Question

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 5.84g AgCl?

Answer 1

• 41 ml

Step-by-step explanation:

All the Ag in the 5.84 g of AgCl precipitate is in the 1.0M solution of AgNO₃.

1. Determine the number of moles of AgCl in 5.84g:

• molar mass of AgCl: 143.32 g/mol

• number of moles = mass in grams / molar mass

• number of moles = 5.84g / 143.32g/mol = 0.04074798mol AgCl

2. The number of moles of Ag is equal to the number of moles of AgCl, since the ratio of atoms is 1 : 1.

• number of moles of Ag: 0.04074798 mol Ag

3. The number of moles of AgNO₃ is equal to the number of moles of Ag because the ratio is 1 : 1 too.

• number of moles of AgNO₃ = 0.04074798 mol AgNO₃

4. The molarity formula allows you to determine the volume of AgNO₃ solution that contains 0.04074798 moles of AgNO₃

• Molarity = number of moles of solute / volume of solution in liters

• 1.0M = 0.04074798 mol / x

• x = 0.04074798mol / 1.0M = 0.04074798 liter

Convert to mililiter:

• 0.04074798 liter × 1,000 ml / liter = 40.7 ml

Round to two significant figures: 41 ml ← answer