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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.

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Answer:

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

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