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If 2M, 200ml HCl, 2M, 100ml `CaCl_(2)` and 5M, 200ml `AlCl_(3)` is mixed then final concentration of `Cl^(-)` will be:A. 2.5MB. 3MC. 3.5MD. 7.6M |
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Answer» `[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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