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

`M=0.362M`

Step-by-step 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*(1molAgCl)/(143.32gAgCl)*(1molAgNO_3)/(1molAgCl)\\\\n_(AgNO_3)=0.03193molAgNO_3`

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

`M=(0.03193mol)/(0.08811L)\\\\M=0.362M`

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