1.0 molal aqueous solution of an electrolyte `A_(2)B_3` is 70% ionized. The boiling point of the solution at 1 atm is: `(K_(b(H_2O))=0.5 "K kg mol"^(-1))`

Options

A. 274.76 K
B. 374.9 K
C. 376.4 K
D. 374.76K

Answer

To solve the problem, we need to find the boiling point of a 1.0 molal aqueous solution of the electrolyte \( A_2B_3 \) that is 70% ionized. We will use the formula for boiling point elevation, which is given by:

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

Where:
- \( \Delta T_b \) = boiling point elevation
- \( i \) = van 't Hoff factor (number of particles the solute breaks into)
- \( K_b \) = ebullioscopic constant of the solvent (water in this case)
- \( m \) = molality of the solution

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

The compound \( A_2B_3 \) dissociates into ions as follows:

\[ A_2B_3 \rightarrow 2A^+ + 3B^{3-} \]

From this dissociation, we can see that for each formula unit of \( A_2B_3 \), it produces a total of \( 2 + 3 = 5 \) ions.

Since the solution is 70% ionized, we can calculate the effective number of particles at equilibrium:

- Initial moles of \( A_2B_3 = 1 \) mol
- Moles dissociated = \( 0.7 \) mol (70% of 1 mol)
- Moles of \( A^+ \) produced = \( 2 \times 0.7 = 1.4 \) mol
- Moles of \( B^{3-} \) produced = \( 3 \times 0.7 = 2.1 \) mol

Total moles at equilibrium:

\[ \text{Total moles} = 1 - 0.7 + 1.4 + 2.1 = 3.8 \text{ mol} \]

Thus, the van 't Hoff factor \( i \) is:

\[ i = \frac{\text{Total moles at equilibrium}}{\text{Initial moles}} = \frac{3.8}{1} = 3.8 \]

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

Given:
- \( K_b = 0.5 \, \text{K kg mol}^{-1} \)
- \( m = 1.0 \, \text{molal} \)

Now, we can substitute the values into the boiling point elevation formula:

\[ \Delta T_b = i \cdot K_b \cdot m = 3.8 \cdot 0.5 \cdot 1.0 = 1.9 \, \text{K} \]

### Step 3: Calculate the new boiling point of the solution

The normal boiling point of water is \( 100 \, \text{°C} \) or \( 373 \, \text{K} \). Therefore, the boiling point of the solution is:

\[ T_{\text{solution}} = T_{\text{boiling point of water}} + \Delta T_b = 373 \, \text{K} + 1.9 \, \text{K} = 374.9 \, \text{K} \]

### Final Answer

The boiling point of the solution at 1 atm is:

\[ \boxed{374.9 \, \text{K}} \]