The molarity of $C l^{-}$ (100 ml) ions in the solution after $0.2$ moles $H C l$ are mixed with $0.1$ moles $C a C l_{2}$ solution?

0.8

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0.04

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0.4

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0.08

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Solution

The correct option is A $0.8$ M
Moles of $C l^{-}$ ions $= 0.2 \times 1 + 0.1 \times 2$
$= 0.4$
If each solution has volume= $250 m l$
Then total volume= $250 + 250 = 500 m l$
$∴ [C l^{-}] = \frac{0.4}{500 \times 10^{- 3}} = 0.8 M$

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