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(B) The solubility $(S)$ of $BaSO_4$ is given by $S = \sqrt{K_{sp}} = \sqrt{1.1 	imes 10^{-10}} = 1.0488 	imes 10^{-5} \ M$. The molar mass of $BaSO

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$410$

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(B) The solubility $(S)$ of $BaSO_4$ is given by $S = \sqrt{K_{sp}} = \sqrt{1.1 	imes 10^{-10}} = 1.0488 	imes 10^{-5} \ M$. The molar mass of $BaSO_4$ is $137 + 32 + (4 	imes 16) = 233 \ g/mol$. The solubility in $g/L$ is $S 	imes 	ext{Molar Mass} = 1.0488 	imes 10^{-5} 	imes 233 = 2.4437 	imes 10^{-3} \ g/L$. The volume of water required to dissolve $1 \ g$ of $BaSO_4$ is $V = \frac{1 \ g}{2.4437 	imes 10^{-3} \ g/L} \approx 409.2 \ L \approx 410 \ L$.

The volume of the water needed to dissolve $1 \ g$ of $BaSO_4$ $(K_{sp} = 1.1 \times 10^{-10})$ at $25 \ ^oC$ is ..... $litre$.

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