$100\text{mL}$ of a $0.1\text{M}NaCl$ solution is added to $300\text{mL}$ of a $0.2\text{M}$ $MgC{l}_2$ solution. $100\text{mL}$ of this solution was further diluted with $400\text{mL}$ of water. The molarity of $C{l}^{-}$ ions in the resultant solution is:

The correct option is D $0.065\text{M}$
Millimoles of $NaCl=M_{NaCl}\times V_{NaCl}=100\times 0.1=10$
Millimoles of $C{l}^{-}=10$
Millimoles of $MgC{l}_2=M_{MgC{l}_2}\times V_{MgC{l}_2}=300\text{mL}\times 0.2M=60$
$1$ millimole of $MgC{l}_2$ contains $2$ millimoles of $C{l}^{-}$
∴ Millimoles of $C{l}^{-}$ in the given $MgC{l}_2$ solution $=2\times 60=120$
Total number of millimoles of $C{l}^{-}=10+120=130$
Total volume after mixing $=(100+300)\text{mL}=400\text{mL}$
Molarity of $C{l}^{-}$ on mixing these solutions $=\frac{130}{400}\text{M}$
Now, $100\text{mL}$ of this solution is diluted with $400\text{mL}$ water,
$\frac{130}{400}\times 100=M_f\times 500,$ where $M_f$ is the final molarity of the resultant solution.
On solving, $M_f=0.065\text{M}$