Final answer:
The molarity of the CaI2 solution is found to be approximately 0.0351 M.
Step-by-step explanation:
To determine the molarity of the solution with CaI2, the first step is to convert the mass of CaI2 from milligrams to grams:
568 mg CaI2 × (1 g / 1000 mg) = 0.568 g CaI2
Next, we find the molar mass of CaI2 by summing the atomic masses of calcium (Ca, approximately 40.08 g/mol) and iodine (I, approximately 126.90 g/mol × 2, since there are two iodine atoms per molecule):
Molar mass of CaI2 = 40.08 g/mol + (126.90 g/mol × 2) = 293.88 g/mol
Now, calculate the number of moles of CaI2 dissolved in the solution:
Number of moles = mass of CaI2 / molar mass of CaI2 = 0.568 g / 293.88 g/mol = 0.001931 moles CaI2
To find the molarity, we divide the number of moles by the volume of the solution in liters:
Volume of solution = 55.0 mL = 0.0550 L
Molarity = number of moles of solute / volume of solution in liters = 0.001931 moles / 0.0550 L = 0.0351 M
The final answer is that the molarity of the CaI2 solution is approximately 0.0351 M.