Which of the following is not correct?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → van’t Hoff factor value for acetic acid = 2, when its concentration is 0.1 m and Ka = 1.8 × 10−6
The van’t Hoff factor i is the ratio of the number of particles present after dissociation (or association) to the number of formula units dissolved. For a weak acid that dissociates to a small extent α:
\( \ce{CH3COOH <=> CH3COO^- + H^+} \qquad i = (1 - \alpha) + \alpha + \alpha = 1 + \alpha \)
\( \displaystyle \alpha \approx \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.8 \times 10^{-6}}{0.1}} = \sqrt{1.8 \times 10^{-5}} \approx 4.2 \times 10^{-3} \)
\( i = 1 + 0.0042 \approx 1.004 \)
Option (1) Acetone and chloroform form a solution with negative deviation from Raoult’s law. — A correct statement. Chloroform forms a hydrogen bond with the oxygen of acetone; the attraction between unlike molecules is stronger, so the vapour pressure is lower than Raoult’s law predicts.
Option (2) HNO3 and H2O form maximum boiling azeotrope at 68% HNO3 and 32% H2O by mass composition. — A correct statement. Solutions with a large negative deviation form a maximum boiling azeotrope; for nitric acid and water it boils at 393.5 K.
Option (3) van’t Hoff factor value for acetic acid = 2, when its concentration is 0.1 m and Ka = 1.8 × 10−6 — Not a correct statement. i = 2 would need complete dissociation. Acetic acid is a weak acid; at 0.1 m it dissociates only about 0.4%, so i is about 1.004.
Option (4) Concentrations of isotonic solutions are same — A correct statement. Isotonic solutions have the same osmotic pressure at a given temperature; since π = CRT, they have the same molar concentration of solute particles.
NCERT (2026-27 edition):
Non-ideal Solutions – “… The solutions that show large negative deviation from Raoult’s law form maximum boiling azeotrope at a specific composition. Nitric acid and water is an example of this class of azeotrope. This azeotrope has the approximate composition, 68% nitric acid and 32% water by mass, with a boiling point of 393.5 K.”
Abnormal Molar Masses – “… In 1880 van’t Hoff introduced a factor i, known as the van’t Hoff factor, to account for the extent of dissociation or association. This factor i is defined as:
\( \displaystyle i = \frac{\text{Normal molar mass}}{\text{Abnormal molar mass}} = \frac{\text{Observed colligative property}}{\text{Calculated colligative property}} \)
\( \displaystyle i = \frac{\text{Total number of moles of particles after association/dissociation}}{\text{Number of moles of particles before association/dissociation}} \)”