Determine the freezing point (nearest integer) of a 1 mole `lkg^(-1)` aqueous solution of a weak electrolyte that is `7.5%` dissociated into two ions (in `.^(@)C`) [Given `K_(f)` of water is `1.86^(@)C//m`].

Explanation

To determine the freezing point of a 1 mole/kg aqueous solution of a weak electrolyte that is 7.5% dissociated into two ions, we can follow these steps:

### Step 1: Understand the Formula for Depression in Freezing Point
The depression in freezing point (\( \Delta T_f \)) can be calculated using the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \( i \) = van 't Hoff factor (number of particles the solute dissociates into)
- \( K_f \) = molal depression constant (given as \( 1.86 \, ^\circ C/m \))
- \( m \) = molality of the solution (given as \( 1 \, mol/kg \))

### Step 2: Calculate the van 't Hoff Factor (\( i \))
The van 't Hoff factor can be calculated using the formula:
\[ i = 1 + \alpha(n - 1) \]
where:
- \( \alpha \) = degree of dissociation (given as \( 7.5\% \) or \( 0.075 \))
- \( n \) = number of ions the electrolyte dissociates into (given as \( 2 \))

Substituting the values:
\[ i = 1 + 0.075(2 - 1) = 1 + 0.075 = 1.075 \]

### Step 3: Substitute Values into the Depression Formula
Now, we can substitute the values into the depression formula:
\[ \Delta T_f = i \cdot K_f \cdot m = 1.075 \cdot 1.86 \cdot 1 \]
Calculating this gives:
\[ \Delta T_f = 1.075 \cdot 1.86 = 1.9995 \approx 2 \, ^\circ C \]

### Step 4: Calculate the Freezing Point of the Solution
The freezing point of the solution can be calculated using:
\[ \Delta T_f = T_f^{\text{pure}} - T_f^{\text{solution}} \]
where \( T_f^{\text{pure}} \) (freezing point of pure water) is \( 0 \, ^\circ C \).

Rearranging gives:
\[ T_f^{\text{solution}} = T_f^{\text{pure}} - \Delta T_f = 0 - 2 = -2 \, ^\circ C \]

### Final Answer
The freezing point of the solution is approximately:
\[ \boxed{-2 \, ^\circ C} \]