Posted by **Jake** on Monday, September 20, 2010 at 5:40pm.

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**, Monday, September 20, 2010 at 6:03pm
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**, Monday, October 17, 2011 at 8:26pm
.369

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