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`.

Options

A. `8.40 g`
B. `16.80 g`
C. `25.20 g`
D. `4.20 g`

Answer

B

Explanation

`N_(NaCl)=(3.78)/(58.5xx(100)/(1000))( because N=(Eq.)/(V "in litre"))`
`=0.646`
Let `w g` of `BaCl_(2)` is dissovled in `250 mL` then
`N_(BaCl_(2)) =(w)/((208)/(2)xx(250)/(1000))=0.0385 w`
`therefore [Cl^(-)]` in both is same,
`:. N_(NaCI)=N_(BaCl_(2))`
`:. 0.6446=0.0385w`
`therefore w=16.80 g`