Determination of the molar mass of acetic acid in benzene using freezing point depression is affected by:

The freezing point depression method for determining molar mass relies on the formula:

$$\Delta T_f = K_f \cdot m \cdot i$$

where $$\Delta T_f$$ is the freezing point depression, $$K_f$$ is the cryoscopic constant, $$m$$ is the molality, and $$i$$ is the van't Hoff factor. The molality $$m$$ is defined as:

$$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{w_{\text{solute}} / M_{\text{solute}}}{w_{\text{solvent}} / 1000}$$

where $$w_{\text{solute}}$$ is the mass of solute, $$M_{\text{solute}}$$ is the molar mass of solute, and $$w_{\text{solvent}}$$ is the mass of solvent in grams. Rearranging for molar mass:

$$M_{\text{solute}} = \frac{K_f \cdot w_{\text{solute}} \cdot 1000 \cdot i}{\Delta T_f \cdot w_{\text{solvent}}}$$

In experiments, $$\Delta T_f$$, $$K_f$$, $$w_{\text{solute}}$$, and $$w_{\text{solvent}}$$ are measured. The van't Hoff factor $$i$$ is assumed to be 1 for non-dissociating and non-associating solutes. However, if $$i \neq 1$$, the calculated molar mass will be incorrect.

For acetic acid ($$CH_3COOH$$) in benzene:

• Acetic acid undergoes association in non-polar solvents like benzene due to hydrogen bonding, forming dimers: $$2CH_3COOH \rightarrow (CH_3COOH)_2$$.
• This reduces the number of particles. For complete dimerization, the van't Hoff factor $$i = \frac{1}{2}$$ because two molecules combine to form one particle.

If we assume $$i = 1$$, the calculated molar mass becomes:

$$M_{\text{calculated}} = \frac{K_f \cdot w_{\text{solute}} \cdot 1000 \cdot 1}{\Delta T_f \cdot w_{\text{solvent}}}$$

But the actual relationship is:

$$\Delta T_f = K_f \cdot m \cdot i = K_f \cdot \frac{w_{\text{solute}} / M_{\text{actual}}}{w_{\text{solvent}} / 1000} \cdot i$$

Rearranging for actual molar mass:

$$M_{\text{actual}} = \frac{K_f \cdot w_{\text{solute}} \cdot 1000 \cdot i}{\Delta T_f \cdot w_{\text{solvent}}}$$

Comparing:

$$M_{\text{calculated}} = \frac{K_f \cdot w_{\text{solute}} \cdot 1000}{\Delta T_f \cdot w_{\text{solvent}}} = \frac{M_{\text{actual}}}{i}$$

Since $$i = \frac{1}{2}$$ for dimerization:

$$M_{\text{calculated}} = \frac{M_{\text{actual}}}{1/2} = 2 \cdot M_{\text{actual}}$$

Thus, the calculated molar mass is twice the actual molar mass, leading to an overestimation due to association.

Evaluating other options:

• A. Complex formation: Acetic acid does not form significant complexes with benzene, so this is not the primary factor.
• C. Partial ionization: Ionization ($$CH_3COOH \rightleftharpoons CH_3COO^- + H^+$$) is negligible in non-polar benzene as ions are unstable, so $$i \approx 1$$ for ionization.
• D. Dissociation: Dissociation would increase $$i$$ above 1, but acetic acid associates in benzene, so dissociation is irrelevant.

Hence, the determination is affected by association.

Hence, the correct answer is Option B.