20 ml of 0.2 M Al₂(SO₄)₃ is mixed with 20 ml of 6.6 M BaCl₂, the concentration of Cl^– ion in solution is

$\text{ Initial molarity of }{\mathrm{Cl}}^{-}{\mathrm{V}}_1\text{ of }\left({\mathrm{Cl}}^{-}\right)=20\mathrm{mL}$
$\text{ So, }{\mathrm{M}}_1{\mathrm{V}}_1={\mathrm{M}}_2{\mathrm{V}}_2$
$2\times 6.6\times 20={\mathrm{M}}_2\times [20+20]\to$
$\mathrm{Mixed} \mathrm{with} 20\mathrm{mL} \mathrm{of} {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$
It increases volume but can't effect concentration of Cl^- because it does not have Cl^- ${\mathrm{M}}_2=\frac{2\times 6.6\times 20}{40}=6.6$