`10.1 g` of `KNO_(3)` is dissolved in `500 mL` of `H_(2) O`. Mass of `Ba(NO_(3))_(2)` that should be added to this solution of get a molality `(m)` of 0.3 with respect to `NO_(3)^(ɵ)` ion is
`(Mw of KNO_(3) = 101 g mol^(-1), Mw of Ba(No_(3))_(2) = 261 g mol^(-1))`
a. `~~ 1.3 g` b. `~~ 13 g` c.`~~ 6.5 g` d. `~~ 65 g`

Explanation

Moles of `KNO_(3) - "moles of" NO_(3)^(ɵ) = (10.1)/(101) = 0.1`
`:' dH_(2) O = 1 g mL^(-1)`.
Weight of `H_(2)O 500 mL xx 1 = 500 g = 0.5 kg`
`m_(NO_(3)^(ɵ) = ("Mas of" NO_(3)^(ɵ)))/("Weight of solvent" (H_(2) O) "in" kg) = (n_(NO_(3)^(ɵ)))/(0.5 kg)`
`implies = (n_(NO_(3)^(ɵ)))/(0.5 kg), :. n_(NO_(3)^(ɵ)) = 0.3 xx 0.5 = 0.15`
Moles of `NO_(3)^(ɵ)` obtained from `KNO_(3) = 0.1`
Moles of `NO_(3)^(ɵ)` required from `Ba (NO_(3))_(2)`
`= 0.15 - 0.1 = 0.5`
`[2 "mol of" NO_(3)^(ɵ))` is obtined from 1 mol of `Ba(NO_(3))_(2)]`
`:.` Moles of `Ba(NO_(3))_(2) = (0.05)/(2)`
Weight of `Ba(NO_(3))_(2) = (0.05)/(2) xx 261 = 6.5 g`