Question

The silver nitrate in 20.00 mL of a certain solution was allowed to react with sodium chloride according to the following equation AgNO3 + NaCl yields AgCl + NaNO3 They AgCl was collected, dried and weighed to .2867g AgCl What was the molarity of the original silver nitrate solution?

Answer 1

Answer:

The answer to your question is: 0.1 M

Explanation:

data

Volume of AgNO3 = 20.00 ml

                   1000 ml --------------  1 l

                       20 ml --------------- x              

          x = 20x 1 /1000 = 0.02

AgCl = 0.2867 g

MW of AgCl = 35.45 + 107.9 = 143.35 g

                       143.35 g -------------- 1 mol

                       0.2867 g -------------  x

  x = 0.2867 x 1 / 143.35 = 0.002 moles of AgCl

   From the balance reaction we see that the proportion of AgNO3 to AgCl is 1:1, then

           1 mol of AgNO3 -------------------- 1 mol of AgCL

                 x                   ---------------------  0.002 moles of AgCl

 x = 0.002 moles of AgNO3

   This moles of AgNO3 are in 20 ml or 0.02 liters

So, Molarity = # moles/liter

      Molarity = 0.002 moles/ 0.02 = 0.1 M