Question

Ag⁺ + Cl^- ↔ AgCl(s)

1.004 g of an impure mixture containing Cl- was dissolved to 100.0 mL. 10.00 mL of this solution required 22.97 mL of 0.05274 M AgNO3 to reach end point. What is the weight percent of Cl- in the unknown solid?

Answer 1

Ans. # Step 1:

# Moles of AgNO3 consumed to reach endpoint = Molarity x Vol. in liters

                                                = 0.05274 M x 0.02297 L = 0.0012114378 mol

# Following stoichiometry of balanced reaction, 1 mol Ag⁺ (from AgNO₃) precipitates 1 mol Cl^-.

So,

            Moles of Cl^- in 10-mL aliquot = Moles of AgNO₃ consumed

                                                            = 0.0012114378 mol

# Total moles of Cl- in original 100-mL aliquot = (0.0012114378 mol / 10 mL) x 100 mL

                                                            = 0.012114378 mol

Now,

            Total mass of Cl^- = Moles x MW = 0.012114378 mol x 35.4527 g mol^-1

                                                = 0.4295 g

# Step 2: % Cl^- in solid sample = (Mass of Cl^- / Mass of sample) x 100

                                                = (0.4295 g / 1.004 g) x 100

                                                = 42.78 %