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Determine the molarity of a solution formed by dissolving 568 mg of CaI2 in enough water to yield 55.0 mL of solution.

User Corina
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1 Answer

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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.

User Alex Lauerman
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