When a solution containing silver ions is mixed with another solution containing chloride ions, a precipitate of silver chloride forms. When 50.00 ml of a silver nitrate solution is mixed with an excess of a sodium dichromate solution, all of the silver ion is precipitated as silver dichromate. The solid is collected, washed, dried, and found to have a mass of 6.5379 g. Calculate the molarity of the original silver nitrate solution.

The molarity of the silver nitrate solution mixed with an excess of sodium dichromate solution is 0.6090 M

Explanation:

1) Find the number of moles of the collected solid.

Name of the collected solid (given) : silver dichromate Chemical formula of silver dichromate: Ag₂Cr₂O₇ Molar mass of Ag₂Cr₂O₇: 431.76 g/mol (you can find this information in the literature or calculate using the atomic masses of each element in the chemical unit formula). Molar mass = mass in grams / number of moles

⇒ number of moles = mass in grams / molar mass

⇒ number of moles = 6.5739 g / 431.76 g/mol = 0.015226 mol of

Ag₂Cr₂O₇.

2) Find the number of moles of Ag atoms

Ag ratio in Ag₂Cr₂O₇: 2 moles Ag : 1 mol Ag₂Cr₂O₇ Then, set the proportion:

2 mol Ag / 1 mol Ag₂Cr₂O₇ = x / 0.015226 mol Ag₂Cr₂O₇

⇒ x = 2 * 0.015226 mol Ag = 0.030452 mol Ag

3) Find the number of moles of silver nitrate

Chemical formula of silver nitrate: Ag (NO₃) Ag ratio in Ag (NO₃) : 1 mol Ag : 1 mol Ag (NO₃) Then, by proportion, 0.030452 mol Ag are contained in 0.030452 mol of Ag (NO₃).

4) Find the molarity of the solution of Ag (NO₃)

M = number of moles of solute / volume in liters of solution M = 0.030452 mol Ag (NO₃) / 0.005000 liter Ag (NO₃) solution = 0.6090 M.

You must use 4 significant digits, because the volume of silver nitrate solution 50.00 ml is indicated with 4 signficant digits.

Conclusion: the molarity of the silver nitrate solution mixed with an excess of sodium dichromate solution is 0.6090 M