Question

When 0.0106 mole of acetic acid was dissolved in 1 kg of water, the freezing point depression for this strength of acid was 0.0205 K. If the calculated freezing point depression is 0.0197 K, Van't Hoff factor (i) and degree of dissociation of acetic acid respectively are

• A. 0.041 and 1.041
• B. 1.041 and 0.1041
• C. 0.041 and 0.041
• D. 1.041 and 0.041

Hint:

Remember the two key formulas for Van't Hoff factor problems: 1. Definition: \( i = \frac{\text{Observed Value}}{\text{Calculated Value}} \) 2. Relation with dissociation: \( i = 1 + (n-1)\alpha \) For association, the formula is \( i = 1 + (\frac{1}{n}-1)\alpha \). Be sure to correctly identify n (number of particles formed per formula unit).

Answer 1

The Correct Option is D

Step 1: Calculate the Van't Hoff factor (i).
The Van't Hoff factor (i) is the ratio of the observed colligative property to the calculated (theoretical) colligative property for a non-electrolyte. \[ i = \frac{\text{Observed colligative property}}{\text{Calculated colligative property}} \] In this case, the colligative property is the depression in freezing point (\(\Delta\)T\(_f\)).

• Observed \(\Delta\)T\(_f\) = 0.0205 K
• Calculated \(\Delta\)T\(_f\) (assuming no dissociation, i=1) = 0.0197 K

\[ i = \frac{0.0205 \text{ K}}{0.0197 \text{ K}} \approx 1.0406 \approx 1.041 \] Step 2: Relate the Van't Hoff factor (i) to the degree of dissociation (\(\alpha\)).
Acetic acid (CH\(_3\)COOH) is a weak electrolyte that partially dissociates in water: \[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \] The relationship between i, \(\alpha\), and the number of ions produced (n) is given by: \[ i = 1 + (n-1)\alpha \] For acetic acid, it dissociates into two ions (CH\(_3\)COO\(^-\) and H\(^+\)), so n = 2. \[ i = 1 + (2-1)\alpha = 1 + \alpha \] Step 3: Calculate the degree of dissociation (\(\alpha\)).
Rearranging the formula: \[ \alpha = i - 1 \] Substitute the value of i we calculated: \[ \alpha = 1.041 - 1 = 0.041 \] Step 4: Final Answer.
The Van't Hoff factor (i) is 1.041, and the degree of dissociation (\(\alpha\)) is 0.041.