A) 49
B) 69
C) 89
D) 98
E) 100
Correct Answer: A
Solution :
\[BaC{{l}_{2}}\] ⇌ \[B{{a}^{2+}}\] + \[2C{{l}^{-}}\] Initially 1 0 0 After dissociation \[a-\alpha \] \[\alpha \] \[2\alpha \] Total = \[1-\alpha +\alpha +2\alpha =1+2\alpha \] \[\alpha =\frac{1.98-1}{\alpha }=\frac{0.98}{\alpha }=0.49\] for a mole \[\alpha =0.49\] For 0.01 mole \[\alpha =\frac{0.49}{0.01}=49\]
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