The degree of association `(alpha)` is given by the expression

Options

A. `alpha=(n(i-1))/(1-n)`
B. `alpha=(i(n-1))/(1+n)`
C. `alpha=(i(n+1))/(1-n)`
D. `alpha=(i(n+1))/(n-1)`

Answer

A

Explanation

`underset("1 mole")(nA)hArrA_(n)`
`1-alpha" "(alpha)/(n)," Total"=1-alpha+(alpha)/(n)`
`therefore i=1-alpha+(alpha)/(n) or 1-i=alpha-(alpha)/(n)=alpha(1-(1)/(n))`
`=alpha((n-1)/(n))`
`or alpha=(n)/((n-1))(i-i)=(n(i-1))/((1-n))`