The molality of `1 M` solution of sodium nitrate is `0.858 mol kg^(-1)`. Determine the density of the solution. How much `BaCl_(2)` would needed to make `250 L` of a solution having same concetration of `Cl^(ɵ)` as the one containing `3.78 g` of `NaCl` per `100 mL`?

Answer

A, B, D

Explanation

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`