M AgNO3 = grams AgNO3/L solution = ?
M AgNO3 = grams AgNO3/0.042 = ?
Therefore, to find M AgNO3 we must find grams AgNO3. We have the volume in L already. The Ag2CrO4 gives us a way to find the mass AgNO3 we started with.
2AgNO3 + Na2CrO4 ==> Ag2CrO4 + 2NanO3
mols Ag2CrO4 = grams/molar mass = ?
mols Ag2CrO4 = 0.520/about 332 = about 0.00157.
Using the coefficients in the balanced equation, convert mols Ag2CrO4 to mols AgNO3. That is
0.00157 x (2 mol AgNO3/1 mol Ag2CrO4) = 0.00157 x 2/1 = about 0.00313 mols AgNO3.
Convert mols AgNO3 to grams. g = mols AgNO3 x molar mass AgNO3. Then substitute that number (in grams) into the M equation at the top and solve for M. Note that I've estimated here and there so you need to go through and recalculate each step more accurately than I did.
Please show me how to do this step by step I would appreciate it
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 0.520 grams. What is the molarity
of the AgNO3 solution?
Answer in units of M.
1 answer
1 answer