The solubility product of ${\mathrm{BaSO}}_4$ is $1\times {10}^{-10}$ at $298 \mathrm{K}$. The solubility of ${\mathrm{BaSO}}_4$ in $0.1\mathrm{M} {\mathrm{K}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)$ solution is    ------$\times {10}^{-9} \mathrm{g} {\mathrm{L}}^{-1}$(nearest integer).
Given: Molar mass of ${\mathrm{BaSO}}_4$ is $233 \mathrm{g} {\mathrm{mol}}^{-1}$

For the insoluble salt barium sulphate, sulphate ion added is the common ion. Due to the common ion effect the solubility of barium sulphate decreases. Let us assume solubility of barium sulphate is $\mathrm{S}$.

${\mathrm{BaSO}}_4\left(\mathrm{s}\right)\rightleftharpoons \underset{\mathrm{S}}{{\mathrm{Ba}}^{2+}}\left(\mathrm{aq}\right)+\underset{\mathrm{S}+0.1}{{\mathrm{SO}}_4^{2-}}\left(\mathrm{aq}\right)$
$\underset{0.1-0.1}{ {\mathrm{K}}_2{\mathrm{SO}}_4}⟶\underset{0.1}{ {\mathrm{K}}^{+}}\left(\mathrm{aq}\right)+\underset{0.1}{{\mathrm{SO}}_4^{2-}}\left(\mathrm{aq}\right)$
${\mathrm{K}}_{\mathrm{SP}}=\left[{\mathrm{Ba}}^{2+}\right]\left[{\mathrm{SO}}_4^{2-}\right]$
$1\times {10}^{-10}=\mathrm{S}(\mathrm{S}+0.1)$
$\simeq 0.1 \mathrm{S}$
$\mathrm{S}={10}^{-9}\mathrm{M}=233\times {10}^{-9}{\mathrm{gL}}^{-1}$