alleryFor 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.)
Keep practising Unit 3B1: Solutions and Concentrations
Browse all Unit 3B1: Solutions and Concentrations questions for JEE Advanced, or explore more Chemistry topics.
Related questions
Hydrogenation of vegetable oils to make margarine uses catalyst:Adsorption isotherm:Freezing-point depression and elevation are colligative properties: depend on:For a 1 m glucose aqueous solution (i = 1), boiling point:For salt CaCl₂ (assume full dissociation), i =For solution of urea (M_w = 60 g/mol) with 6 g in 100 g water, molality:Osmotic pressure of 0.01 M sucrose solution at 27°C (R = 0.0821 L atm/mol K):Two volatile liquids A (P°_A = 100) and B (P°_B = 200) form ideal solution. For x_A = 0.4,