Answer:
[tex]M=0.362M[/tex]
Explanation:
Hello!
In this case, according to the following chemical reaction:
[tex]AgNO_3(aq)+NaCl(aq)\rightarrow AgCl(s)+NaNO_3(aq)[/tex]
It is possible to compute the moles of silver nitrate via stoichiometry that produced 4.576 g of silver chloride as shown below:
[tex]n_{AgNO_3}=4.576gAgCl*\frac{1molAgCl}{143.32gAgCl}*\frac{1molAgNO_3}{1molAgCl}\\\\n_{AgNO_3}=0.03193molAgNO_3[/tex]
Thus, since the molarity is obtained by dividing moles by volume, we obtain:
[tex]M=\frac{0.03193mol}{0.08811L}\\\\M=0.362M[/tex]
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