Q.2 g of benzoic acid ($C_6H_5COOH$) dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg mol$^{-1}$. What is the percentage association of acid if it forms dimer in solution?

Concept: Colligative properties (freezing-point depression) combined with association equilibrium. When benzoic acid dimerizes in benzene, the effective number of particles decreases, reducing the observed depression below the theoretical value.

Step 1: Calculate theoretical molality (assuming no association)

Molar mass of benzoic acid = $122 \, \text{g mol}^{-1}$

Step 2: Find observed molality from experimental data

Using $\Delta T_f = K_f \cdot m_{\text{observed}}$:

Step 3: Apply van't Hoff factor and association formula

For dimerization ($2A \rightleftharpoons A_2$), if $\alpha$ is the degree of association:

Percentage association = $0.992 \times 100 = 99.2\%$

Why colligative properties reveal molecular association

Freezing-point depression depends only on the number of solute particles, not their identity. When benzoic acid molecules associate into dimers through hydrogen bonding, the total particle count drops below what we'd expect from isolated molecules. The van't Hoff factor $i$ captures this deviation: $i < 1$ signals association, and by measuring how much smaller $i$ is, we can calculate the fraction of molecules that have paired up.

The key relationship is:

where $m$ is the molality calculated as if no association occurred.

Step-by-step solution

1. Calculate the theoretical molality (assuming no association)

The molar mass of benzoic acid $C_6H_5COOH$ is:

Moles of benzoic acid dissolved:

Molality (moles per kg of solvent):

2. Find the van't Hoff factor from observed depression

The observed freezing-point depression is $\Delta T_f = 1.62$ K. Rearranging the colligative property equation:

This matches the book's own route: the experimentally observed molar mass is $M_{obs} = \frac{4.9 \times 2 \times 1000}{1.62 \times 25} = 241.98 \text{ g mol}^{-1}$, so $i = \frac{122}{241.98} = 0.504$.

3. Relate the van't Hoff factor to the degree of association

When benzoic acid forms dimers:

Let $\alpha$ be the degree of association (fraction of molecules that dimerize). Starting with 1 mole:

• Monomers remaining: $1 - \alpha$
• Dimers formed: $\frac{\alpha}{2}$
• Total particles: $(1 - \alpha) + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}$

The van't Hoff factor is:

4. Solve for the degree of association

5. Convert to percentage

Percentage association:

Method: Van't Hoff Factor Approach for Association

This problem uses the Van't Hoff factor ($i$) to account for the association of benzoic acid into dimers in benzene.

Step-by-step solution

Step 1: Calculate the theoretical (expected) molality

If no association occurred, the molality would be:

• Molar mass of benzoic acid ($C_6H_5COOH$) = $7 \times 12 + 6 \times 1 + 2 \times 16 = 122$ g/mol
• Moles of acid = $\frac{2}{122} = 0.01639$ mol
• Mass of benzene = 25 g = $0.025$ kg

Theoretical molality:

Step 2: Calculate the observed (actual) molality from freezing point depression

Using $\Delta T_f = K_f \times m_{\text{observed}}$:

Step 3: Find the Van't Hoff factor

(This is the same $i$ the book obtains via the observed molar mass: $i = \frac{122}{241.98} = 0.504$.)

Step 4: Relate $i$ to degree of association ($\alpha$)

For dimerization: $2A \rightleftharpoons A_2$

If $\alpha$ fraction of acid associates into dimers:

• Moles of monomer left = $1 - \alpha$
• Moles of dimer formed = $\alpha/2$
• Total moles after association = $(1 - \alpha) + \alpha/2 = 1 - \alpha/2$

The Van't Hoff factor is:

Step 5: Solve for $\alpha$

Step 6: Express as percentage

Final Answer:

99.2% of benzoic acid molecules associate into dimers in benzene solution.

Here are the common mistakes students make on this exact problem, along with how to avoid each one.

1. Confusing Observed Molar Mass with Normal Molar Mass

The Mistake:

Students often plug the given mass (2 g) and mass of solvent (25 g) directly into the formula for molar mass without first calculating the observed molar mass from the freezing point depression.

How to Avoid:

Always separate the two steps:

1. Find observed molar mass ($M_{obs}$) using:

where $w_{solute} = 2 \text{ g}$, $w_{solvent} = 25 \text{ g}$, $K_f = 4.9 \text{ K kg mol}^{-1}$, $\Delta T_f = 1.62 \text{ K}$.

2. Find normal molar mass ($M_{normal}$) from the molecular formula:

Only then compare them.

2. Using the Wrong Formula for Association

The Mistake:

Students sometimes use the dissociation formula ($i = 1 + \alpha$) for association problems.

How to Avoid:

For dimer formation (2 molecules → 1 dimer), the van’t Hoff factor is:

where $\alpha$ is the fraction of molecules that associate.

• If all molecules dimerize ($\alpha = 1$), then $i = 0.5$.
• If none dimerize ($\alpha = 0$), then $i = 1$.

3. Forgetting the Relationship Between $i$ and Molar Mass

The Mistake:

Students calculate $i$ but then don’t connect it correctly to $\alpha$.

How to Avoid:

Remember:

Then set:

Solve for $\alpha$:

4. Calculation Errors in $M_{obs}$

The Mistake:

Mixing up units — especially forgetting to convert grams of solvent to kilograms.

How to Avoid:

Always write the formula with units:

• $K_f$ is in K kg mol⁻¹ → solvent mass must be in kg (hence ×1000).
• Double-check arithmetic: Numerator = $4.9 \times 2 \times 1000 = 9800$ Denominator = $1.62 \times 25 = 40.5$ $M_{obs} = \frac{9800}{40.5} \approx 241.98 \text{ g/mol}$

5. Interpreting the Percentage Incorrectly

The Mistake:

Stopping at $\alpha = 0.5$ and writing “50%” without checking if it’s the percentage of molecules associated.

How to Avoid:

• $\alpha$ is the fraction of molecules that associate.
• Percentage association = $\alpha \times 100\%$.
• In this problem:

Percentage association = 99.2% (nearly complete dimerization).

Quick Checklist to Avoid Mistakes

Final Answer: