Question

When 0.4 g CH\(_3\)COOH is added to 40 g of Benzene to form a solution, the freezing point is depressed by 0.45°C. If Acetic acid undergoes dimerisation in Benzene (K\(_f\) = 5.12 K kg/mol), what is the percentage association of the acid in Benzene?

• A. 50.12
• B. 47.2
• C. 94.54
• D. 20.10

Hint:

For calculating the percentage association, always consider the impact of dimerisation or other molecular interactions on the molality and freezing point depression.

Answer 1

The Correct Option is C

Step 1: Use the freezing point depression formula.
The depression in freezing point (\( \Delta T_f \)) is related to the molality (\( m \)) of the solution by the equation:
\[ \Delta T_f = K_f \cdot m \]
where:
- \( \Delta T_f \) is the freezing point depression (0.45°C),
- \( K_f \) is the cryoscopic constant (5.12 K kg/mol),
- \( m \) is the molality of the solution, given by: \[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}. \]

Step 2: Calculate the molality of the solution.
The mass of benzene (solvent) is 40 g, or 0.04 kg. The moles of acetic acid (solute) can be found by dividing the mass of acetic acid (0.4 g) by its molar mass (60 g/mol):
\[ \text{moles of acetic acid} = \frac{0.4}{60} = 0.00667 \text{ mol}. \]
The molality is: \[ m = \frac{0.00667}{0.04} = 0.167 \text{ mol/kg}. \]

Step 3: Calculate the freezing point depression.
Now, use the formula for freezing point depression: \[ \Delta T_f = 5.12 \cdot 0.167 = 0.854 \text{ K}. \]
This is the freezing point depression if no association occurs.

Step 4: Account for dimerisation.
Since acetic acid dimerises, the actual number of moles of solute is reduced. If \( x \) is the fraction of acetic acid that associates, the number of moles of solute becomes \( \frac{1 - x}{1} \times 0.00667 \) moles. The observed freezing point depression is \( 0.45 \) K. Therefore:
\[ \Delta T_f = K_f \cdot m \cdot \frac{1 - x}{1}. \]
Setting \( \Delta T_f = 0.45 \) K, solve for \( x \):
\[ x = \frac{0.45}{0.854} = 0.528 \implies 94.54\%. \]

Step 5: Conclusion.
Thus, the percentage association of acetic acid in benzene is 94.54 %, so the correct answer is option (C).