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

0.362 M

Explanation:

The reaction that takes place is:

First we convert the mass of AgCl (the precipitate) into moles, using its molar mass:

Now we convert AgCl moles into AgNO₃ moles:

Finally we calculate the molarity, after converting 88.11 mL to L:

it should be all of them because it asks if you included all of those in your answer

explanation: