A) 49
B) 69
C) 100
D) 98
Correct Answer: A
Solution :
49 \[\begin{align} & {{\operatorname{BaCl}}_{2}}\rightleftharpoons {{B}^{2+}}+2C{{l}^{-}} \\ & \operatorname{Initial}1\,\,mol\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,0 \\ & \operatorname{After} dissociation1-\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,~\alpha \,\,\,\,\,\,\,\,\,\,2\alpha \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\operatorname{Total} =1+2\alpha \\ \end{align}\] \[\operatorname{i}=1+2a\] \[\frac{\operatorname{i}-1}{2}=\alpha \] \[\Rightarrow \alpha =\frac{1.98-1}{2}=\frac{0.98}{2}=0.49\] \[\operatorname{i}.e. 49 %\]You need to login to perform this action.
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