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150 mL 0.08 M BaCl2 is added to 100 mL 0...

150 mL 0.08 M `BaCl_2` is added to 100 mL 0.1 M `Al_(2)(SO_(4))_(3)` and it is allowed to complete the precipitation reaction. Calculate the molarity of `AICI_3` in the final solution.

A

0.032 M

B

0.040 M

C

0.120 M

D

2.240 M

Text Solution

Verified by Experts

The correct Answer is:
A

`N_(Al_(2)(SO_(4))_(3)) = 36/250`
`M_(Al_(2)(SO_(4))_(3)) = 36/250 xx 1/6 = 6/250 = 0.024 M`
`N_(AlCl_(3)) = 24/250` N
`M_(AlCl_(3)) = 24/250 xx 1/3 = 8/250 = 0.032 M`
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