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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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The correct answer is:
2
2
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,
$$
\begin{aligned}
& \left(1-\frac{1}{n}\right)=\frac{1}{2} \\
& \frac{1}{n}=1-\frac{1}{2}=\frac{1}{2} \\
& \therefore n=2
\end{aligned}
$$
$$
\begin{aligned}
& i=1-\alpha\left(1-\frac{1}{n}\right) \\
& 0.9=1-0.2\left(1-\frac{1}{n}\right)
\end{aligned}
$$
On solving,
$$
\begin{aligned}
& \left(1-\frac{1}{n}\right)=\frac{1}{2} \\
& \frac{1}{n}=1-\frac{1}{2}=\frac{1}{2} \\
& \therefore n=2
\end{aligned}
$$
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