Van't Hoff factor of $0.01 M$ $B a {C l}_{2}$ is $1.98$ , percentage dissociation of $B a {C l}_{2}$ on this concentration will be:

69

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100

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49

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98

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Solution

The correct option is C $49$
$B a {C l}_{2} ⇌ {B a}^{2 +} + 2 {C l}^{-}$

Initially $1$ $0$ $0$
After disso. $1 - α$ $α$ $2 α$

Total moles after dissociation

$= 1 - α + α + 2 α$

$= 1 + 2 α$

$i = \frac{o b s e r v e d m o l e s o f s o l u t e}{n o r m a l m o l e s o f s o l u t e} = \frac{1 + 2 α}{1}$

$∴ α = \frac{i - 1}{2} = \frac{1.98 - 1}{2} = \frac{0.98}{2} = 0.49$

For $1$ mole $α = 0.49$

For $0.01$ mole $α = \frac{0.49}{0.01} = 49$ %

Hence, the correct option is $C$

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