The freezing point depression of 0.1 molal solution of acetic acid in benzene is 0.256 K , `K_(f)` for benzene is 5.12 K Kg `mol^(-1)`. What conclusion can you draw about the molecular state of acetic acid in benzence ?

Options

A. Acetic acid is doubly associated
B. Benzene is doubly associated
C. Both are equally associated
D. None of the above.

Answer

To solve the problem, we need to determine the molecular state of acetic acid in benzene based on the provided data. Here’s a step-by-step solution:

### Step 1: Understand the Given Data
We have the following information:
- Molality (m) of acetic acid solution = 0.1 mol/kg
- Freezing point depression (ΔTf) = 0.256 K
- Freezing point depression constant (Kf) for benzene = 5.12 K kg/mol

### Step 2: Use the Freezing Point Depression Formula
The formula for freezing point depression is given by:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \( \Delta T_f \) is the freezing point depression,
- \( i \) is the van 't Hoff factor,
- \( K_f \) is the freezing point depression constant,
- \( m \) is the molality of the solution.

### Step 3: Rearrange the Formula to Solve for \( i \)
We can rearrange the formula to find the van 't Hoff factor \( i \):
\[ i = \frac{\Delta T_f}{K_f \cdot m} \]

### Step 4: Substitute the Values into the Formula
Now, substitute the values into the equation:
\[ i = \frac{0.256}{5.12 \cdot 0.1} \]

### Step 5: Calculate \( i \)
Now, calculate \( i \):
\[ i = \frac{0.256}{0.512} = 0.5 \]

### Step 6: Analyze the Result
The calculated value of \( i \) is 0.5. The van 't Hoff factor \( i \) indicates the number of particles the solute breaks into when dissolved.

### Step 7: Conclusion About Molecular State
Since \( i < 1 \), this suggests that acetic acid tends to associate in the solution, likely forming dimer molecules. In this case, acetic acid molecules are likely forming pairs (dimers) in benzene.

### Final Conclusion
Thus, the conclusion we can draw is that acetic acid in benzene is doubly associated, which means it forms dimer molecules.