Question

10.0 mL of unknown CL- solution requires 22.0 mL of 0.050 M AgNO3 to reach the end point. What is the concentration of Cl- in the unknown? If the solution was prepared by dissolving 1.00 g of solid in 100 mL, what is the weight % CL- in the solid? Ag+ + CL- -------> AgCl

Answer 1

Part (i)

The no. of mmol of AgNO3 required for the Cl^- solution to reach the endpoint can be calculated as follows.

n_AgNO3 = 22 mL * 0.05 mmol/mL

i.e. n_AgNO3 = 1.1 mmol (Since 1 M = 1 mmol/mL)

The concentration of Cl^- in the unknown can be calculated as follows.

[Cl^-] = 1.1 mmol/10 mL

i.e. [Cl^-] = 0.11 mmol/mL = 0.11 M

Part (ii)

The molar mass of AgCl = 143.318 g/mol

The no. of moles of AgCl in 1 g = 1 g/(143.318 g/mol)

i.e. n_AgCl = 0.00698 mol

The molar mass of Cl^- = 35.450 g/mol

The weight of Cl- in the solid AgCl = 0.00698 mol * 35.450 g/mol = 0.2474 g

The weight % of Cl^- in AgCl = (0.2474/1)*100 = 24.74 %