HELPPP PLS IM DESPRATE MAJOR POINTS NEED ASAP We add excess NaCl solution (58.44 g/mol) to 38 mL of a solution of silver nitrate (AgNO3 169.88 g/mol), to form insoluble solid AgCl. When it has been dried and weighed, the mass of AgCl (143.32 g/mol) is found to be 1.11 grams.
What is the molarity of the original AgNO3 solution? The formula weight of NaNO3 is 85.00 g/mol.
Answer in units of M.

The original solution was 0.20 M AgNO₃.

Explanation:

We can first write the balanced chemical equation of the double-replacement reaction between the NaCl solution and AgNO₃ solution:

Recall that molarity is given by:

To find the molarity of the original AgNO₃ solution, then, we will need to find the amount of AgNO₃ in moles. We are given that the volume of the original solution was 38 mL.

To determine the moles of AgNO₃, we can convert the mass of AgCl to moles of AgCl and then to moles of AgNO₃.

We are given that the formula weight of AgCl is 143.32 g/mol.

From the equation, we can see that one mole of AgNO₃ yields one mole of AgCl.

This yields two ratios:

Using the above ratios, multiply with the given mass of AgCl:

Therefore, the molarity of the original AgNO₃ solution is:

In conclusion, the original solution was 0.20 M AgNO₃.