Kavya, the reaction occurring is as follows:-

Na2SO4+BaCl2---------> BaSO4+2NaCl

Here, 1 mol of Na2SO4 reacts with with 1 mol of BaCl2 to give 1 mol of BaSO4

Also, Molar mass of Na2SO4= 142 g/mol

Molar mass of BaSO4= 233.3 g/mol

Molar mass of BaCl2= 208.3 g/mol

Now, in BaSO4, no of moles of sulphate ion= no of moles of Barium ion

and no of moles of Barium ion in 10 g BaSO4= 10/233.3 =  0.0428

therefore, the no of moles of Sulphate ions= 0.0428

Also, the no of moles of sulphate ion here, is equal to no of moles of Na2SO4.

We know that Molarity= no of moles/volume in litres

or, no of moles= Molarity* volume of Na2SO4

Therefore, 0.0428= 5 * volume of Na2SO4

or, the volume of Na2SO4 required= 0.00856 Litres or 8.57 mL

Hope this helps you in further calculations related to this topic.