A molecule ‘M’ associates in a given solvent according to the equation $\mathrm{M}\rightleftharpoons {\left(\mathrm{M}\right)}_{\mathrm{n}}$ 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

Let the degree of association be $\alpha$

$$\begin{gathered} \mathrm{M}\rightleftharpoons {\left(\mathrm{M}\right)}_{\mathrm{n}} \\ \mathrm{Initially} 1 0 \\ \mathrm{After} \mathrm{time} \mathrm{t} 1-\mathrm{\alpha } \frac{\mathrm{\alpha }}{\mathrm{n}} \end{gathered}$$

Total moles after association = $1-\mathrm{\alpha }+\frac{\mathrm{\alpha }}{\mathrm{n}}=1+\left(\frac{1}{\mathrm{n}}-1\right)\mathrm{\alpha }$

                        $$\begin{gathered} \mathrm{i}=\frac{\mathrm{moles} \mathrm{after} \mathrm{association}}{\mathrm{initial} \mathrm{moles}}=\frac{1+\left({\displaystyle \frac{1}{\mathrm{n}}}-1\right)\mathrm{\alpha }}{1} \\ \mathrm{i}-1=\left(\frac{1}{\mathrm{n}}-1\right); \\ \mathrm{i}=1-\mathrm{\alpha }+\frac{\mathrm{\alpha }}{\mathrm{n}}; \\ \mathrm{By} \mathrm{substituting} \mathrm{i}=0.9 , \mathrm{\alpha }=0.2 \mathrm{we} \mathrm{get} \\ \mathrm{n}=2 \end{gathered}$$