Question

What is the percent of chloride ion in a sample if 1.324g of sample produces 1.0881g of AgCl when treated with excess Ag⁺\(A{g^ + } + C{l^ - } \to AgCl\)

Short answer

23.76% of chloride ion is present.

Step-by-step solution

Determine no of moles

Moles of AgCl=1.0881/143.3= 0.00759 mol or 7.59 x 10^-3 mol

From the reaction, we can see that 1mol of Cl^- gives 1 mol of AgCl

So, 7.59 x 10^-3 mol of AgCl= 7.59 x 10^-3 of Cl^-

Also, 1 mol of Cl^-=35.453g

7.59 x 10^-3 moles of Cl^-

= 7.59 x 10^-3 x 35.453 g

=0.269g of Cl^-

Mass percent

\(\begin{array}{l} = \frac{{Mass\,of\,C{l^ - }}}{{Total\,mass}} \times 100\\ = \frac{{0.2691}}{{1.1324}} \times 100\\ = 23.76\% \end{array}\)