A weak electrolyte, AB is 5% dissociated in aqueous solution. What is the freezing point of 0.1 111 aqueous solution of AB? `(K_f of water = 1.86 Km^-1)`

Options

A. -2.7°C
B. - 0.195°C
C. -0.534°C
D. 0.15°C

Answer

To solve the problem step by step, we will follow the concepts of freezing point depression and the dissociation of weak electrolytes.

### Step 1: Understand the dissociation of the electrolyte
Given that AB is a weak electrolyte and is 5% dissociated, we can denote the degree of dissociation (α) as:
\[ \alpha = \frac{5}{100} = 0.05 \]

### Step 2: Determine the van 't Hoff factor (i)
For a weak electrolyte that dissociates into two ions (A⁺ and B⁻), the van 't Hoff factor (i) can be calculated as:
\[ i = 1 + \alpha = 1 + 0.05 = 1.05 \]

### Step 3: Calculate the molality of the solution
The molality (m) of the solution is given as 0.1 mol/kg.

### Step 4: Use the freezing point depression formula
The formula for freezing point depression (ΔTf) is given by:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \( K_f \) is the freezing point depression constant of water, which is 1.86 °C kg/mol.
- \( m \) is the molality of the solution.

Substituting the known values:
\[ \Delta T_f = 1.05 \cdot 1.86 \cdot 0.1 \]

### Step 5: Calculate ΔTf
Calculating ΔTf:
\[ \Delta T_f = 1.05 \cdot 1.86 \cdot 0.1 = 0.1953 \, °C \]

### Step 6: Determine the freezing point of the solution
The freezing point of pure water is 0 °C. The freezing point of the solution (Tf) can be calculated as:
\[ T_f = T_{solvent} - \Delta T_f \]
\[ T_f = 0 - 0.1953 = -0.1953 \, °C \]

### Final Answer
The freezing point of the 0.1 molal aqueous solution of AB is approximately:
\[ T_f \approx -0.1953 \, °C \]

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