What volume of 5 molar Na2So4 must be added to 25 ml of 1 molar of bacl2 to produce 10 gram of b a s o4
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.
