Question

A `"0.4596 g"` sample of a pure soluble chloride compound is dissolved in water, and all of the chloride ion is precipitated as `"AgCl"` by the addition of an excess of silver nitrate. The mass of the resulting `"AgCl"` is found to be `"0.6326 g"`...?

Answer 1

`%Cl=34.1%` by mass..........

Explanation:

Good question......we first determine the mass of chloride present, we gots.....

`Ag^+ + Cl^(-) rarr AgCl(s)darr`

`"Moles of AgCl"=(0.6326*g)/(143.32*g*mol^-1)=4.414xx10^-3*mol`.

Note that it would not be terribly practical to use silver chloride gravimetrically; it would tend to photo-oxidize, and at any rate it is a very curdy material that would be hard to isolate.......

Now here, chloride anion was the material in deficiency; i.e. silver ion was present in excess.....there was thus a mass of `4.414xx10^-3*molxx35.45*g*mol^-1=0.1565*g` in the original sample.

And so `%Cl=(0.1565*g)/(0.4595*g)xx100%=34.1%`