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.

       ⇒ 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

        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

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

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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