A molecule M associates in a given solvent according to the equation M⇌$$(M)n$$.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

Question

A molecule M associates in a given solvent according to the equation M⇌$$(M)n$$.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

Infinity Learn · Accepted Answer

Let the degree of association be α                                               M⇌(M)n$$  Initially               $$1$$          $$0$$ After time         $$1-$$α     αnTotal moles after association $$=1-$$α$$+$$α$$n=1+1n-1$$α                        $$i=moles$$ after association intial moles                                       $$=1+1n-1$$α$$1$$ or $$i-1=1n-1$$α                         .....$$(i) We have, $$i=0.9$$ and α$$=0.2On$$ putting values of i and α in Eq. (i) we get n $$=$$ $$2$$

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