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The solubility of $\mathrm{BaSO}_4$ in water is $2.33 \times 10^{-3} \mathrm{gm} /$ litre. $\left.\mathrm{BaSO}_4=233\right)$
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The correct answer is:
$1 \times 10^{-10}$
The solubility of $\mathrm{BaSO}_4$ in $g /$ /litre is given $2.33 \times 10^{-3}$
$\because$ in mole/litre. $n=\frac{W}{m . w t}=1 \times 10^{-5}=\frac{2.33 \times 10^{-3}}{233}$
Because $\mathrm{BaSO}_4$ is a compound
$K_{s p}=S^2=\left[1 \times 10^{-5}\right]^2=1 \times 10^{-10}$
$\because$ in mole/litre. $n=\frac{W}{m . w t}=1 \times 10^{-5}=\frac{2.33 \times 10^{-3}}{233}$
Because $\mathrm{BaSO}_4$ is a compound
$K_{s p}=S^2=\left[1 \times 10^{-5}\right]^2=1 \times 10^{-10}$
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