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How much BaCl(2) would be needed to make...

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

Text Solution

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The correct Answer is:
A, B, D
i. `d_(sol) = M ((Mw_(2))/(1000) + (1)/(m))` `(Mw_(2) NaNO_(3) = 85 g mol^(-1))`
`d = 1 xx ((85)/(1000) + (1)/(0.858))`
`= 0.085 + 1.165 = 1.25 g mL^(-1)`
ii. `M (BaCl_(2)) = (x xx 1000)/(208 xx 250) = (4 x)/(208) M of Ca^(2+)` and
`(2 xx 4x)/(208) M of Cl^(ɵ)`
`M(NaCl) = (3.78 xx 1000)/(58.5 xx 100) = 0.646 M`
`(2 xx 4x)/(208) = 0.646, x = (0.646 xx 208)/(8) = 16.8 g`
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