(a) `100 g NaCl` solution `-= 10 g NaCl`
`(100)/(1.07)mL NaCl` solution `-=10 g NaCl`
`:. 1000mLNaCl` solution `=(10xx1000xx1.07)/(100)g NaCl`
`=(10xx1000xx1.07)/(100)xx(35.5)/(58.5)= 64.93 Cl^(-1) "ions"`
(b) Similarly of `BaCl_(2)` wt. of `Cl^(-)` ion
`=(10xx1000xx1.10)/(100)xx(2xx35.5)/(208)= 37.55 g`