1.

How much `BaCl_(2)` would be needed to make `250 mL` of a solution having same eoncentration of `Cl^(o+)` as the one containing `3.78 g` of `NaCl` per `100 mL`.

Answer» `N_(NaCl) = (3.78 xx 1000)/(58.5 xx 100)`
`= 0.646`
Let `W g` of `BaCl_(2)` is dissolved in `250 mL`, then
`N_(BaCl_(2)) = (W)/((208)/(2) xx (250)/(1000)) = 0.038 W`
`:. [Cl^(ɵ)]` in both is same
`:. N_(NaCl) = N_(BaCl_(2))`
`:. 0.646 = 0.085 W`
`: W = 16.08 g`


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