50 mL of 0.1(M) Ba${Cl}_2$ solution and 150 mL 0.1 (M) AgCl solution are mixed. Calculate the molarity of ${Cl}^{-}$ ions in the final solution.

The correct option is B 0.125
${BaCl}_2{BaCl}_2⟶{Ba}^{+2}+2{Cl}^{-}$

$50ml,0.1M$ $0.1M$ $0.1M$ $0.2M$

$Molarity=\frac{n}{{Vol}^m}$

$AgClAgCl⟶{Ag}^{+}+{Cl}^{-}$

$150ml,0.1M$ $0.1M$ $0.1M$ $0.1M$

${BaCl}_2⟶2{Cl}^{-}$

No. of moles=MxV

=$0.2$x$50$

=$10$

${AgCl}^{}⟶1{Cl}^{-}$

No. of moles=MxV

=$0.2$x$150$

=$15$

Total no. moles of ${Cl}^{-}=10+15=25$

Total volume=$200$

$Molarity=\frac{25}{200}=\frac{1}{8}=0.125$

The molarity of ${Cl}^{-}$ ions in the final solution is $0.125M$