Question

In a freezing point experiment, aqueous acetic acid solution gave $\Delta T_f$ (observed) = 0.02 K. Calculated $\Delta T_f$ for the same solution was found to be 0.018 K.
What is the van't Hoff factor of acetic acid?

• A. $\dfrac{9}{10}$
• B. $\dfrac{1}{10}$
• C. $\dfrac{10}{9}$
• D. $\dfrac{10}{1}$

Hint:

Van’t Hoff factor $i$ accounts for the degree of dissociation or association in solution.

Answer 1

The Correct Option is C

The van't Hoff factor $i$ is calculated as: $$ i = \frac{\text{Observed } \Delta T_f}{\text{Calculated } \Delta T_f} = \frac{0.02}{0.018} = \frac{10}{9} $$

Answer 2

In a freezing point experiment, aqueous acetic acid solution gave ΔT_f (observed) = 0.02 K. Calculated ΔT_f for the same solution was found to be 0.018 K. What is the van't Hoff factor of acetic acid?

Step 1: Understand the concept of van't Hoff factor (i):
The van’t Hoff factor (i) is defined as the ratio of observed colligative property to the theoretical (calculated) colligative property assuming no association or dissociation:
\[ i = \frac{\Delta T_f \ (\text{observed})}{\Delta T_f \ (\text{calculated})} \]

Step 2: Plug in the values:
Given:
- Observed ΔT_f = 0.02 K
- Calculated ΔT_f = 0.018 K

\[ i = \frac{0.02}{0.018} = \frac{20}{18} = \frac{10}{9} \]

Step 3: Interpretation:
Since the value of i is slightly greater than 1, it indicates that acetic acid undergoes partial ionization or association is less likely. In water, acetic acid partially ionizes as:
CH₃COOH ⇌ CH₃COO⁻ + H⁺
This partial ionization increases the number of particles in solution, which raises the observed depression in freezing point compared to the calculated value.

Final Answer:
\[ \boxed{\dfrac{10}{9}} \]