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Chapter 4: 4.15 (page 208)

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

Expert verified

23.76% of chloride ion is present.

Step by step solution

01

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-

02

Mass percent

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

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