Posted by
**Shoba** on
.

If your solution’s final volume of 32.55 mL yields 3.6 grams of AgCl, determine the Molarity of the final solution. Also, how many milliliters of a 1.85 M solution of AgNO3 did you add to an excess of

aqueous NaCl to provide the 3.6 grams of AgCl?

Actually this question is two parts.

The first part is to find the Final Molarity which is

(3.6 g * 1 mole /143.35 g) / (32.55 ml * 1 L /1000 ml) = 0.77 mol/L or M

The second part is to find how many millilitres of AgNo3 which has 1.85 M.

I don't understand how to do this second part.

Please help me this.