We add excess Na2CrO4 solution to 42.0 mL
of a solution of silver nitrate (AgNO3) to form
insoluble solid Ag2CrO4. When it has been
dried and weighed, the mass of Ag2CrO4 is
found to be 0.590 grams. What is the molarity
of the AgNO3 solution?
Answer in units of M
See your post below.
To find the molarity of the AgNO3 solution, we need to use the concept of stoichiometry.
1. First, let's write the balanced chemical equation for the reaction between Na2CrO4 and AgNO3:
2 AgNO3 + Na2CrO4 → Ag2CrO4 + 2 NaNO3
2. From the equation, we can see that the mole ratio between Ag2CrO4 and AgNO3 is 2:1. This means that for every 2 moles of Ag2CrO4, there must be 1 mole of AgNO3.
3. Next, we need to calculate the number of moles of Ag2CrO4 formed:
Mass of Ag2CrO4 = 0.590 grams
Molar mass of Ag2CrO4 = (2 x atomic mass of Ag) + atomic mass of Cr + (4 x atomic mass of O) = (2 x 107.87 g/mol) + 51.996 g/mol + (4 x 16.00 g/mol) = 332.74 g/mol
Number of moles of Ag2CrO4 = Mass of Ag2CrO4 / Molar mass of Ag2CrO4
= 0.590 g / 332.74 g/mol
4. Now, using the mole ratio from step 2, we can determine the number of moles of AgNO3:
Number of moles of AgNO3 = 1/2 x Number of moles of Ag2CrO4
= 1/2 x (0.590 g / 332.74 g/mol)
5. Finally, we can calculate the molarity (M) of the AgNO3 solution:
Molarity (M) = Number of moles of AgNO3 / Volume of AgNO3 solution (in liters)
= (0.500 x (0.590 g / 332.74 g/mol)) / (42.0 mL / 1000 mL/L)
= (0.500 x (0.590 g / 332.74 g/mol)) / 0.0420 L
Simplifying the equation gives us the answer in units of M.