A molecule $M$ associates in a given solvent according to the equation $\mathrm{M}…

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:
  1. 3
  2. 5
  3. 2
  4. 4

Solution

van't Hoff factor (i) and the degree of association are related as below : $ \begin{aligned} & i=1-\alpha\left(1-\frac{1}{n}\right) \\ & 0.9=1-0.2\left(1-\frac{1}{n}\right) \end{aligned} $ On solving, $ \begin{aligned} & \left(1-\frac{1}{n}\right)=\frac{1}{2} \\ & \frac{1}{n}=1-\frac{1}{2}=\frac{1}{2} \\ & \therefore n=2 \end{aligned} $

Asked in: JEE Main 2013 (22 Apr Online)

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