1mL of 10mM $KCl$ solution was mixed with 4mL of 0.05 M $BaC{{l}_{2}}$ solution. The final $C{{l}^{-}}$ (chloride anion) concentration is:
(A)- 41 mM
(B)- 42 mM
(C)- 67 mM
(D)- 82 mM

Hint: From, the moles of the ion present in each solution along with the total volume of the solution. The concentration of ions can be obtained from the formula Molarity formula.

Complete step by step solution:
It is given that two solutions, that is, sodium chloride and barium chloride are mixed together. Such that, 1 ml of 10 mM of $KCl$ solution mixed with 4 mL of 0.05 M of $BaC{{l}_{2}}$ solution.
So, in order to find the chloride concentration in the final solution, we will find the moles of chloride ion in the total volume of the solution as follows:
In $KCl$ solution, given the concentration (10 mM) and volume (1 mL) the moles of $KCl$ present will be:
$Molarity=\dfrac{Moles\,of\, KCl }{Volume\,of\,solution\,in\,L}$
$10\,\times {{10}^{-3}}M=\dfrac{Moles\,of\, KCl }{1\times {{10}^{-3}}L}$
$Moles\,of KCl=10\times {{10}^{-3}}M\times 1\times {{10}^{-3}}L=10\times {{10}^{-6}}moles$
As we know that one molecule of $KCl$ has one potassium and one chloride ion. Then, from the stoichiometric ratio of $KCl$, we have, one mole of $KCl$ has one mole of $C{{l}^{-}}$. So, $10\times {{10}^{-6}}\,moles$ of $KCl$ will have $10\times {{10}^{-6}}\,moles$of $C{{l}^{-}}$ions.
Similarly, in $BaC{{l}_{2}}$, one molecule of barium chloride has one barium and two chloride ions. From the stoichiometric ratio, one mole $BaC{{l}_{2}}$ has two moles of chloride ion.
Also, the moles of $BaC{{l}_{2}}$ are $=Molarity\,of\,BaC{{l}_{2}}\times Volume\,of\,solution$
$=0.05M\times 4\times {{10}^{-3}}L=0.2\times {{10}^{-3}}moles$
The moles of chloride ion in $0.2\times {{10}^{-3}}\,moles$ of $BaC{{l}_{2}}$will be $(2\times 0.2\times {{10}^{-3}})=0.4\times {{10}^{-3}}\,moles=400\times {{10}^{-6}}moles$
Therefore, the moles of chloride ion in the final solution will be obtained by adding the moles in $KCl$ and $BaC{{l}_{2}}$ solution. So, we get $(10\times {{10}^{-6}}+400\times {{10}^{-6}})=410\times {{10}^{-6}}\,moles$.
Also, the volume of the total solution will be $(1+4)=5\times {{10}^{-3}}L$
Thus, the molarity of the solution will be $=\dfrac{Moles}{Volume}=\dfrac{410\times {{10}^{-6}}}{5\times {{10}^{-3}}}=82\times {{10}^{-3}}M=82mM$

Therefore, the final $C{{l}^{-}}$ (chloride anion) concentration in the solution is option (D)- 82 mM.

Note: The unit conversion must be taken care of, as the concentration (in molarity) is expressed in SI units of moles and litres. Also, the stoichiometric ratio of the molecule should also be considered while calculating the moles of ions in the solution.