A weak electrolyte, AB, is 5% dissociated in aqueous solution. What is the freezing point of a 0.100 molal aqueous solution of AB? `K_f` for water is 1.86 deg/molal.

Options

A. `-3.8^@C`
B. `-0.1953^@C`
C. `-1.7^@C`
D. `-0.78^@C`

Explanation

To find the freezing point of a 0.100 molal aqueous solution of the weak electrolyte AB that is 5% dissociated, we can follow these steps:

### Step 1: Determine the degree of dissociation (α)
Given that AB is 5% dissociated, we can express this as:
\[ \alpha = \frac{5}{100} = 0.05 \]

### Step 2: Calculate the van 't Hoff factor (i)
The van 't Hoff factor (i) can be calculated using the formula:
\[ i = 1 + \alpha(n - 1) \]
For the electrolyte AB, which dissociates into two ions (A⁺ and B⁻), we have:
\[ n = 2 \]
Substituting the values:
\[ i = 1 + 0.05(2 - 1) = 1 + 0.05 = 1.05 \]

### Step 3: Calculate the depression in freezing point (ΔTf)
The depression in freezing point can be calculated using the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \( K_f \) for water = 1.86 °C/m
- \( m \) (molality) = 0.100 molal

Substituting the values:
\[ \Delta T_f = 1.05 \cdot 1.86 \cdot 0.100 \]
Calculating this gives:
\[ \Delta T_f = 1.05 \cdot 1.86 \cdot 0.100 = 0.1953 \, °C \]

### Step 4: Calculate the freezing point of the solution
The freezing point of the solution can be found using the formula:
\[ \text{Freezing point of solution} = \text{Freezing point of solvent} - \Delta T_f \]
The freezing point of pure water (solvent) is 0 °C. Therefore:
\[ \text{Freezing point of solution} = 0 - 0.1953 = -0.1953 \, °C \]

### Final Answer
The freezing point of the 0.100 molal aqueous solution of AB is approximately:
\[ \text{Freezing point} = -0.1953 \, °C \]

---