For acetic acid dissolved in water, the observed freezing point depression is 0.0205 K while the value calculated assuming no dissociation is 0.0197 K. The van't Hoff factor is:
A0.041
B2.041
C1.041
D0.961
Answer & Solution
Correct answer: C. 1.041
1. $i = $ observed colligative property / calculated colligative property.
2. = observed $\Delta T_f$ / calculated $\Delta T_f$.
3. = $0.0205 / 0.0197 = 1.041$. (Option D is the degree of dissociation, $i - 1$.)
_Source: NCERT Class 12 Chemistry Ch 1 "Solutions", p.25_
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