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

The answer to your question is: 0.1 M

Step-by-step 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