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Molarity

0.2 mole of H...

Question

$0.2 m o l e$ of $H C l$ and $0.1 m o l e$ of barium chloride were dissolved in water to produce a $500 m L$ solution. The molarity of the $C l^{-}$ ions is:

0.80 M

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0.09 M

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0.06 M

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0.12 M

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Solution

The correct option is A $0.80 M$
Number of moles of $H C l$ dissolved $(n_{H C l}) = 0.2 m o l$

And number of moles of $C l^{-}$ obtained from $H C l,$

$N_{c l^{-}} = n_{H C l} = 0.2 m o l$

Number of moles of $B a C l_{2}$ dissolved
$(n_{B a C l_{2}}) = 0.1 m o l$

And number of moles of $C l^{-}$ obtained from $B a C l_{2},$

$n_{c l^{-}} = 2 (n_{B a C l_{2}}) = 2 \times 0.1 = 0.2 m o l$

Thus, total number of moles of $C l^{-}$ dissolved
$= 0.2 + 0.2 = 0.4 m o l$

Now $m o l a r i t y = \frac{n o . o f m o l e s}{V o l u m e (i n m L)} \times 1000$

$= \frac{0.4}{500} \times 1000 = 0.80 M$

Thus, molarity of $C l^{-}$ ion is $0.80 M .$

So, option (D) is the correct answer.

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