posted by Shoba .
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?
I don't understand the question.
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.