$100 mL$ of a $0.1 M N a C l$ solution is added to $300 mL$ of a $0.2 M$ $M g C l_{2}$ solution. $100 mL$ of this solution was further diluted with $400 mL$ of water. The molarity of $C l^{-}$ ions in the resultant solution is:

0.725 M

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0.055 M

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0.045 M

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0.065 M

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Solution

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

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