A) \[{{\text{V}}_{\text{final}}}\text{=4}{{\text{V}}_{\text{intial}}}\]
B) \[{{V}_{final}}=6{{V}_{\operatorname{int}ial}}\]
C) \[Ca{{F}_{2}}({{K}_{sp}}=1.7\times {{10}^{-10}})\]
D) \[{{10}^{-4}}\,M\,C{{a}^{2+}}+{{10}^{-4}}M{{F}^{-}}\]
Correct Answer: B
Solution :
\[\underset{1-\alpha }{\mathop{4A}}\,\underset{\frac{\alpha }{4}}{\mathop{{{(A)}_{4}}}}\,\] \[i=\frac{1-\alpha +\frac{\alpha }{4}}{1}=1-\frac{3}{4}\alpha \] \[\alpha \] = degree of association \[=1=100%\] \[\therefore \] \[i=1-\frac{3}{4}=0.25\]
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