Answer:
The molarity of the silver nitrate solution mixed with an excess of sodium dichromate solution is 0.6090 M
Explanation:
1) <u>Find the number of moles of the collected solid</u>.
- Name of the collected solid (given): <em>silver dichromate</em>
- 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) <u>Find the number of moles of Ag atoms</u>
<u />
- 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)<u> Find the number of moles of silver nitrate</u>
<u />
- 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) <u>Find the molarity of the solution of Ag(NO₃)</u>
- 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 s<em>ilver nitrate solution 50.00 ml</em> 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