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 . 5 xx 100 // 1000) = 0. 646(because N = ( Eq)/("V in litre"))`
Let g of `BaCl _(2)` is dissolved in 250 mL then `N _(BaCl _(2)) = ( w )/(( 208)/(2) xx (250)/(1000))= -0. 0 385 w`
`therefore [ C l ^(-)]` in both are same. `therefore N _(NaCl) = N _(Ba Cl _(2))`
`therefore 0.646 = 0.038w " " therefore w = 17.00 g`