Initially 22.2 g of calcium chloride are dissolved in water to make a 500 mL solution. Then the solution is again diluted upto 2000 mL. Calculate the molarity of $C l^{-}$ ions in the final solution if the 80% of the given salt dissociates in the final solution:

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Solution

We know,
Molar mass of the $C a C l_{2} = (40 + 71) g / m o l = 111 g / m o l$
Given, mass of $C a C l_{2}$ = 22.2 g
So, moles of $C a C l_{2}$ present $= \frac{22.2}{111} = 0.2 m o l$
Again,
one formula unit of $C a C l_{2}$ contains 2 $C l^{-}$ ions.
So, 0.2 moles of $C a C l_{2}$ should produce 0.4 moles of $C l^{-}$ ions.
But in the question, we are given that only 80% of $C a C l_{2}$ dissociates in the final solution.
Thus, moles of $C l^{-}$ ions produced $= 0.4 \times 0.8 = 0.32 m o l$

Hence molarity of the $C l^{-}$ ions in the final solution
$= \frac{0.32}{2000} \times 1000 = 0.16 M$
(Since volume of the final solution is 2000 mL)

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