \( m \) is the molality of the solution, which is \( 1.0 \, \text{mol/kg} \)

Now substituting the values:
\[ \Delta T_b = 3.4 \cdot 0.52 \cdot 1.0 = 1.768 \, \text{K} \]

**Step 3: Calculate the new boiling point**
The normal boiling point of water is \( 100 \, \text{°C} \) or \( 373 \, \text{K} \). The new boiling point will be:
\[ \text{New boiling point} = 373 \, \text{K} + 1.768 \, \text{K} = 374.768 \, \text{K} \]

### Final Answer:
The boiling point of the solution is approximately \( 374.77 \, \text{K} \).

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