A molecule \(X\) associates in a given solvent as per the following equation:
\(nX \rightleftharpoons (X)_n\)
For a given concentration of \(X\), the van’t Hoff factor was found to be 0.80 and the fraction of associated molecules was \(0.3\). The correct value of ‘n’ is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 3
When a solute associates, n of its molecules join to form one larger particle, so the number of particles in the solution goes down and the van’t Hoff factor i becomes less than 1. Take 1 mol of X, of which a fraction α associates:
\( \displaystyle \ce{nX <=> (X)_n} \)
\( \displaystyle \text{moles at equilibrium: }(1-\alpha)\ \text{of X}+\frac{\alpha}{n}\ \text{of }\ce{(X)_n} \)
\( \displaystyle i=\frac{\text{particles after association}}{\text{particles before}}=1-\alpha+\frac{\alpha}{n} \)
With i = 0.80 and α = 0.3:
\( \displaystyle 0.80=1-0.3+\frac{0.3}{n} \)
\( \displaystyle \frac{0.3}{n}=0.10 \)
\( \displaystyle n=3 \)
So three molecules of X associate to form one particle, and n = 3.
Shortcut: The drop in i equals the fraction associated times (1 − 1/n): 0.20 = 0.3 × (1 − 1/n), so 1 − 1/n = 2/3 and n = 3.
Note: The same/similar question has been asked in different years by NTA as per the details given below:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| – | – | 15-05 S1 | 07-06 S3 | – |