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"#...?

1 Answer
anor277
Sep 5, 2017

#%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%#

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