Question

The van's Hoff factor for $0.1$ molar $Ba(N{O}_3{)}_2$ solutions is $2.74$. Its degree of dissociation is:

• A. $0.74$
• B. $0.87$
• C. $0.91$
• D. $1.0$

Answer 1

The correct option is B $0.87$

The relationship between van't Hoff factor and the degree of dissociation is

$i=1+\alpha (n-1)$

$i=$ van't Hoff factor $\alpha =$ degree of dissociation

$n=$ no. of ions formed

$i=2.74$ (given) $\Rightarrow \alpha =?$

From one unit compound

$n=3$

$Ba(N{O}_3{)}_2⟶{Ba}^{2+}+2{NO}_3^{-}$

$\Rightarrow 2.74=1+\alpha (3-1)$

$2\alpha =2.74-1$ $\therefore$ degree of dissociation is $0.87$

$\alpha =\frac{1.74}{2}$

$\alpha =0.87$

Ans :- Option B