Question

In an experiment, a 88.11 mL sample of unknown silver nitrate solution was treated with 9.753 g of sodium chloride, resulting in 4.576 g of precipitate. Calculate the molarity of the silver nitrate solution

Answer 1

Answer:

$M=0.362M$

Explanation:

Hello!

In this case, according to the following chemical reaction:

$AgNO_3(aq)+NaCl(aq)\rightarrow AgCl(s)+NaNO_3(aq)$

It is possible to compute the moles of silver nitrate via stoichiometry that produced 4.576 g of silver chloride as shown below:

$n_{AgNO_3}=4.576gAgCl*\frac{1molAgCl}{143.32gAgCl}*\frac{1molAgNO_3}{1molAgCl}\\\\n_{AgNO_3}=0.03193molAgNO_3$

Thus, since the molarity is obtained by dividing moles by volume, we obtain:

$M=\frac{0.03193mol}{0.08811L}\\\\M=0.362M$

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