Answer
About $0.095 mol/l$.
Work Step by Step
$752 mg=0.752g$
$n_{Na_{2}SO_{4}}=m\div M=0.752g\div142g/mol\approx0.0053(mol)$
Because $1$ mol of $Na_{2}SO_{4}$ contains $1$ mol of sulfate ion, $n_{SO_{4}^{2-}}=0.0053(mol)$
$BaSO_{4}$ is a precipitate in water, so, the net ionic reaction is:
$Ba^{2+}(aq)+SO_{4}^{2-}(aq) -> BaSO_{4}(s)$
$n_{Ba^{2+}}=n_{SO^{2-}_{4}}=0.0053(mol)$
Because $1$ mol of $BaCl_{2}$ contains $1$ mol of barium ion, $n_{BaCl_{2}}=0.0053(mol)$
$C=n_{BaCl_{2}}\div V=0.0053(mol)\div 0.0558l=0.095(mol/l)$
