A saturated solution of AgCl was found to have a concentration of 1.30 x 10-5 mol dm-3. The solubility product of AgCl therefore is?

The concentration of a saturated solution of AgCl is given as 1.30 x 10^-5 mol/dm^3. We can use this information to calculate the solubility product of AgCl. The solubility product (Ksp) of a sparingly soluble salt is the product of the concentrations of its ions in a saturated solution, with each concentration raised to a power equal to the number of ions produced by one formula unit of the salt. For AgCl, the dissociation reaction is: AgCl(s) ⇌ Ag+(aq) + Cl-(aq) The concentration of Ag+ and Cl- in a saturated solution of AgCl is the same and can be represented as x mol/dm^3. Therefore, the Ksp for AgCl can be calculated as follows: Ksp = [Ag+][Cl-] = x * x = x^2 Since the concentration of Ag+ and Cl- in a saturated solution of AgCl is the same, we can assume that x is equal to the given concentration of 1.30 x 10^-5 mol/dm^3. Therefore, Ksp = (1.30 x 10^-5)^2 = 1.69 x 10^-10 mol^2/dm^6. So, the correct answer is option (C): 1.69 x 10^-10 mol^2/dm^6.