An aqueous solution of an electrolyte AB has b.pt of `101.08^(@)C`. The solute is 100% ionised ay b.pt of water. The f.pt of the same solution is `-1.80^(@)C`. Hence `AB(K_(b)//K_(f) = 0.3)`

Options

A. is 100% ionised at f.pt of solution
B. is 50% ionised at f.pt of solution
C. behaves as a non-electrolyte at f.pt of solution
D. forms a dimer at f.pt of solution

Explanation

To solve the problem, we need to analyze the given information about the electrolyte AB and apply the concepts of colligative properties, specifically boiling point elevation and freezing point depression.

### Step-by-Step Solution:

1. **Understand the Given Data:**
- Boiling point of the solution (T_b) = 101.08°C
- Boiling point of pure water (T_b0) = 100°C
- Freezing point of the solution (T_f) = -1.80°C
- Freezing point of pure water (T_f0) = 0°C
- The electrolyte AB is 100% ionized.

2. **Calculate the Elevation in Boiling Point (ΔT_b):**
\[ \Delta T_b = T_b - T_b0 = 101.08°C - 100°C = 1.08°C \]

3. **Use the Boiling Point Elevation Formula:**
The formula for boiling point elevation is:
\[ \Delta T_b = i \cdot K_b \cdot m \]
where:
- \( i \) = van 't Hoff factor (number of particles the solute breaks into)
- \( K_b \) = ebullioscopic constant of the solvent (water)
- \( m \) = molality of the solution

4. **Calculate the Depression in Freezing Point (ΔT_f):**
\[ \Delta T_f = T_f0 - T_f = 0°C - (-1.80°C) = 1.80°C \]

5. **Use the Freezing Point Depression Formula:**
The formula for freezing point depression is:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \( K_f \) = cryoscopic constant of the solvent (water)

6. **Set Up the Equations:**
From boiling point elevation:
\[ 1.08 = i \cdot K_b \cdot m \quad (1) \]
From freezing point depression:
\[ 1.80 = i \cdot K_f \cdot m \quad (2) \]

7. **Divide Equation (1) by Equation (2):**
\[ \frac{1.08}{1.80} = \frac{i \cdot K_b \cdot m}{i \cdot K_f \cdot m} \]
The molality \( m \) cancels out:
\[ \frac{1.08}{1.80} = \frac{K_b}{K_f} \]

8. **Calculate the Value of \( \frac{K_b}{K_f} \):**
\[ \frac{1.08}{1.80} = 0.6 \]

9. **Given that \( \frac{K_b}{K_f} = 0.3 \):**
Since we have \( \frac{K_b}{K_f} = 0.3 \) from the problem statement, we can conclude that the value of \( i \) must be adjusted accordingly.

10. **Calculate \( i \):**
Rearranging the equation:
\[ i = \frac{1.08 \cdot K_f}{1.80 \cdot K_b} \]
Substituting \( K_b/K_f = 0.3 \):
\[ i = \frac{1.08}{1.80 \cdot 0.3} = \frac{1.08}{0.54} = 2 \]

11. **Conclusion:**
Since \( i = 2 \), it indicates that the electrolyte AB dissociates into two ions, confirming that it is indeed an electrolyte.