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Chemistry Question on Ionic Equilibrium In Solution

The correct equation for the degree of association ′α′'\alpha' of an associating solute, ′n′'n' molecules of which undergoes association in solution, is

A

α=n(i−1)1−n\alpha = \frac{n(i -1)}{ 1- n}

B

α=i(n−1)1+n\alpha = \frac{i(n -1)}{ 1 + n}

C

α=i(1+n)1−n\alpha = \frac{i( 1 + n)}{ 1- n}

D

α=i(n+1)1−n\alpha = \frac{i(n + 1)}{ 1- n}

Answer

α=n(i−1)1−n\alpha = \frac{n(i -1)}{ 1- n}

Explanation

Solution

nA⇔Ann A \quad \Leftrightarrow A_{n}

1 mole

1−ααn,1-\alpha \quad \frac{\alpha}{n}, \quad Total =1−α+αn=1-\alpha+\frac{\alpha}{n}
∴i=1−α+αn\therefore i=1-\alpha+\frac{\alpha}{n} or 1−i=α−αn=α(1−1n)1-i=\alpha-\frac{\alpha}{n}=\alpha\left(1-\frac{1}{n}\right)
=α(n−1n)=\alpha\left(\frac{n-1}{n}\right)

or α=n(n−1)(i−i)=n(i−1)(1−n)\alpha=\frac{n}{(n-1)}(i-i)=\frac{n(i-1)}{(1-n)}