Question

The degree of dissociation $\alpha$ of a weak electrolyte, $A_2{B}_3$ is related to van`t Hoff factor $\left(i\right)$ by the expression:

• A. $\alpha =\frac{i-1}{4}$
• B. $\alpha =\frac{i+1}{4}$
• C. $\alpha =\frac{1-i}{4}$
• D. $\alpha =\frac{i-4}{4}$

Answer 1

The correct option is A $\alpha =\frac{i-1}{4}$

$$\begin{array}{ccccc}& A_2{B}_3& \to 2A^{3+}\left(g\right)& +& 3B^{2-}\\ \text{Initial}& 1& 0& & 0\\ \text{After dissociation}& 1-\alpha & 2\alpha & & 3\alpha \end{array}$$
Therefore, van't Hoff factor, $i=1-\alpha +2\alpha +3\alpha =1+4\alpha$
$\Rightarrow \alpha =\frac{i-1}{4}$