We add excess Na2CrO4 solution to 56.0 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 0.450 grams. What is the molarity
of the AgNO3 solution?
Answer in units of M
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2AgNO3 + Na2CrO4 ==> Ag2CrO4 + 2NaNO3
mols Ag2CrO4 = g/molar mass = 0.450/331.7 = 0.00136. Now convert mols Ag2CrO4 to mols AgNO3.
0.00136 mols Ag2CrO4 = 0.00136 x (2 mols AgNO3/1 mol Ag2CrO4) = 0.00136 x 1/2 = ?
Then M AgNO3 = mols AgNO3/L AgNO3 = mols AgNO3/0.056 = ?
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