To find the solubility of calcium chloride CaCl in water at 25C, we can use the solubility product constant Ksp and set up an equilibrium expression.When CaCl dissolves in water, it dissociates into its ions:CaCl s Ca aq + 2Cl aq Let the solubility of CaCl be represented by 's' moles per liter. Then, the concentration of Ca ions will be 's' and the concentration of Cl ions will be '2s'. The Ksp expression for this equilibrium is:Ksp = [Ca] * [Cl]Substitute the concentrations of the ions in terms of 's':Ksp = s * 2s Given that Ksp = 1.98 10, we can solve for 's':1.98 10 = s * 4s 1.98 10 = 4sNow, divide both sides by 4:s = 1.98 10 / 4s = 4.95 10Now, take the cube root of both sides to find 's':s = 4.95 10 ^1/3 s 1.7 10 MSo, the solubility of calcium chloride in water at 25C is approximately 1.7 10 moles per liter.