Final answer:
The molarity of a solution formed by dissolving 568 mg of CaI₂ in water is 0.00193 M, assuming the final solution volume is 1 liter.
Step-by-step explanation:
The student has asked to determine the molarity of a solution formed by dissolving 568 mg of CaI₂. To find the molarity, we first need to convert the mass of CaI₂ from milligrams to grams. There are 1,000 milligrams in a gram, so 568 mg is equal to 0.568 grams. The molar mass of CaI₂ is found by adding the atomic masses of calcium (Ca, 40.08 g/mol) and iodine (I, 126.90 g/mol x 2), giving a total molar mass of 293.88 g/mol for CaI₂. Next, we calculate the number of moles of CaI₂ by dividing the mass of the compound by its molar mass: 0.568 g / 293.88 g/mol = 0.00193 moles.
To determine the molarity, we divide the number of moles by the volume of the solution in liters. Since the question does not specify the final volume of the solution, we will assume it is 1 liter for calculation purposes. Therefore, the molarity is 0.00193 M. If a different volume is used, the molarity can be recalculated accordingly.