Posted by Jake on .
We add excess NaCl solution (58.44 g/mol) to
64 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
1.91 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.

chemistry 
DrBob222,
NaCl + AgNO3 ==> AgCl + NaNO3.
You have 1.91 g AgCl. How many moles is that?
1.91/143.32 = ??
How many moles AgNO3 is that? Since 1 mole AgCl is produced for every mole AgNO3 initially, moles AgCl must equal moles AgNO3. Then what is the molarity.
M = moles/L. You have moles and you have L. Solve for M. 
chemistry 
Lola,
.369