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0.2 mole of HCI and 0.2 mole of barium c...

`0.2` mole of `HCI and 0.2` mole of barium chloride were dissolved in water to produce a `500mL` solution. The molarity of the `CI^-` ions is :

A

.04 M

B

0.8 M

C

0.4 M

D

0.08 M

Text Solution

Verified by Experts

The correct Answer is:
B
`0.2 "mole" HCI = 0.2 "mole" CI^(-)`
`0.1 "mole" CaCI_(2) = 0.2 "mole" CI^(-)`
`"Totel"CI^(-) = 0.4 "mole".`
`"Hence,Molarity of " CI^(-) =(n xx 1000)/(V(mL))`
`(0.4 xx 1000)/500=0.8`
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