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For 0.1 M acetic acid (Ka = 1.8 × 10⁻⁵), van't Hoff factor i is:
A2
BSlightly > 1 (partial dissociation, ~1.013)
C0
D1 (no dissociation)
Answer & Solution
Correct answer: B. Slightly > 1 (partial dissociation, ~1.013)
Acetic acid is weak: only fraction dissociates. α ≈ sqrt(Ka/c) = sqrt(1.8e-5/0.1) ≈ 0.0134. i = 1 + α ≈ 1.013. Colligative effects only slightly larger than for non-electrolyte. (Strong electrolytes like NaCl have i closer to 2.)
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