$100mL$ solution containing $\text{'}x\text{'}MS{O}_3^{2-}and\text{'}y\text{'}M{S}_2{O}_3^{2-}$ exactly requires $80mL$ of $0.05MCr{O}_4^{2-}$ in alkaline medium for complete oxidation. $Cr{O}_4^{2-}$ reduces to $Cr{O}_2^{1-}$ and the only sulphur containing product formed is $S{O}_4^{2-}.$ After the completion of reaction, the solution is treated with excess $BaC{l}_2$ and all the $S{O}_4^{2-}$ is precipitated as $BaS{O}_4.$ The weight of $BaS{O}_4$ formed is found to be 0.9336g. (Molecular weight of  $BaS{O}_4$ is 233.4 g/mol)