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

Answer 1

Answer: (A)

49

Answer 2

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