A weak electrolyte, AB is 5% dissociated in aqueous solution. What is the freezing point of 0.1 111 aqueous solution of AB? (K1 of water = 1.86 K m-1)

Options

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

Answer

To solve the problem, we need to determine the freezing point depression of a 0.1 molal solution of the weak electrolyte AB, which is 5% dissociated. We will use the formula for freezing point depression:

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

Where:
- \(\Delta T_f\) = freezing point depression
- \(i\) = van 't Hoff factor (degree of dissociation)
- \(K_f\) = freezing point depression constant of the solvent (water in this case)
- \(m\) = molality of the solution

### Step 1: Calculate the van 't Hoff factor (i)

Since AB is a weak electrolyte and is 5% dissociated, we can express the degree of dissociation (\(\alpha\)) as:

\[ \alpha = \frac{5}{100} = 0.05 \]

The van 't Hoff factor \(i\) for a dissociating electrolyte can be calculated as:

\[ i = 1 + \alpha \]

Substituting the value of \(\alpha\):

\[ i = 1 + 0.05 = 1.05 \]

### Step 2: Substitute values into the freezing point depression formula

We know:
- \(K_f = 1.86 \, \text{K kg}^{-1}\)
- \(m = 0.1 \, \text{molal}\)

Now we can calculate \(\Delta T_f\):

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

Substituting the known values:

\[ \Delta T_f = 1.05 \cdot 1.86 \cdot 0.1 \]

Calculating this:

\[ \Delta T_f = 1.05 \cdot 1.86 \cdot 0.1 = 0.1953 \, \text{K} \]

### Step 3: Calculate the freezing point of the solution

The freezing point of pure water is \(0 \, \text{°C}\). The freezing point of the solution can be calculated as:

\[ \text{Freezing point of solution} = \text{Freezing point of solvent} - \Delta T_f \]

Substituting the values:

\[ \text{Freezing point of solution} = 0 \, \text{°C} - 0.1953 \, \text{K} \]

Converting to Celsius:

\[ \text{Freezing point of solution} = -0.1953 \, \text{°C} \]

### Final Answer

Thus, the freezing point of the 0.1 molal aqueous solution of AB is approximately:

\[ \text{Freezing point of solution} \approx -0.1953 \, \text{°C} \]