A student has a mixture of salt (NaCl) and sugar (C12, H22O11). To determine the percent composition, the student measures out 5.84 g of the mixture and dissolves the sample in 100.0 mL of water. The student knows aqueous NaCl will react with aqueous silver nitrate (AgNO3) to form solid silver chloride (AgCl). The student determines the sugar will not react with silver nitrate. How many mL of 1.0 M AgNO3 will be required to precipitate 5.84 g of AgCl?

41 ml

Explanation:

Every atom of Ag in AgNO₃ will be present in the AgCl precipitate. Thus, you can determine the amount of Ag in 5.84g of AgCl, and then the volume of AgNO₃ that contains that amount.

1. Ag in 5.84g of AgCl

You can set a proportion using the mass of Ag in one mole of AgCl:

atomic mass of Ag: 107.868g molar mass of AgCl: 143.32 g/mol

x/5.84g of AgCl = 107.868g of Ag / 143.32 g of AgCl

x = 4.395 g of Ag

2. Number of moles of AgNO₃ that contain 4.395 g of Ag

Set a proportion, too:

107.868 g of Ag / 1 mol of AgNO₃ = 4.395 g of Ag / x

x = 0.0407 mol of AgNO₃

3. Volume of AgNO₃

Use the molarity of the AgNO₃ solution:

Molarity = number of moles of solute / volume of solution in liters 1.0M = 0.0407 mol / V V = 0.0407 mol / 1.0M = 0.0407 liter V = 0.0407 liter * 1,000 ml / liter = 40.7 ml ≈ 41 ml ← answer