1 molal aqueous solution of an electrolyte `AB_(2)` is `80%` ionized into `A^(+2)` and `B^(-1)` ions . The boiling point of the solution at 1 atm is :
`[K_(b) (H_(2)O) = 0.5` kg `mol^(-1)`]

Options

A. `274.3K`
B. `373.5K`
C. `374.3 K`
D. `100.5K`

Answer

To find the boiling point of a 1 molal aqueous solution of the electrolyte \( AB_2 \) that is 80% ionized, we will use the formula for boiling point elevation, which is given by:

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

where:
- \( \Delta T_b \) is the boiling point elevation,
- \( i \) is the van 't Hoff factor (number of particles the solute breaks into),
- \( K_b \) is the ebullioscopic constant of the solvent (for water, \( K_b = 0.5 \, \text{kg mol}^{-1} \)),
- \( m \) is the molality of the solution.

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

The electrolyte \( AB_2 \) dissociates into \( A^{2+} \) and \( 2B^{-} \). Therefore, the dissociation can be represented as:

\[ AB_2 \rightarrow A^{2+} + 2B^{-} \]

If 1 mole of \( AB_2 \) dissociates completely, it produces \( 1 + 2 = 3 \) moles of ions. However, since the solution is 80% ionized, we need to calculate the effective number of particles in the solution.

\[ \text{Effective number of moles of ions} = 0.8 \times 3 + 0.2 \times 1 = 2.4 + 0.2 = 2.6 \]

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

\[ i = 2.6 \]

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

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

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

Substituting the known values:

\[ \Delta T_b = 2.6 \cdot 0.5 \cdot 1 = 1.3 \, \text{°C} \]

### Step 3: Calculate the new boiling point

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

\[ \text{Boiling point of the solution} = 100 \, \text{°C} + \Delta T_b = 100 \, \text{°C} + 1.3 \, \text{°C} = 101.3 \, \text{°C} \]

### Final Answer

The boiling point of the solution at 1 atm is \( 101.3 \, \text{°C} \).

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