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What volume of 0.8M $A l C l_{3}$ solution should be mixed with $50 m l$ of $0 .2 M CaC l_{2}$ solution to get a solution of chloride ion concentration equal to 0.6M

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$100 m l$

$50 m l$

$5 .56 m l$

$4 .89 m l$

answer is C.

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Detailed Solution

Use $[C 𝓁^{-}] = \frac{t o t a l n o . o f m o l e s o f C 𝓁^{Θ}}{t o t a l v o l u m e}$

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