The solubility of $B a S O_{4}$ in water is 0.00233 g per litre at $30^{o} C$ (atomic weight of Ba = 137). The solubility of $B a S O_{4}$ in $0.1 M (N H_{4})_{2} S O_{4}$ solution at the same temperature is -

10^{- 5} m o l e / l i t r e

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10^{- 6} m o l e / l i t r e

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10^{- 8} m o l e / l i t r e

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10^{- 9} m o l e / l i t r e

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Solution

The correct option is A $10^{- 5} m o l e / l i t r e$
solution:

$M_{X} X_{Y} \Rightarrow x M^{P +} (a q) + y X^{2 -} (a q)$
$s o l u b i i t y = \frac{2.42 \times 10^{-} 3}{233} = 1.038 \times 10^{- 5} m o l e / L$

hence the correct option:A

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