Question

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

• A. $100\%$
• B. $92\%$
• C. $87\%$
• D. $74\%$

Answer 1

The correct option is C $87\%$

$Ba(N{O}_3{)}_2$ dissociate as:

Initial at equilibrium,
$\underset{\underset{(0.1-x)M}{0.1M}}{Ba(N{O}_3{)}_2}\leftrightharpoons \underset{\underset{aM}{x}}{B{a}_0^{2+}}+\underset{\underset{2aM}{2x}}{\underset{0}{2N{O}_3^{-}}}$.
Thus, n$=3$
$i=n\alpha +(1-\alpha )=3\alpha +1-\alpha$ (per mole)
Van't Hoff factor $i=1+2\alpha \Rightarrow 2.74=1+2\alpha$
$2\alpha =1.74$
$\alpha =0.87$
$\alpha \left(\%\right)=87\%$.