An aqueous solution of a solute AB has b.p. of `101.08^(@)C( AB` is `100%` ionised at boiling point of the solution ) and freezes at `-1.80 ^(@)C`. Hence, `AB(K_(b)//K_(f)=0.3)`

Options

A. is `100%` ionised at the `f.p.` of the solution
B. behaves as non`-` electrolyte at the `f.p.` of the solution
C. forms dimer
D. none of the above

Explanation

To solve the problem, we need to analyze the boiling point elevation and freezing point depression of the solution containing the solute AB.

### Step 1: Calculate the boiling point elevation (ΔT_b)
The boiling point of the solution is given as 101.08 °C, while the boiling point of pure water is 100 °C. Thus, we can calculate the boiling point elevation (ΔT_b):

\[ \Delta T_b = T_{b, \text{solution}} - T_{b, \text{pure water}} = 101.08 °C - 100 °C = 1.08 °C \]

**Hint:** The boiling point elevation is the difference between the boiling point of the solution and the boiling point of the pure solvent.

### Step 2: Calculate the freezing point depression (ΔT_f)
The freezing point of the solution is given as -1.80 °C, while the freezing point of pure water is 0 °C. Thus, we can calculate the freezing point depression (ΔT_f):

\[ \Delta T_f = T_{f, \text{pure water}} - T_{f, \text{solution}} = 0 °C - (-1.80 °C) = 1.80 °C \]

**Hint:** The freezing point depression is the difference between the freezing point of the pure solvent and the freezing point of the solution.

### Step 3: Use the van 't Hoff factor (i)
Since AB is 100% ionized at the boiling point, it dissociates into two ions (A⁺ and B⁻). Therefore, the van 't Hoff factor (i) at the boiling point is:

\[ i = 2 \]

**Hint:** The van 't Hoff factor represents the number of particles the solute dissociates into in solution.

### Step 4: Set up the equations for boiling point elevation and freezing point depression
The equations for boiling point elevation and freezing point depression are given by:

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

Where:
- \( K_b \) is the ebullioscopic constant,
- \( K_f \) is the cryoscopic constant,
- \( m \) is the molality of the solution.

### Step 5: Relate the two equations
From the two equations, we can derive a relationship:

\[ \frac{\Delta T_b}{\Delta T_f} = \frac{i \cdot K_b \cdot m}{i \cdot K_f \cdot m} = \frac{K_b}{K_f} \]

Since \( K_b/K_f = 0.3 \) is given in the problem, we can substitute the values of ΔT_b and ΔT_f:

\[ \frac{1.08}{1.80} = 0.3 \]

### Step 6: Solve for the van 't Hoff factor (i)
Now, substituting the known values into the equation:

\[ \frac{1.08}{1.80} = \frac{0.3 \cdot 2}{i} \]

Cross-multiplying gives:

\[ 1.08 \cdot i = 0.3 \cdot 2 \cdot 1.80 \]

Calculating the right side:

\[ 1.08 \cdot i = 1.08 \]

Now, solving for \( i \):

\[ i = \frac{1.08}{1.08} = 1 \]

### Conclusion
The van 't Hoff factor (i) at the freezing point is 1, indicating that the solute behaves as a non-electrolyte at this temperature.

**Final Answer:** The correct option is B, indicating that AB behaves as a non-electrolyte at the freezing point of the solution.