We add excess Na2CrO4 solution to 27.0 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 0.420 grams. What is the molarity
of the AgNO3 solution?
Answer in units of M
See the AgNO3 problem before.
9090
To find the molarity of the AgNO3 solution, we need to use the balanced chemical equation and the stoichiometry of the reaction.
The balanced chemical equation for the reaction between Na2CrO4 and AgNO3 is:
2 AgNO3 + Na2CrO4 → Ag2CrO4(s) + 2 NaNO3
From the equation, we can see that 2 moles of AgNO3 react with 1 mole of Ag2CrO4. We have the mass of Ag2CrO4, which is 0.420 grams, and we know the molar mass of Ag2CrO4 is 331.74 g/mol.
First, we need to calculate the number of moles of Ag2CrO4:
moles of Ag2CrO4 = mass of Ag2CrO4 / molar mass of Ag2CrO4
moles of Ag2CrO4 = 0.420 g / 331.74 g/mol
Next, we can use the stoichiometry of the reaction to find the number of moles of AgNO3:
moles of AgNO3 = (moles of Ag2CrO4) / (2 moles of AgNO3 / 1 mole of Ag2CrO4)
moles of AgNO3 = (0.420 g / 331.74 g/mol) / (2/1)
moles of AgNO3 = (0.420 g / 331.74 g/mol) * (1/2)
Finally, to find the molarity (M) of the AgNO3 solution, we need to divide the moles of AgNO3 by the volume of the AgNO3 solution in liters:
Molarity of AgNO3 = (moles of AgNO3) / (volume of AgNO3 solution in liters)
Molarity of AgNO3 = (moles of AgNO3) / (27.0 mL / 1000 mL/L)
Molarity of AgNO3 = ((0.420 g / 331.74 g/mol) * (1/2)) / (27.0 mL / 1000 mL/L)
Calculating this will give you the molarity of the AgNO3 solution in units of M.