Question

A molecule $M$ associates in a given solvent according to the equation $M \rightleftharpoons (M)_n$. 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 :

• A. $3$
• B. $5$
• C. $2$
• D. $4$

Answer 1

Answer: (C)

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