Question

Relationship between van't Hoffs factor $(i)$ and degree of dissociation $(\alpha )$ is

• A. $i=\frac{\alpha -1}{n^{\prime }-1}$
• B. $i=\frac{\alpha -1}{1-n^{\prime }}$
• C. $\alpha =\frac{1-i}{n^{\prime }-1}$
• D. $\alpha =\frac{i-1}{n^{\prime }-1}$

Answer 1

Answer: (D)

Relationship between vant Hoff factor (i) and degree of dissociation $(\alpha )$ is given by
$\alpha =\frac{1-i}{n^{\prime }-1}$
where, n is the number of ions formed after dissociation
The relationship can be obtained as follows:
$\underset{\text{Initially}}{\text{For the reaction,}}\underset{1mole}{A}\rightleftharpoons \underset{0}{n^{\prime }B}$
After dissociation $(1-\alpha )$ mole n' $\alpha$
Total number of moles present in the solution
$=(1-\alpha )+n^{\prime }\alpha =1+(n^{\prime }-1)\alpha$
van't Hoff factor, $i=1+\left(n^{\prime }-1\right),\alpha >1$ if $n^{\prime }\ge 2$
$\therefore \alpha =\frac{i-1}{n^{\prime }-1}$