How much `BaCl_2` would be needed to make 250mL of a solution having same concentration of `Cl^(-)` as the one containing 3.78 g of NaCl per 100 mL?

Options

A. `15.30 g`
B. `12.80 g`
C. `18.30 g`
D. `17.00 g`

Answer

D

Explanation

`N_(NaCl)=(3.78)/(58.5xx100"/"1000)=0.646(:. N=(Eq.))/("V in litre")`
Let w g of `BaCl_2` is dissolved in 250 mL then `" "N_(BaCl_2)= (w)/((208)/(2)xx(250)/(1000))= 0.0385w`
`therefore [Cl^(-)]" in both are same. "" "therefore N_(NaCl)= N_(BaCl_2)`
`therefore " "0.646= 0.038w" "therefore w= 17.00g`.