The volume of water needed to dissolve 1 g of ${\mathrm{BaSO}}_4({\mathrm{K}}_{\mathrm{sp}}=1.1\times {10}^{-10})$ at ${25}^{\circ }\mathrm{C}$ is $\left\{\mathrm{M}.\mathrm{W} \mathrm{o}{\text{⨍ BaSO}}_4\text{ is 233 }\right\}$ 

${\mathrm{BaSO}}_4-\to {\mathrm{Ba}}^{2+}+{\mathrm{SO}}_4^{-2} {\mathrm{K}}_{\mathrm{sp}} ={\mathrm{S}}^2 \mathrm{S}=\sqrt{{\mathrm{K}}_{\mathrm{sp}}}$

$\mathrm{S}=\sqrt{{\mathrm{K}}_{\mathrm{sp}}}=\sqrt{1.1\times {10}^{-10}}=1.05\times {10}^{-5}$ Mass of ${\mathrm{BaSO}}_4=$

$1.05\times {10}^{-5}\times 233=244.65\times {10}^{-5}\mathrm{g}$ Volume needed=

$\frac{1}{244.65\times {10}^{-5}}=0.0040874\times {10}^{-5}=408.74$