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 solve the problem, we will follow these steps:

### Step 1: Understand the dissociation of the weak electrolyte
The weak electrolyte AB dissociates into A⁺ and B⁻ ions. Given that it is 5% dissociated, we can express the degree of dissociation (α) as:
\[ \alpha = 0.05 \]

### Step 2: Calculate the Van't Hoff factor (i)
The Van't Hoff factor (i) is given by the formula:
\[ i = 1 + \alpha \]
Substituting the value of α:
\[ i = 1 + 0.05 = 1.05 \]

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

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

### Step 4: Calculate the freezing point of the solution
The freezing point of the solution (Tf) can be found by subtracting ΔTf from the freezing point of pure water (0 °C):
\[ T_f = 0 °C - 0.1953 °C = -0.1953 °C \]

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

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