Question

The vant Hoff factor of $ BaC{{l}_{2}} $ at $ 0.01 \,M $ concentration is $ 1.98 $ . The percentage of dissociation of $ BaC{{l}_{2}} $ at this concentration is:

• A. 49
• B. 69
• C. 89
• D. 98
• E. 100

Answer 1

Answer: (A)

$ BaC{{l}_{2}} \rightleftharpoons B{{a}^{2+}}+2C{{l}^{-}} $
Initial 0.01M
At equilibrium $ (0.01-x)MxM2xM $
$ i=\frac{(0.01-x)+x+2x}{0.01} $
$ =\frac{0.01+2x}{0.01}=1.98 $
$ x=0.0049 $
% $ \,\alpha=\frac{x}{0.01}\times 100=\frac{0.0049\times 100}{0.01}=49 $ %