A) 49
B) 69
C) 89
D) 98
E) 100
Correct Answer: A
Solution :
\[BaC{{l}_{2}}B{{a}^{2+}}+2C{{l}^{-}}\] Initial 0.01M At equilibrium \[(0.01-x)MxM2xM\] \[i=\frac{(0.01-x)+x+2x}{0.01}\] \[=\frac{0.01+2x}{0.01}=1.98\] \[x=0.0049\] \[%\,a=\frac{x}{0.01}\times 100=\frac{0.0049\times 100}{0.01}=49%\]You need to login to perform this action.
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