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A 15mL sample of 0.20 M" " MgCl(2) is ad...

A `15mL` sample of `0.20 M" " MgCl_(2)` is added to `45ML` of `0.40 M AlCl_(3)` What is the molarity of `Cl` ions in the final solution

A

`1.0M`

B

`0.60M`

C

`0.35M`

D

`0.30M`

Text Solution

Verified by Experts

The correct Answer is:
A
`"Molarity of" Cl^(-)`
`=(M_(1)V_(1)+M_(2)V_(2))/("Total vol")`
`=(15 xx .2 xx 2 + 45 xx .45 xx 3)/(15 + 45) = (60)/(60) = 1M` .
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