# 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"...?

Sep 5, 2017

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

#### Explanation:

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

$A {g}^{+} + C {l}^{-} \rightarrow A g C l \left(s\right) \downarrow$

$\text{Moles of AgCl} = \frac{0.6326 \cdot g}{143.32 \cdot g \cdot m o {l}^{-} 1} = 4.414 \times {10}^{-} 3 \cdot m o l$.

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 $4.414 \times {10}^{-} 3 \cdot m o l \times 35.45 \cdot g \cdot m o {l}^{-} 1 = 0.1565 \cdot g$ in the original sample.

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