A) 49
B) 69
C) 89
D) 98
Correct Answer: A
Solution :
\[\begin{align} & \underset{(0.01-x)M}{\mathop{\underset{0.01\,M}{\mathop{BaC{{l}_{2}}}}\,}}\,\,\,\,\,\,\,\,\,\underset{xM}{\mathop{\underset{{}}{\mathop{B{{a}^{2+}}}}\,}}\,\,\,+\,\,\,\underset{2xM}{\mathop{\underset{{}}{\mathop{2C{{l}^{-}}}}\,}}\,Initial \\ & At\,\,equilibrium \\ \end{align}\] \[i=\frac{(0.01-x)+x+2x}{0.01}\] \[=\frac{0.01+2x}{0.01}=1.98\] \[x=0.0049\] \[%\,\,\alpha =\frac{x}{0.01}\times 100\] \[=\frac{0.0049\times 100}{0.01}=49%\]
You need to login to perform this action.
You will be redirected in
3 sec
