$0.4 m o l e s$ of HCl and $0.2 m o l e s$ of $C a C l_{2}$ were dissolved in water to have $500 m L$ of solution, the molarity of $C l^{-}$ ion is:

0.8 M

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1.6 M

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1.2 M

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10.0 M

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Solution

The correct option is A $1.6 M$
$H C l 0.4 m o l e s ⇌ H^{+} + C l^{-} 0.4 m o l e s$
$C a C l_{2} 0.2 m o l e s ⇌ C a^{2 +} + 2 C l^{-} 2 \times 0.2 = 0.4 m o l e s$
Total $C l^{-} m o l e s = 0.4 + 0.4 = 0.8 m o l e s$
$M o l a r i t y = \frac{m o l e s}{V o l . i n L}$
$∴$ Molarity of $C l^{-} = \frac{0.8}{0.5} = 1.6 M$ .

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