This question has a sting in the tail. It expects you (i) to do the tedious arithmetic involved in the percentage mass; and (ii) it tries to distract students who know that BaSO_4BaSO4 is insoluble in aqueous solution. Of course, the insolubility of BaSO_4BaSO4 is a distractor.
Ba^(2+) + SO_4^(2-) rarr BaSO_4(s)darrBa2++SO2−4→BaSO4(s)⏐⏐↓
But we only need to assess the concentration of chloride ion WHICH REMAINS IN SOLUTION.
"Moles of"Moles of BaCl_2=(50*mLxx20.8%*g*mL^-1)/(208.23*g*mol^-1)BaCl2=50⋅mL×20.8%⋅g⋅mL−1208.23⋅g⋅mol−1 =0.050*mol=0.050⋅mol.
And thus there are 0.100*mol0.100⋅mol with respect to "chloride ion"chloride ion. And of course this is present in a COMBINED VOLUME of 100*mL+50*mL=150*mL=0.150*L100⋅mL+50⋅mL=150⋅mL=0.150⋅L
Now concentration is simply "moles of solute"/"volume of solution"=(0.100*mol)/(150xx10^-3L)moles of solutevolume of solution=0.100⋅mol150×10−3L
[Cl^-]=0.667*mol*L^-1[Cl−]=0.667⋅mol⋅L−1
