Answer:
0.0714 M for the given variables
Explanation:
The question is missing some data, but one of the original questions regarding this problem provides the following data:
Mass of copper(II) acetate: 
Volume of the sodium chromate solution: 
Molarity of the sodium chromate solution: 
Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

Moles of the sodium chromate solution would be found by multiplying its volume by molarity:

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.
Write the net ionic equation for this reaction:

Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:

According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:

The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:
