A molecule \(M\) associates in a given solvent according to the equation \(\mathrm{M} \rightleftharpoons (\mathrm{M})_n\). For a certain concentration of M, the van't Hoff factor was found to be 0.9 and the fraction of associated molecules was 0.2. The value of \(n\) is :
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- 3
- 5
- 2
- 4
The Correct Option is C
Solution and Explanation
Moles of M left = \(1-\alpha\); moles of \((\mathrm{M})_n\) formed = \(\alpha/n\).
Total moles = \(1 - \alpha + \alpha/n\), so \(i = 1 - \alpha + \alpha/n\).
Put \(i = 0.9\), \(\alpha = 0.2\): \(0.9 = 1 - 0.2 + 0.2/n\).
\(0.2/n = 0.9 - 0.8 = 0.1\), hence \(n = 2\).
Answer: (3) 2
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