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: Determine the dissociation of the electrolyte
The electrolyte \( A_2B_3 \) dissociates in water as follows:
\[ A_2B_3 \rightarrow 2A^+ + 3B^- \]
From this dissociation, we can see that for every 1 mole of \( A_2B_3 \), 5 moles of ions are produced (2 moles of \( A^+ \) and 3 moles of \( B^- \)).

### Step 2: Calculate the number of ions produced
The total number of ions produced from the dissociation is:
\[ n = 2 + 3 = 5 \]

### Step 3: Calculate the van 't Hoff factor (i)
The van 't Hoff factor \( i \) can be calculated using the formula:
\[ i = 1 + (n - 1) \cdot \alpha \]
where \( \alpha \) is the degree of ionization (given as 50%, or 0.5). Substituting the values:
\[ i = 1 + (5 - 1) \cdot 0.5 = 1 + 4 \cdot 0.5 = 1 + 2 = 3 \]

### Step 4: Calculate the elevation in boiling point (\( \Delta T_b \))
The elevation in boiling point is given by the formula:
\[ \Delta T_b = i \cdot m \cdot K_b \]
where:
- \( m \) is the molality of the solution (3.0 molal),
- \( K_b \) for water is given as 0.52 K kg mol\(^{-1}\).

Substituting the values:
\[ \Delta T_b = 3 \cdot 3.0 \cdot 0.52 = 4.68 \, \text{K} \]

### Step 5: Calculate the boiling point of the solution
The boiling point of the solution can be calculated using:
\[ T_b = T_{b0} + \Delta T_b \]
where \( T_{b0} \) (the boiling point of pure water) is 100°C or 373 K. Therefore:
\[ T_b = 373 + 4.68 = 377.68 \, \text{K} \]

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

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