The molar concentration of chloride ion in a solution made by addding $300\text{ml}$ of $3.0\text{M}KCl$ solution to a $200\text{ml}$ of $4.0{\text{M BaCl}}_2$ solution will be:

The correct option is A $5.0\text{M}$
Given, $300\text{ml}$ of $3\text{M}KCl$ and, $200\text{ml}$ of $4\text{M}BaC{l}_2$ solution.
Now,
$\begin{array}{cccc}& KCl\iff & K^{+}+& C{l}^{-}\\ & 3\text{M}& & 3\text{M}\end{array}$

And, $\begin{array}{cccc}& BaC{l}_2\iff & B{a}^{2+}+& 2C{l}^{-}\\ & 4\text{M}& & 2\times 4\text{M}\end{array}$
Therefore, molar concentration of $C{l}^{-}=\frac{M_1{V}_1+M_2{V}_2}{V_1+V_2}=\frac{(3\times 300)+(2\times 4\times 200)}{300+200}=\frac{900+1600}{500}=5.0\text{M}$
Therefore, option (A) $5\text{M}$ will be the correct answer.