Question

The van't Hoff factor for $0.1\text{M},Ba(N{O}_3{)}_2$ solution is $2.74$. The degree of dissociation is:

• A. $91.3\%$
• B. $87\%$
• C. $100\%$
• D. $74\%$

Answer 1

The correct option is B $87\%$

$Ba(N{O}_3{)}_2$ dissociates as,
$Ba(N{O}_3{)}_2\rightleftharpoons B{a}^{2+}+2N{O}_3^{-}$
$\therefore \text{Degree of dissociation,}\alpha =\frac{i-1}{n-1}\Rightarrow i=1+\alpha (n-1)$
Where
$n$ = number of particles in solution after dissociation of one molecule or a compound
$\alpha =$degree of dissociation
$i=$ van't Hoff factor
Now $i=1+\alpha (3-1)$
$i=1+2\alpha$ $\{n=3\}$
On Substituting the value of $i$ in equation, we get
$2.74=1+2\alpha$
$\Rightarrow 1.74=2\alpha$
$\Rightarrow \alpha =\frac{1.74}{2}=0.87$ or
$\alpha =0.87\times 100=87\text{\%}.$
So the degree of dissociation is $87\%$.