If 2M, 200ml HCl, 2M, 100ml `CaCl_(2)` and 5M, 200ml `AlCl_(3)` is mixed then final concentration of `Cl^(-)` will be:
If 2M, 200ml HCl, 2M, 100ml `CaCl_(2)` and 5M, 200ml `AlCl_(3)` is mixed then final concentration of `Cl^(-)` will be:
A. 2.5M
B. 3M
C. 3.5M
D. 7.6M
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`[Cl^(-)]=(M_(1)V_(1)+2M_(2)V_(2)+3M_(3)V_(3))/(V_(1)+V_(2)+V_(3))`
`(2xx200+2xx2xx100+3xx5xx200)/(200+100+200)`
`=(400+400+3000)/(500)=(3800)/(500)=7.6M`
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