Question:medium

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

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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).
Updated On: Apr 28, 2026
  • 0.041 and 1.041
  • 1.041 and 0.1041
  • 0.041 and 0.041
  • 1.041 and 0.041
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
The Van't Hoff factor ($i$) corrects colligative property equations for solutes that dissociate or associate. The degree of dissociation ($\alpha$) relates directly to $i$ for weak electrolytes.
Step 2: Key Formula or Approach:
1. Calculate $i$: $i = \frac{\Delta T_{f} (\text{observed})}{\Delta T_{f} (\text{calculated})}$

2. Relate $i$ to $\alpha$: For acetic acid dissociating into 2 ions ($CH_{3}COOH \rightleftharpoons CH_{3}COO^{-} + H^{+}$), $n=2$. The formula is $i = 1 + \alpha(n - 1)$, which simplifies to $i = 1 + \alpha$.
Step 3: Detailed Explanation:
1. Calculate Van't Hoff factor ($i$):
Given observed depression = 0.0205 K.
Given calculated (theoretical) depression = 0.0197 K.
\[ i = \frac{0.0205}{0.0197} \approx 1.0406 \approx 1.041. \]

2. Calculate degree of dissociation ($\alpha$):
Using the relation $i = 1 + \alpha$:
\[ 1.041 = 1 + \alpha \]
\[ \alpha = 1.041 - 1 = 0.041. \]
Step 4: Final Answer:
The Van't Hoff factor is 1.041 and the degree of dissociation is 0.041.
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